I need help with this integral:

$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx.$$

The integrand graph looks like this:

$\hspace{1in}$The integrand graph

The approximate numeric value of the integral: $$I\approx8.372211626601275661625747121...$$

Neither Mathematica nor Maple could find a closed form for this integral, and lookups of the approximate numeric value in WolframAlpha and ISC+ did not return plausible closed form candidates either. But I still hope there might be a closed form for it.

I am also interested in cases when only numerator or only denominator is present under the logarithm.

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    $\begingroup$ Do you have any reason to believe there is a closed form for that horrid-looking thing? $\endgroup$
    – dfeuer
    Commented Nov 11, 2013 at 17:12
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    $\begingroup$ In the meantime, I have been able to manipulate the integral into the following form: $$8 \int_0^{\infty} du \frac{(u^2-1)(u^4-6 u^2+1)}{u^8+4 u^6+70 u^4+4 u^2+1} \log{u}$$ from which I may deduce that there is in fact a closed form (because the roots of the denominator are expressible in closed form, a little messy but not bad). But because there are eight roots, a residue-based approach will be very very messy and almost hopeless without some computer algebra. $\endgroup$
    – Ron Gordon
    Commented Nov 12, 2013 at 0:21
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    $\begingroup$ A related problem. $\endgroup$ Commented Nov 19, 2013 at 18:46
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    $\begingroup$ @MhenniBenghorbal: Related how? $\endgroup$
    – Ron Gordon
    Commented Nov 19, 2013 at 21:02
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    $\begingroup$ Out of curiosity, is there an application for this integral? $\endgroup$ Commented Nov 16, 2014 at 0:35

14 Answers 14


I will transform the integral via a substitution, break it up into two pieces and recombine, perform an integration by parts, and perform another substitution to get an integral to which I know a closed form exists. From there, I use a method I know to attack the integral, but in an unusual way because of the 8th degree polynomial in the denominator of the integrand.

First sub $t=(1-x)/(1+x)$, $dt=-2/(1+x)^2 dx$ to get

$$2 \int_0^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} $$

Now use the symmetry from the map $t \mapsto 1/t$. Break the integral up into two as follows:

\begin{align} & 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_1^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\ &= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_0^{1} dt \frac{t^{1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\ &= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \end{align}

Sub $t=u^2$ to get

$$4 \int_0^{1} \frac{du}{1-u^2} \log{\left (\frac{5-2 u^2+u^4}{1-2 u^2 +5 u^4} \right )}$$

Integrate by parts:

$$\left [2 \log{\left (\frac{1+u}{1-u} \right )} \log{\left (\frac{5-2 u^2+u^4}{1-2 u^2 +5 u^4} \right )}\right ]_0^1 \\- 32 \int_0^1 du \frac{\left(u^5-6 u^3+u\right)}{\left(u^4-2 u^2+5\right) \left(5 u^4-2 u^2+1\right)} \log{\left (\frac{1+u}{1-u} \right )}$$

One last sub: $u=(v-1)/(v+1)$ $du=2/(v+1)^2 dv$, and finally get

$$8 \int_0^{\infty} dv \frac{(v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v}$$

With this form, we may finally conclude that a closed form exists and apply the residue theorem to obtain it. To wit, consider the following contour integral:

$$\oint_C dz \frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1} \log^2{z}$$

where $C$ is a keyhole contour about the positive real axis. This contour integral is equal to (I omit the steps where I show the integral vanishes about the circular arcs)

$$-i 4 \pi \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} + 4 \pi^2 \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1}$$

It should be noted that the second integral vanishes; this may be easily seen by exploiting the symmetry about $v \mapsto 1/v$.

On the other hand, the contour integral is $i 2 \pi$ times the sum of the residues about the poles of the integrand. In general, this requires us to find the zeroes of the eight degree polynomial, which may not be possible analytically. Here, on the other hand, we have many symmetries to exploit, e.g., if $a$ is a root, then $1/a$ is a root, $-a$ is a root, and $\bar{a}$ is a root. For example, we may deduce that

$$z^8+4 z^6+70z^4+4 z^2+1 = (z^4+4 z^3+10 z^2+4 z+1) (z^4-4 z^3+10 z^2-4 z+1)$$

which exploits the $a \mapsto -a$ symmetry. Now write

$$z^4+4 z^3+10 z^2+4 z+1 = (z-a)(z-\bar{a})\left (z-\frac{1}{a}\right )\left (z-\frac{1}{\bar{a}}\right )$$

Write $a=r e^{i \theta}$ and get the following equations:

$$\left ( r+\frac{1}{r}\right ) \cos{\theta}=-2$$ $$\left (r^2+\frac{1}{r^2}\right) + 4 \cos^2{\theta}=10$$

From these equations, one may deduce that a solution is $r=\phi+\sqrt{\phi}$ and $\cos{\theta}=1/\phi$, where $\phi=(1+\sqrt{5})/2$ is the golden ratio. Thus the poles take the form

$$z_k = \pm \left (\phi\pm\sqrt{\phi}\right) e^{\pm i \arctan{\sqrt{\phi}}}$$

Now we have to find the residues of the integrand at these 8 poles. We can break this task up by computing:

$$\sum_{k=1}^8 \operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1) \log^2{z}}{z^8+4 z^6+70z^4+4 z^2+1}\right ]=\sum_{k=1}^8 \operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1}\right ] \log^2{z_k}$$

Here things got very messy, but the result is rather unbelievably simple:

$$\operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1}\right ] = \text{sgn}[\cos{(\arg{z_k})}]$$


Actually, this is a very simple computation. Inspired by @sos440, one may express the rational function of $z$ in a very simple form:

$$\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1} = -\left [\frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right ]$$


$$p(z)=z^4+4 z^3+10 z^2+4 z+1$$

The residue of this function at the poles are then easily seen to be $\pm 1$ according to whether the pole is a zero of $p(z)$ or $p(-z)$.


That is, if the pole has a positive real part, the residue of the fraction is $+1$; if it has a negative real part, the residue is $-1$.

Now consider the log piece. Expanding the square, we get 3 terms:

$$\log^2{|z_k|} - (\arg{z_k})^2 + i 2 \log{|z_k|} \arg{z_k}$$

Summing over the residues, we find that because of the $\pm1$ contributions above, that the first and third terms sum to zero. This leaves the second term. For this, it is crucial that we get the arguments right, as $\arg{z_k} \in [0,2 \pi)$. Thus, we have

$$\begin{align}I= \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} &= \frac12 \sum_{k=1}^8 \text{sgn}[\cos{(\arg{z_k})}] (\arg{z_k})^2 \\ &= \frac12 [2 (\arctan{\sqrt{\phi}})^2 + 2 (2 \pi - \arctan{\sqrt{\phi}})^2 \\ &- 2 (\pi - \arctan{\sqrt{\phi}})^2 - 2 (\pi + \arctan{\sqrt{\phi}})^2]\\ &= 2 \pi^2 -4 \pi \arctan{\sqrt{\phi}} \\ &= 4 \pi \, \text{arccot}{\sqrt{\phi}}\\\end{align}$$

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    $\begingroup$ +1. $\text{arccot}(\sqrt{\phi})$-definitely one of the most weirdest closed form solutions to have ever been obtained! $\endgroup$
    – user17762
    Commented Dec 2, 2013 at 6:31
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    $\begingroup$ @ShikariShambu: which makes me wonder if we can generate a list of the oddest closed-form solutions to integrals, sums, products, diff eq'ns, and the like. What makes a closed form "odd"? What makes this one odd? The juxtaposition of the arccotangent and $\phi$? This could lead to...a weird discussion, but a fun one nonetheless. $\endgroup$
    – Ron Gordon
    Commented Dec 18, 2013 at 23:41
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    $\begingroup$ I know that $\phi$ has some nice properties, but at the end of the day it's just another algebraic number: $$\phi = \frac{1+\sqrt{5}}{2},$$ and all we've done is taken the arc cotangent of the square root of it and multiplied by $4\pi$. - just wondering why this result would be seen as "odd". It most likely appears because it happens to be a root of some polynomial which has $(1+\sqrt{5})/2$ as one of its roots. $\endgroup$
    – pshmath0
    Commented Feb 12, 2014 at 18:11
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    $\begingroup$ @RonGordon When I see things like this, I don't consider it odd so much as a pleasant surprise. Phi is known as a number of beauty for many reasons, and to see it emerge from, to quote our friend dfeuer, "that horrid-looking thing", is quite a pleasant surprise indeed! No tin-foil hats, no cube-root-of-31-conspiracies, just running into an old friend at the grocery =) $\endgroup$
    – corsiKa
    Commented Mar 13, 2014 at 19:28
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    $\begingroup$ @corsiKa: $\phi$ alone, sure. But now imagine your meeting your old friend, but dressed in drag with a Kaiser-era military helmet on, spike and all. That's sort of the feeling you get when you see, not regular old $\phi$, but ARCCOT SQRT $\phi$. $\endgroup$
    – Ron Gordon
    Commented Mar 14, 2014 at 17:08


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    $\begingroup$ Where are all the replies to this answer? $\endgroup$ Commented May 14, 2023 at 18:20
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    $\begingroup$ This question got popular on social media recently so the moderators probably removed all comments because people we probably interacting too much @SarthikGupta $\endgroup$
    – wjmccann
    Commented Jul 3, 2023 at 1:52
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    $\begingroup$ @wjmccann Yea, here's some videos for context. None of that is my work, I just stumbled upon it today. $\endgroup$
    – Mast
    Commented Jul 7, 2023 at 18:09

NEW ANSWER. I found yet another way of solving this problem. My new solution does not use contour integration, and is based on the following observation: for $|z| \leq 1$,

$$ - \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log(1 - zx) \, dz= \pi \sin^{-1} z - \pi \log \left( \tfrac{1}{2}+\tfrac{1}{2}\sqrt{1-z^{2}} \right) . $$

As I want to keep both the old answer and the new answer, I posted my new solution to other page. You can check it here.

OLD ANSWER. Okay here is another solution. It is also related to my generalization.

We claim the following proposition:

Proposition. If $0 < r < 1$ and $r < s$, then $$ I(r, s) := \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx = 4\pi \arcsin r. \tag{1} $$

Assuming this proposition, all that we have to do is to solve the non-linear system of equations

$$ 2rs = 2 \quad \text{and} \quad r^{2} + s^{2} - 1 = 2. $$

The unique solution satisfying the condition of the proposition is $r = \phi - 1$ and $s = \phi$. So by $\text{(1)}$ we have

\begin{align*} \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log \left( \frac{1 + 2x + 2x^{2}}{1 - 2x + 2x^{2}} \right) \, dx & = I(\phi-1, \phi) \\ &= 4\pi \arcsin (\phi - 1) = 4\pi \operatorname{arccot} \sqrt{\phi}. \end{align*}

Thus it remains to prove the proposition.

Proof of Proposition. We divide the proof into several steps.

Step 1. (Case reduction by analytic continuation) We first remark that given $r$ and $s$, we always have

$$ \min_{-1 \leq x \leq 1} \{ 1 \pm 2rsx + (r^{2} + s^{2} - 1)x^{2} \} > 0. \tag{2} $$

Indeed, it is not hard to check if we utilize the following equality

$$ 1 \pm 2rsx + (r^{2} + s^{2} - 1)x^{2} = (1 \pm rsx)^{2} - (1 - r^{2})(1 - s^{2}) x^{2}. $$

Then $\text{(2)}$ shows that the integrand of $I(r, s)$ remains holomoprhic under small perturbation of $s$ in $\Bbb{C}$. So it allows us to extend $s \mapsto I(r, s)$ as a holomorphic function on some open set containing the line segment $(r, \infty) \subset \Bbb{C}$. Then by the principle of analytic continuation, it is sufficient to prove that $\text{(1)}$ holds for $r < s < 1$.

Step 2. (Integral representation of $I$) Put $r = \sin \alpha$ and $s = \sin \beta$, where $ 0 < \alpha < \beta < \frac{\pi}{2}$. Then

\begin{align*} I(r, s) &= \int_{-1}^{1} \frac{1+x}{x\sqrt{1-x^{2}}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx \\ &= \int_{0}^{1} \frac{2}{x\sqrt{1-x^{2}}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx \qquad (\because \text{ parity}) \\ &= \int_{1}^{\infty} \frac{2}{\sqrt{x^{2}-1}} \log \left( \frac{x^{2} + 2rsx + (r^{2} + s^{2} - 1)}{x^{2} - 2rsx + (r^{2} + s^{2} - 1)} \right) \, dx \qquad (x \mapsto x^{-1}) \\ &= \int_{0}^{1} \frac{2}{t} \log \left( \frac{\left(t+t^{-1}\right)^{2} + 4rs\left(t+t^{-1}\right) + 4(r^{2} + s^{2} - 1)}{\left(t+t^{-1}\right)^{2} - 4rs\left(t+t^{-1}\right) + 4(r^{2} + s^{2} - 1)} \right) \, dt, \end{align*}

where in the last line we utilized the substitution $x = \frac{1}{2}(t + t^{-1})$. If we introduce the quartic polynomial \begin{align*} p(t) = t^{4} + 4rst^{3} + (4r^{2}+4s^{2}-2)t^{2} + 4rst + 1, \end{align*}

then by the property $p(1/t) = t^{-4}p(t)$, we can simplify

\begin{align*} I(r, s) &= 2 \int_{0}^{1} \frac{\log p(t) - \log p(-t)}{t} \, dt = \int_{0}^{\infty} \frac{\log p(t) - \log p(-t)}{t} \, dt \\ &= - \int_{0}^{\infty} \left( \frac{p'(t)}{p(t)} + \frac{p'(-t)}{p(-t)} \right) \log t \, dt = - \frac{1}{2} \Re \int_{-\infty}^{\infty} \left( \frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right) \log z \, dz, \end{align*}

where we choose the branch cut of $\log$ in such a way that it avoids the upper-half plane

$$\Bbb{H} = \{ z \in \Bbb{C} : \Im z > 0 \}.$$

Step 3. (Residue calculation) Since

$$ f(z) := \left( \frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right) \log z = O\left(\frac{\log z}{z^{2}} \right) \quad \text{as } z \to \infty, $$

by replacing the contour of integration by a semicircle of sufficiently large radius, it follows that

\begin{align*} I(r, s) = - \frac{1}{2} \Re \left\{ 2 \pi i \sum_{z_{0} \in \Bbb{H}} \operatorname{Res}_{z = z_{0}} f(z) \right\} = \pi \Im \sum_{z_{0} \in \Bbb{H}} \operatorname{Res}_{z = z_{0}} f(z). \end{align*}

(It turns out that $f(z)$ has only logarithmic singularity at the origin. So it does not account for the value of $I(r, s)$.) But by a simple calculation, together with the condition $ 0 < \alpha < \beta < \frac{\pi}{2}$, we easily notice that the zeros of $p(z)$ are exactly

$$ e^{\pm i(\alpha + \beta)} \quad \text{and} \quad -e^{\pm i(\alpha - \beta)}. $$

Now let $Z_{+}$ be the set of zeros of $p(z)$ in $\Bbb{H}$ and $Z_{-}$ be the set of zeros of $p(z)$ in $-\Bbb{H}$. Then

$$ Z_{+} = \{ e^{i(\beta+\alpha)}, -e^{-i(\beta - \alpha)} \} \quad \text{and} \quad Z_{-} = \{ e^{-i(\beta+\alpha)}, -e^{i(\beta- \alpha)} \}. $$

This in particular shows that

$$ \frac{p'(z)}{p(z)}\log z = \sum_{z_{0} \in Z_{+}} \frac{\log z}{z - z_{0}} + \text{holomorphic function on } \Bbb{H} $$


$$ \frac{p'(-z)}{p(-z)}\log z = -\sum_{z_{0} \in -Z_{-}} \frac{\log z}{z - z_{0}} + \text{holomorphic function on } \Bbb{H}. $$

So it follows that

\begin{align*} I(r, s) &= \pi \Im \left\{ \sum_{z_{0} \in Z_{+}} \log z_{0} - \sum_{z_{0} \in -Z_{-}} \log z_{0} \right\} \\ &= \pi \Im \left\{ \log e^{i(\beta+\alpha)} + \log e^{i(\pi-\beta+\alpha)} - \log e^{i(\pi-\beta-\alpha)} - \log e^{i(\beta-\alpha)} \right\} \\ &= \pi \Im \left\{ i(\beta+\alpha) + i(\pi-\beta+\alpha) - i(\pi-\beta-\alpha) - i(\beta-\alpha) \right\} \\ &= 4\pi \alpha = 4\pi \arcsin r. \end{align*}

This completes the proof.

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    $\begingroup$ Very nice, and it helps me simplify a piece of my proof as well (I think our solutions have more in common than not). One question, though: what about the branch point of the log in the residue calculation? I know it doesn't seem to matter as you do end up with the correct solution, but you may want to say something about avoiding the branch point at the origin and defining a branch of the log (which I think you do anyway with your restrictions on $\alpha$ and $\beta$) in the complex plane. $\endgroup$
    – Ron Gordon
    Commented Nov 17, 2013 at 8:45
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    $\begingroup$ @RonGordon, As written in the solution, the branch cut of the log is chosen so that it avoids the upper half plane. So it would be safe if we choose it as the negative y-axis, but I think the standard branch cut $(-\infty, 0)$ would also works. $\endgroup$ Commented Nov 17, 2013 at 15:44
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    $\begingroup$ Oh my, there it is. My bad, so sorry. In any case, reading your solution helped me simplify a small part of mine, so thanks. $\endgroup$
    – Ron Gordon
    Commented Nov 17, 2013 at 15:53

Our aim is to give an elementary proof of Proposition formula (1) in the answer of @sos440. We first note that $$ \min_{-1\leq x\leq1}\{1\pm2rsx+(r^{2}+s^{2}-1)x^{2}\}>0. $$ Indeed, if $x=\pm1$ then $$ 1\pm2rsx+(r^{2}+s^{2}-1)x^{2}\geq(r-s)^{2}>0, $$ if $x=0$ then $$ 1\pm2rsx+(r^{2}+s^{2}-1)x^{2}=1>0, $$ if $-1<x<1$, $x\neq0$ then the equations \begin{eqnarray*} \frac{\partial}{\partial s}(1\pm2rsx+(r^{2}+s^{2}-1)x^{2}) & = & 0,\\ \frac{\partial}{\partial r}(1\pm2rsx+(r^{2}+s^{2}-1)x^{2}) & = & 0, \end{eqnarray*} give $\pm r=sx$, $\pm s=rx$, which is impossible.

In the second step we show that $I(r,s)$ is independent of $s$. $$ \frac{\partial}{\partial s}I(r,s)=\int_{-1}^{1}\sqrt{\frac{1+x}{1-x}}\cdot\frac{4r(1+(r^{2}-s^{2}-1)x^{2})}{(1-2rsx+(r^{2}+s^{2}-1)x^{2})(1+2rsx+(r^{2}+s^{2}-1)x^{2}}\, dx. $$ Substituting $x:=-x$ and adding them we obtain $$ 2\frac{\partial}{\partial s}I(r,s)=\int_{-1}^{1}\frac{2}{\sqrt{1-x^{2}}}\cdot\frac{4r(1+(r^{2}-s^{2}-1)x^{2})}{(1-2rsx+(r^{2}+s^{2}-1)x^{2})(1+2rsx+(r^{2}+s^{2}-1)x^{2}}\, dx, $$ that is, $$ \frac{\partial}{\partial s}I(r,s)=\int_{-1}^{1}\frac{1}{\sqrt{1-x^{2}}}\cdot\frac{4r(-s^{2}+r^{2}-1)x^{2}+4r}{1+(r^{2}+s^{2}-1)^{2}x^{4}+(2s^{2}-4r^{2}s^{2}+2r^{2}-2)x^{2}}\, dx. $$ Substituting $x:=\sin(t)$ we have $$ \frac{\partial}{\partial s}I(r,s) = \int_{-\pi/2}^{\pi/2}\frac{4r(-s^{2}+r^{2}-1)\sin(t)^{2}+4r}{1+(r^{2}+s^{2}-1)^{2}\sin(t)^{4}+(2s^{2}-4r^{2}s^{2}+2r^{2}-2)\sin(t)^{2}}\, dt $$ $$ =\int_{-\pi/2}^{\pi/2}-\frac{8r((-s^{2}+r^{2}-1)\cos(2t)+s^{2}-r^{2}-1)}{(r^{2}+s^{2}-1)^{2}\cos(2t)^{2}-2(r^{2}-s^{2}-1)(r^{2}+1-s^{2})\cos(2t)+r^{4}+(2-6s^{2})r^{2}+(s^{2}+1)^{2}}\, dt $$ $$ = \int_{-\pi}^{\pi}-\frac{4r((-s^{2}+r^{2}-1)\cos(y)+s^{2}-r^{2}-1)}{(r^{2}+s^{2}-1)^{2}\cos(y)^{2}-2(r^{2}-s^{2}-1)(r^{2}+1-s^{2})\cos(y)+r^{4}+(2-6s^{2})r^{2}+(s^{2}+1)^{2}}\, dy. $$ Introducing the new variable $T:=\tan\frac{y}{2}$ we obtain \begin{eqnarray*} \frac{\partial}{\partial s}I(r,s) & = & \int_{-\infty}^{\infty}-\frac{4r(s^{2}-r^{2})T^{2}-4r}{(r-s)^{2}(r+s)^{2}T^{4}+((2-4s^{2})r^{2}+2s^{2})T^{2}+1}\, dT\\ & = & -\frac{4r(s^{2}-r^{2})}{(r-s)^{2}(r+s)^{2}}\int_{-\infty}^{\infty}\frac{T^{2}+a}{T^{4}+bT^{2}+b^{2}/4+d}\, dT\\ & = & -\frac{4r(-s^{2}+r^{2})}{(r-s)^{2}(r+s)^{2}}\cdot\frac{(2a(b^{2}+4d)+(b^{2}+4d)^{3/2})\pi}{(b^{2}+4d)^{3/2}\sqrt{\sqrt{b^{2}+4d}+b}}, \end{eqnarray*} where $$ a=-\frac{1}{s^{2}-r^{2}}, $$ $$ b=\frac{(2-4s^{2})r^{2}+2s^{2}}{(r-s)^{2}(r+s)^{2}}, $$ $$ b^{2}+4d=\frac{4}{(r-s)^{2}(r+s)^{2}}. $$ It gives $2ab^{2}+8da+(b^{2}+4d)^{3/2}=0$.

Since $\frac{\partial}{\partial s}I(r,s)=0$ we have $$ I(r,s)=I(r,1)=\int_{-1}^{1}\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\log\left(\frac{(1+rx)^{2}}{(1-rx)^{2}}\right)dx. $$ From this $$ \frac{\partial}{\partial r}I(r,1)=\int_{-1}^{1}\sqrt{\frac{1+x}{1-x}}\frac{4}{1-r^{2}x^{2}}\, dx. $$ Similarly as above we get $$ \frac{\partial}{\partial r}I(r,1)=\int_{-1}^{1}\frac{4}{\sqrt{1-x^{2}}(1-r^{2}x^{2})}\, dx=\frac{4\pi}{\sqrt{1-r^{2}}}=4\pi(\arcsin r)'. $$ It implies $$ I(r,1)=4\pi\arcsin r+C. $$ Taking the limit $\lim_{r\to0+}$ we obtain $C=0$, that is, $I(r,s)=4\pi\arcsin r$.

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    $\begingroup$ @sos440 Thanks. It was not a trivial task. How could you discover your general Proposition? $\endgroup$
    – vesszabo
    Commented Mar 9, 2014 at 19:28
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    $\begingroup$ I first tried variants of the original integral by changing coefficients. Using inverse symbolic calculators, I found some patterns. Then I tried choose a nice parameters that makes the (conjectured) result look simple. $\endgroup$ Commented Mar 9, 2014 at 20:22

For the purposes of alternative methods, it may be of interest to note that the integrand

$$f(x)=\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\log\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)$$ may be rewritten in terms of hyperbolic trigonometric functions. Using $$\tanh^{-1}(z) = \frac{1}{2}\log\left(\frac{1+z}{1-z}\right),$$ and we obtain

$$f(x)=\frac{1}{x}e^{\tanh^{-1}x}\log\left(\frac{1+\frac{2x}{1+2x^2}}{1-\frac{2x}{1+2x^2}}\right) = e^{\tanh^{-1} x}\left(\frac{2\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)}{x}\right).$$

The rational function in the bracket, which we will denote $s(x)$, is symmetric about $x=0$.

The desired integral is

$$I=\int_{-1}^1 f(x)dx = \int_{-1}^1e^{\tanh^{-1}x}s(x)dx,$$

which, by adding the indicated useful definite integral to both side, gives

$$I + \int_{-1}^1 e^{-\tanh^{-1}x}s(x)dx = 2\int_{-1}^1 \frac{s(x)dx}{\sqrt{1-x^2}}.$$

Now using the change of variable $x=-y$ we have $$\int_{-1}^1 e^{-\tanh^{-1} x}s(x)dx = -\int_1^{-1} e^{\tanh y}s(-y)dy = \int_{-1}^1 e^{\tanh y}s(y)dy = I,$$ by the symmetry of $s(x)$. Hence, we finally obtain

$$I = \int_{-1}^1\frac{s(x)dx}{\sqrt{1-x^2}} = 2\int_{-1}^1\frac{1}{x\sqrt{1-x^2}}\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx.$$

This integral is symmetric about $x=0$, so we have

$$I=4\int_0^1\frac{1}{x\sqrt{1-x^2}}\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx,$$ which can be rewritten $$I=-4\int_0^1\left(\frac{d}{dx}\text{sech}^{-1}x\right)\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx.$$

Using integration by parts this results in


We could also make the change of variable $y=\text{sech}^{-1}x$ to obtain

$$I=8\int_0^\infty\frac{y(\cosh^2(y)-2)\sinh y}{\cosh^4(y)+4}dy= 8\int_0^\infty\frac{y\sinh^3 y}{\cosh^4y+4}dy-8\int_0^\infty\frac{y\sinh y}{\cosh^4 y+4}dy.$$

  • 8
    $\begingroup$ How have I not seen this one prior 0_0 gorgeous solution $\endgroup$ Commented May 15, 2017 at 22:54

This answer provides a way to find $I=\displaystyle\int_0^1\dfrac{\ln\left(x^4-2x^2+5\right)-\ln\left(5x^4-2x^2+1\right)}{1-x^2}\ dx$ (which @RonGordon obtained above) with differentiating under the integral sign. A $u$-substitution of $u=\dfrac{1+x^2}{1-x^2}$ yields this.

$$I=\dfrac{1}{2}\displaystyle\int_1^\infty\dfrac{\ln\left(\frac{u^2+2u+2}{u^2-2u+2}\right)}{\sqrt{u^2-1}}\ du.$$ Now integrate by parts with $a=\ln\left(\frac{u^2+2u+2}{u^2-2u+2}\right)$ and $db=\dfrac{du}{\sqrt{u^2-1}}.$ $$I=\left.\ln\left(\dfrac{u^2+2u+2}{u^2-2u+2}\right)\ln(u+\sqrt{u^2-1})\right]^\infty_1+2\displaystyle\int_1^\infty\dfrac{u^2-2}{u^4+4}\ln\left(u+\sqrt{u^2-1}\right)\ du$$ The first term is equal to $0$, so we are left with this. $$I=2\displaystyle\int_1^\infty\dfrac{u^2-2}{u^4+4}\ln\left(u+\sqrt{u^2-1}\right)\ du$$ We now begin the step of differentiating under the integral. Consider the following integral: $$f(a)=a\displaystyle\int_1^\infty\dfrac{x^2-a^2}{x^4+a^4}\ln\left(x+\sqrt{x^2-1}\right)\ dx$$ Note that trivially, $f(0)=0.$ A quick $u=\dfrac{x}{a}$ yields this. $$f(a)=\displaystyle\int_{\frac{1}{a}}^\infty\dfrac{u^2-1}{u^4+1}\ln\left(au+\sqrt{(au)^2-1}\right)\ du$$ Differentiating with respect to $a$ and using the Chain Rule, we get this. $$f'(a)=-1\times\dfrac{-1}{a^2}\times\dfrac{\left(\frac{1}{a}\right)^2-1}{\left(\frac{1}{a}\right)^4+1}\ln\left(a\left(\dfrac{1}{a}\right)+\sqrt{\left(a\left(\dfrac{1}{a}\right)\right)^2-1}\right)+\displaystyle\int_{\frac{1}{a}}^\infty\dfrac{x^2-1}{x^4+1}\times\dfrac{x}{\sqrt{(ax)^2-1}}\ dx$$ Luckily, the first term cancels, so we are left with this. $$f'(a)=\displaystyle\int_{\frac{1}{a}}^\infty\dfrac{x^2-1}{x^4+1}\times\dfrac{x}{\sqrt{(ax)^2-1}}\ dx$$ A $u$-substitution of $u=\sqrt{(ax)^2-1}$ yields this. $$f'(a)=\displaystyle\int_0^\infty\dfrac{u^2+1-a^2}{(u^2+1)^2+a^4}\ du$$ Consider the integral with $u\mapsto\dfrac{\sqrt{a^4+1}}{u}$ $$f'(a)=\dfrac{1}{\sqrt{a^4+1}}\displaystyle\int_0^\infty\dfrac{(1-a^2)u^2+(a^4+1)}{u^4+2u^2+(a^2+1)}\ du$$ If we add these two versions of the integral and divided the numerator and denominator of the integrand by $u^2$, we get the following. $$f'(a)=\dfrac{(1-a^2)+\sqrt{a^4+1}}{2\sqrt{a^4+1}}\times\displaystyle\int_0^\infty\dfrac{1+\frac{\sqrt{a^4+1}}{u^2}}{\left(u-\frac{\sqrt{a^4+1}}{u}\right)^2+2\left(1+\sqrt{a^4+1}\right)}\ du$$ We can finally perform a very nice substitution of $w=u-\dfrac{\sqrt{a^4+1}}{u}$ to solve this integral. $$f'(a)=\dfrac{(1-a^2)+\sqrt{a^4+1}}{2\sqrt{a^4+1}}\times\displaystyle\int_{-\infty}^\infty\dfrac{dw}{w^2+2\left(1+\sqrt{a^4+1}\right)}\ dw$$ Thus, we can finally say that $f'(a)=\dfrac{(1-a^2)+\sqrt{a^4+1}}{2\sqrt{a^4+1}}\times\dfrac{\pi}{\sqrt{2\left(1+\sqrt{a^4+1}\right)}}.$ After a bit of considerable algebra, we can simply that to obtain this. $$f'(a)=\dfrac{\pi}{2}\sqrt{\dfrac{\sqrt{a^4+1}-a^2}{a^4+1}}$$ Integrating, we can now say this about the value of $f(a).$ $$f(a)=\dfrac{\pi}{2}\displaystyle\int_0^a\sqrt{\dfrac{\sqrt{x^4+1}-x^2}{x^4+1}}\ dx$$ Only one $u$-substitution of $u=\sqrt{x^4+1}-x^2$ is required here to obtain this. $$f(a)=\dfrac{\pi}{2\sqrt{2}}\displaystyle\int_{\sqrt{a^4+1}-a^2}^1\dfrac{du}{\sqrt{1-u^2}}$$ This, of course, is equal to $\dfrac{\pi\arccos\left(\sqrt{a^4+1}-a^2\right)}{2\sqrt{2}}.$

We will now manipulate this result to a function with $\arctan$ in it.


Our desired value for our original integral is $\sqrt{2}f\left(\sqrt{2}\right).$

$$\boxed{\displaystyle\int_0^1\dfrac{\ln\left(x^4-2x^2+5\right)-\ln\left(5x^4-2x^2+1\right)}{1-x^2}\ dx=\pi\arctan\left(\sqrt{\dfrac{\sqrt{5}-1}{2}}\right)=\pi\text{arccot}\sqrt{\phi}}$$

So the final answer to the original problem is $4\pi\text{arccot}\sqrt{\phi}.$

  • 1
    $\begingroup$ Credit goes to Ronak's answer on Brilliant.org (Pulished on Dec 31, 2015). $\endgroup$
    – MathGod
    Commented Dec 27, 2016 at 6:56
  • 4
    $\begingroup$ @IshanSingh I didn't copy from that answer; I was given the original integral and subsequently the initial $u=\dfrac{1+x^2}{1-x^2}$ by a friend who posed this question to me. $\endgroup$
    – Arcturus
    Commented Dec 28, 2016 at 7:58
  • $\begingroup$ You can remove the square root with $4\pi\,\text{arccosec}\,{\phi}$ $\endgroup$
    – Henry
    Commented Sep 17, 2020 at 1:22
  • $\begingroup$ I'm glad my answer is represented here as well :) $\endgroup$ Commented May 27 at 20:06

Noteworthy, RIES (http://mrob.com/pub/ries/index.html) finds closed form from numerical value in the form of an equation: $$ \cos{\left( \frac{x}{\pi} \right)}+1=\frac{2}{\phi^6}. $$

Simplifying above, we get another form of the result: $$ I = \pi \arccos{(17-8\sqrt{5})}. $$

  • 3
    $\begingroup$ I used RIES also but made some simple adjustments to get a simpler form of the answer $I=8\pi \arcsin(\phi-1)$. $\endgroup$
    – Somos
    Commented Jun 25, 2017 at 19:09
  • $\begingroup$ @Somos Your answer (about $16.74$) seems to be double that of Andrzej Odrzywolek and of Arcturus (about $8.37$) $\endgroup$
    – Henry
    Commented Sep 17, 2020 at 1:19
  • $\begingroup$ @Henry Oops! Thanks. The correct answer is $\,I=4\pi \arcsin(\phi-1)$. Somehow I doubled it. $\endgroup$
    – Somos
    Commented Sep 17, 2020 at 2:19

Figured I would contribute and add a self-contained real analytic method:

I will use the following representations that are fairly straight-forward to prove: $$2\int_{0}^{\infty}\frac{\cos(x)-\cos(t \, x)}{x}\left(e^{-x\sqrt{-a-bi}}+e^{-x\sqrt{-a+bi}}\right) \, dx=\ln \left((t^2 -a)^2 + b^2\right)-\ln \left((1-a)^2+b^2\right)$$ $$\int_{0}^{\infty}\frac{\sin(x)}{x}\left(e^{-x\sqrt{-a-bi}}+e^{-x\sqrt{-a+bi}}\right) \, dx=\arctan\left(\frac{1}{\sqrt{-a-bi}}\right)+\arctan\left(\frac{1}{\sqrt{-a+bi}}\right)$$ $$\int_{0}^{\infty}\frac{\cos(t)-\cos(t \, x)}{x^2-1} \, dx=\frac{\pi}{2}\sin(t),\> \text{for} \, \> t\geq 0$$

Now we can begin evaluating $I$: $$I=\int_{-1}^{1}\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right) \, dx$$

Enforce the substitution $\sqrt{\frac{1+x}{1-x}} = u$: $$\implies I = \int_{0}^{\infty}\frac{4u^2}{u^4-1}\ln\left(\frac{5u^4-2u^2+1}{u^4-2u^2+5}\right)\, du \stackrel{u \, \mapsto \frac{1}{u}}{=}\int_{0}^{\infty}\frac{4}{u^4-1}\ln\left(\frac{5u^4-2u^2+1}{u^4-2u^2+5}\right) \, du$$

From adding the two, we can deduce: $$I = \int_{0}^{\infty}\frac{2}{u^2-1}\ln\left(\frac{5u^4-2u^2+1}{u^4-2u^2+5}\right) \, du$$

Since $$\ln\left(5 u^4-2 u^2+1\right)-\ln(4)=\ln\left(\left(u^2-\frac{1}{5}\right)^2+\frac{4}{25}\right)-\ln\left(\frac{4}{5}\right)$$ $$\ln(u^4-2 u^2+5)-\ln(4)=\ln((u^2-1)^2+4)-\ln(4)$$

We can write $$A=\int_{0}^{\infty}\frac{2}{u^2-1}\left(\ln\left(\left(u^2-\frac{1}{5}\right)^2+\frac{4}{25}\right)-\ln\left(\frac{4}{5}\right)\right)\,du\\ B=\int_{0}^{\infty}\frac{2}{u^2-1}\left(\ln((u^2-1)^2+4)-\ln(4)\right)\, du$$ such that $I = A-B$

Now from the starting representations we deduce: $$\begin{align} \implies A &= \int_{0}^{\infty}\frac{4}{u^2-1}\int_{0}^{\infty}\frac{\cos(t)-\cos(u \, t)}{t}\left(e^{-t\sqrt{(-1-2i)/5}}+e^{-t\sqrt{(-1+2i)/5}}\right)\,dt\,du \\ &= 2\pi\int_{0}^{\infty}\frac{\sin(t)}{t}\left(e^{-t\sqrt{(-1-2i)/5}}+e^{-t\sqrt{(-1+2i)/5}}\right)\, dt \\ &= 2\pi\arctan\left(\frac{\sqrt{5}}{\sqrt{-1-2i}}\right)+2\pi\arctan\left(\frac{\sqrt{5}}{\sqrt{-1+2i}}\right) \\ &= \Re\left(4\pi \arctan\left(\frac{\sqrt{5}}{\sqrt{-1-2i}}\right)\right)\end{align}$$

Similarly, one finds $$B=\Re\left(4\pi\arctan\left(\frac{1}{\sqrt{-1-2i}}\right)\right)$$

By using the identity: $$\arctan (x)+\arctan (y)=\arctan \left(\frac{x+y}{1-x y}\right)$$

One deduces $$\boxed{I=4\pi \operatorname{arccot} \left(\sqrt{\phi}\right)}$$


Well, 9 years later, this answer is going to utilize an approach via a contour integral. Denote $$I = \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \ln \left(\frac{2x^2+2x+1}{2x^2-2x+1}\right) \,\mathrm{d}x \in \mathbb{R}$$ our integral of interest.

Complex function

Let $$f(z) = \frac{1}{z} \sqrt{\frac{z+1}{z-1}}\left(\ln \left({\frac{2z+1+i}{2z-1+i}}\right) + \frac{\pi i}{2}\right),$$ where $\sqrt{w}$ and $\ln w$ are chosen such that they have a branch cut at the negative real axis (the so-called principal branches) in $w$-plane, that is $\arg w \in \left(-\pi,\pi\right]$. We denote $$h(z) = \frac{1}{z} \sqrt{\frac{z+1}{z-1}}, \qquad g(z) = \ln \left({\frac{2z+1+i}{2z-1+i}}\right) + \frac{\pi i}{2}$$ so $f(z) = h(z)g(z)$. To find the branch cuts of $f(z)$, we need to find which points on the $z$-plane are mapped onto the negative real axis in $w$. That is to solve

$$ \boxed{\sqrt{w}}_{B.C.}:\quad w=\frac{z+1}{z-1} = -t,\qquad t>0; \qquad\qquad \boxed{\ln w}_{B.C.}:\quad w=\frac{2z+1+i}{2z-1+i} = -t, \qquad t>0.$$ After simple manipulations, we arrive at $$ \boxed{h(z)}_{B.C.}:\quad z=\frac{t-1}{t+1} \in (-1,1); \qquad\qquad \boxed{g(z)}_{B.C.}:\quad z = \frac{t-1}{2(t+1)} - \frac{i}{2} \in \left(\frac{-1-i}{2},\frac{1-i}{2}\right).$$ where $(z_1,z_2)$ means here a line segment from $z_1$ to $z_2$ for any in general complex $z_1,z_2$.

Poles and residues

At $z\to \infty$, we have $f(z) = \frac{\pi i}{2z} + O(1/z^2)$, a simple pole. So $$\operatorname{Res}_\infty f(z) = - \lim_{z\to\infty} z f(z) = - \frac{\pi i}{2}.$$ Note that there is no pole at $z=0$ since $$\frac{2z+1+i}{2z-1+i} = -i-2iz+O(z^2),$$ so $$g(z) = -\frac{\pi i}{2}+2z + \frac{\pi i}{2} +O(z^2) = 2z +O(z^2),$$ which cancels the simple pole of $h(z)$ at $0 \pm 0i$.

Branch jumps

  • $(-1,1):$ Let $t \in (-1,1)$. Since the complex square root function changes sign on the branch discontinuity, $h(t+i0) = -h(t-i0)$ and so $f(t+i0) = -f(t-i0)$.
  • $\left(\frac{-1-i}{2},\frac{1-i}{2}\right):$ Let $t \in (-1/2,1/2)$. For $z = t - 1/2 i \pm i0,$ $$ \frac{2z+1+i}{2z-1+i} \bigg{|}_{t - i/2 \pm i0} = -r + \left(\frac{2z+1+i}{2z-1+i}\right)' \bigg{|}_{t - i/2} (\pm i0) = - r - \frac{4}{(1-2t)^2} (\pm i 0) = -r \mp i0.$$ where $-r$ is some negative number (branch cut mapped on negative numbers $-r$). Hence, $$g(t+i0) - g(t-i0) = \ln(-r-i0)-\ln(-r+i0) = -2\pi i.$$

Residue theorem

We integrate $f(z)$ along contour $C$ defined in Figure below. The contour is also split as $C = C_1 + \ldots + C_6$.

Contour of integration

According to the Residue theorem, $$\oint_C f(z) \, \mathrm{d}x = - 2\pi i \operatorname{Res}_\infty f(z) = -\pi^2.$$

On the other hand, $$ \oint_C f(z) \, \mathrm{d}z = \sum_{k=1}^6 \int_{C_k} f(z) \, \mathrm{d}z. \tag{1}$$


For $C_1, \ldots C_6$, we have the following parametrisation:

  • $ \circleddash C_1: z = t + i0, t\in (-1,1), dz = dt$; Since $h(z)$ has a branch cut here, $f(t+i0) = h(t+i0)g(t)$. To obtain $h(t+i0)$, we first compute $$ \frac{z+1}{z-1}\bigg{|}_{t+i0} = \frac{z+1}{z-1}\bigg{|}_{t}+\left(\frac{z+1}{z-1}\right)'\bigg{|}_{t} i0 = -\frac{1+t}{1-t} - \frac{2}{(1-t)^2}i0 = -\frac{1+t}{1-t}-i0,$$ so $h(t+i0) = -\frac{i}{t} \sqrt{\frac{1+t}{1-t}}$. For $g(z)$, we can write $$ g(t) = \ln\left|\frac{2t+1+i}{2t-1+i}\right| + \arg\left(\frac{2t+1+i}{2t-1+i}\right) + \frac{\pi i}{2} = \frac12 \ln\left(\frac{2t^2+2t+1}{2t^2-2t+1}\right) + i \xi(t)$$ for some real function $\xi(t)$ which vanishes at $t=0$. Although we don't need this function, we express it anyway for completeness: Since $-\pi < \arg(2t+1+i)-\arg (2t-1+i) < \pi$, we have $\arg\left(\frac{2t+1+i}{2t-1+i}\right) = \arg(2t+1+i)-\arg (2t-1+i)$ and so $$ \xi(t) = \arctan\left(2t-1\right)-\arctan\left(2t+1\right) + \frac{\pi}{2}.$$ Overall, $$ f(t+i0) = - \frac{i}{t}\sqrt{\frac{1+t}{1-t}} \left( \frac12 \ln\left(\frac{2t^2+2t+1}{2t^2-2t+1}\right) + i \xi(t) \right) .$$ Hence $$ \int_{C_1} f(z) \, \mathrm{d}z = -\int_{\circleddash C_1} f(z) \, \mathrm{d}z = - \int_{-1}^{1} f(t+i0) \, \mathrm{d}t = \frac{i}{2}I - I_0,$$ where we denote $$I_0 = \int_{-1}^{1} \frac{\xi(t)}{t}\sqrt{\frac{1+t}{1-t}} \, \mathrm{d}t \in \mathbb{R}.$$
  • $C_2$: Using knowledge of branchjumps, $$\int_{C_2} f(z) \, \mathrm{d}z = -\int_{\circleddash C_2} f(z) \, \mathrm{d}z = -\int_{C_1} (-f(z)) \, \mathrm{d}z = \frac{i}{2}I - I_0.$$
  • $C_3 = \circleddash C_6$, so $$\int_{C_3} f(z) \, \mathrm{d}z + \int_{C_6} f(z) \, \mathrm{d}z = \int_{C_3} f(z) \, \mathrm{d}z - \int_{\circleddash C_6} f(z) \, \mathrm{d}z = 0$$
  • $C_4$: Using knowledge of branchjumps, $$\int_{C_4} f(z) \, \mathrm{d}z = -\int_{\circleddash C_4} h(z)g(z) \, \mathrm{d}z = -\int_{C_5} h(z)(g(z)-2\pi i) \, \mathrm{d}z = -\int_{C_5} f(z) \, \mathrm{d}z + 2\pi i\int_{C_5} h(z) \, \mathrm{d}z.$$

Comparison & Antiderivative

Comparing the terms, that is by using $(1)$, we get $$-\pi^2 = iI - 2I_0 +2\pi i \int_{C_5} h(z) \, \mathrm{d}z.\tag{2}$$

We now solve the remaining integral $\int_{C_5} h(z) \, \mathrm{d}z$. For brevity, we denote $a = \frac{-1-i}{2}, b =\frac{1-i}{2}$, so $C_5 = (a,b)$. Notice $h(z)$ is now holomorphic in the vicinity of $C_5$. If we find its antiderivative $H(z)$ also holomorphic there, then

$$ \int_{C_5} h(z) \, \mathrm{d}z = H(b) - H(a).\tag{3}$$

This task is rather simple. First, we start by integrating $h(z)$ as if it was a real function. For $x>1$, we have by substitution $(x-1)/(x+1)=u^2$,

$$\int h(x) \, \mathrm{d}x = \int \frac{1}{x}\sqrt{\frac{x+1}{x-1}} \, \mathrm{d}x = \int \frac{4u^2}{u^4-1} \, \mathrm{d}u = 2\operatorname{argcoth}\left(\sqrt{\frac{x+1}{x-1}}\right) + 2\operatorname{arccot}\left(\sqrt{\frac{x+1}{x-1}}\right).$$

Hence, a suitable antiderivative in the complex plane is the following: $$H(z) = \ln \left(\frac{\sqrt{\frac{z+1}{z-1}}+1}{\sqrt{\frac{z+1}{z-1}}-1}\right) - i\ln \left(\frac{\sqrt{\frac{z+1}{z-1}}+i}{\sqrt{\frac{z+1}{z-1}}-i}\right).$$

We know the branch of $H(z)$ due to $\nu = \sqrt{\frac{z+1}{z-1}}$ is $z \in (-1,1)$. Assuming the principal branches, we now check where are the other possible branches of $H(z)$ due to $\ln$'s.

$$ \boxed{\ln{w}}_{B.C.}:\quad w=\frac{\nu+1}{\nu-1} = -t,\qquad t>0; \qquad\qquad \boxed{\ln w}_{B.C.}:\quad w=\frac{\nu+i}{\nu-i} = -t, \qquad t>0.$$ $$ \boxed{w(\nu)}_{B.C.}:\quad \nu = \frac{t-1}{t+1} = \in (-1,1) ; \qquad\qquad \boxed{w(\nu)}_{B.C.}:\quad \nu=i\frac{t-1}{t+1} \in (-i,i).$$ $$ \boxed{H(z)}_{B.C.}:\quad z = \frac{\nu^2+1}{\nu^2-1} \in (-\infty,0).$$

Hence, the B.C. of $H(z)$ lies at $(-\infty,1)$, safely away from $C_5$. Substituting $(3)$ into $(2)$, we get the value of $I$. And as a bonus, also $I_0$. For $I$,

$$ I = 2\pi\operatorname{Re} (H(a) - H(b)).\tag{4}$$

Substituing $a$ and $b$ into $\frac{z+1}{z-1}$, we get

$$\frac{a+1}{a-1} = -\frac{1}{5} + \frac{2i}{5} = \frac{1}{q^2}e^{2\alpha i}, \qquad \frac{b+1}{b-1}=-1+2i = q^2 e^{2\alpha i}$$ with $q = 5^{1/4}>1$ and $\alpha = \frac{\pi}{2} - \frac{1}{2}\arctan 2$. Note that $\pi/4 < \alpha < \pi/2$, so

$$\sqrt{\frac{a+1}{a-1}} = \frac{1}{q} e^{\alpha i}, \qquad \sqrt{\frac{b+1}{b-1}} = q e^{\alpha i}.$$

In the case of $b$, substituing into $\nu = \sqrt{\frac{z+1}{z-1}}$ and after simple manipulations, $$\frac{\nu + 1}{\nu - 1} \bigg{|}_{\nu = q e^{\alpha i}} = \frac{q^2-2 i q \sin\alpha-1}{q^2-2 q \cos\alpha+1}, \qquad \frac{\nu + i}{\nu - i} \bigg{|}_{\nu = q e^{\alpha i}} \frac{q^2+2 i q \cos\alpha-1}{q^2-2 q \sin\alpha+1}.$$

Hence, since the denominators are real, $$H(b) = \ln\left(q^2-2iq\sin\alpha-1\right)-\ln\left(q^2-2q\cos\alpha+1\right)-i\ln\left(q^2+2iq\cos\alpha-1\right)+i \ln\left(q^2-2q\sin\alpha+1\right),$$

expanding the logarithms into real and imaginary part, $$H(b) = \frac{1}{2} \ln\left(q^4-2 q^2 \cos (2 \alpha )+1\right)+i \arg \left(q^2-2 i q\sin\alpha-1\right)-\ln\left(q^2-2q\cos\alpha+1\right)-\frac{1}{2} i \ln\left(1+2q^2\cos (2 \alpha )+q^4\right)+\arg \left(q^2+2iq\cos\alpha-1\right)+i\ln\left(q^2-2q \sin\alpha+1\right)$$

Since we are interested only in real part (just $I$), $$\operatorname{Re} H(b) = \frac{1}{2} \ln\left(q^4-2 q^2 \cos (2 \alpha )+1\right)-\ln\left(q^2-2q\cos\alpha+1\right)+\arg \left(q^2+2iq\cos\alpha-1\right)$$

Similarly for $a$ (change $q$ to $1/q$). \begin{align*} \operatorname{Re} H(a) & = \frac{1}{2} \ln\left(\frac{1}{q^4}\!-\!\frac{2}{q^2} \cos (2 \alpha)\!+\!1\right)-\ln\left(\frac{1}{q^2}\!-\!\frac{2}{q}\cos\alpha\!+\!1\right)+\arg \left(\frac{1}{q^2}\!+\!\frac{2i}{q}\cos\alpha\!-\!1\right)\\ &= \frac{1}{2} \ln\left(q^4\!-\!2 q^2 \cos (2 \alpha )\!+\!1\right)-\ln\left(q^2\!-\!2q\cos\alpha\!+\!1\right)+\pi - \arg \left(q^2\!+\!2iq\cos\alpha\!-\!1\right). \end{align*}

Therefore, we have ourselves the grand finale, $$I = 4\pi \left(\frac{\pi}{2} - \arg \left(q^2\!+\!2iq\cos\alpha\!-\!1\right)\right) = 4\pi \operatorname{arccot} \frac{2q\cos\alpha}{q^2-1}.$$

To show this result is the same as the ones found already by others, note that $$\left(\frac{2q\cos\alpha}{q^2-1}\right)^2 = \frac{4q^2\sin^2 \left(\frac{\pi}{2} -\alpha\right)}{(q^2-1)^2} = \frac{4\sqrt5\sin^2 \left(\frac12 \arctan 2\right)}{(\sqrt5-1)^2} = 2\sqrt5\,\frac{1-\frac{1}{\sqrt5}}{(\sqrt5-1)^2} = \frac{1+\sqrt5}{2}.$$

  • 1
    $\begingroup$ This is THE HARDCORE way. $\endgroup$
    – oO_ƲRF_Oo
    Commented Sep 14, 2023 at 21:45

This is not really an answer, but grossly too long for an comment. I didn't know how to simplify it beyond the final solution.

$$I=\int_{-1}^1 \frac{1}{x}\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)\text{d}{x}$$

Begin with the substitution of $x=-\cos2a$ $$I=\int_{-1}^1 \frac{1}{-\cos2a}\sqrt{\frac{1-\cos2a}{1+\cos2a}}\ln\left(\frac{2\cos^2 2a-2\cos 2a+1}{2\cos^2 2a-2\cos2a+1}\right)\text{d}{x}$$

By the tangent and cos double angle properties

$$I=\int_{-1}^1 -\sec2a|\tan a|\ln\left(\frac{-2\cos^22a+\cos 4a+2}{2\cos2a+\cos4a+2}\right)\text{d}{a}$$

Were just getting started. Now replace $a=\frac{1}{2}\text{gd}(b)$ where $\text{gd}$ is the Gudermannian function.

$$I=\int_{-1}^1 -\sec(\text{gd}(b))|\tan(\text{gd}(\frac{b}{2}))|\ln\left(\frac{-2\cos^2(\text{gd}(b))+\cos (2\text{gd}(b))+2}{2\cos^2(\text{gd}(b))+\cos (2\text{gd}(b))+2}\right)\text{d}{a}$$

Hehe. Now we get to simplify a bit. This is under the definition of Gudermannian properties.

$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{-2\text{sech}^2 b+(\text{sech}^2b+\tanh^2b)+2}{2\text{sech}^2 b+(\text{sech}^2b+\tanh^2b)+2}\right)$$

Now, use properties of $\tanh$ and $\text{sech} $ to simplify even further

$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{(1-\text{sech}^2 b)+2}{(1+\text{sech}^2 b)+2}\right)$$

Our goal is to create an $\text{arctanh}$ function, but that will obviously take some serious effort. Factor out a $3$ to generate that $1$ needed even if it makes an ugly factoring.

$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{3(1-\frac{\text{sech}^2 b}{3})}{3(1+\frac{\text{sech}^2 b}{3})}\right)$$

And now cut out all of the 3's. After this cut, use a property of $\ln$'s to reciprocate the argument of $\ln$. And multiply 2 and 1/2

$$I=\int_{-1}^1 2\text{cosh}\space b|\sinh\frac{b}{2}|\frac{1}{2}\ln\left(\frac{(1+\frac{\text{sech}^2 b}{3})}{(1-\frac{\text{sech}^2 b}{3})}\right)$$

And what do you know! You're there! Use a property of $\ln$ and $\text{arctanh}$ to generate a much CLEANER form (also by throwing the 2 in front).

$$I=2\int_{-1}^1 \text{cosh}\space b|\sinh\frac{b}{2}|\text{arctanh}(\frac{\text{sech}^2b}{3})$$

This function is even, and we can know that because all parts of what is above, $\cosh b,|\sinh b|, $ etc. all even. So we can do the following.

$$I=4\int_{0}^1 \text{cosh}\space b|\sinh\frac{b}{2}|\text{arctanh}(\frac{\text{sech}^2b}{3})$$

This is just an idea, and like I said not a real solution. I have no idea where to continue beyond this, but I thought it may help to come up with a new idea to solve.

  • 2
    $\begingroup$ After further inspection, I messed up my work here. I will leave this post here howver becUse the purpose of the post still holds (ideas to solve) $\endgroup$
    – user285523
    Commented Nov 22, 2015 at 17:42
  • $\begingroup$ Don't you need to change the limits after you make the first change of variable $x = -\cos 2a$? $\endgroup$
    – r9m
    Commented Dec 2, 2015 at 16:59
  • $\begingroup$ @user23055 not really. There are lots of mistakes and only consists of only substitution $\endgroup$
    – user311151
    Commented Apr 2, 2016 at 4:48

$\color{green}{\textbf{Version of 17.09.23.}}$

Let $\;y=\sqrt{\dfrac{1+x}{1-x}},\;$ then $\;x=\dfrac{1-y^2}{1+y^2},\;$ $$I=\int\limits_{-1}^1\dfrac1x\,\sqrt{\dfrac{1+x}{1-x}}\,\ln\,\dfrac{2x^2+2x+1}{2x^2-2x+1}\,\text dx = 4\int\limits_0^\infty \ln\,\dfrac{y^4-2y^2+5}{5y^4-2y^2+1}\,\dfrac{\text dy}{1-y^4}=I_0-I_1,$$ where $$I_0=2\int\limits_0^\infty \ln\,\dfrac{y^4-2y^2+5}{5y^4-2y^2+1}\,\dfrac{\text dy}{1-y^2},\quad I_1=2\int\limits_0^\infty \ln\,\dfrac{y^4-2y^2+5}{5y^4-2y^2+1}\,\dfrac{\text dy}{1+y^2}.\tag1$$ By the substitution $\;y=\dfrac1z\;$ easily to get $$\int\limits_1^\infty \ln\,\dfrac{y^4-2y^2+5}{5y^4-2y^2+1}\,\dfrac{\text dy}{1+y^2}=-\int\limits_0^1 \ln\,\dfrac{z^4-2z^2+5}{5z^4-2z^2+1}\,\dfrac{\text dz}{1+z^2}.$$ I.e. $$I_1=0.\tag2$$

On the other hand, $$I_{0}=I_{00}-I_{01},$$ where $$I_{00}=2\int\limits_0^\infty \ln\,\dfrac{y^4-2y^2+5}{(y^2+1)^2}\,\dfrac{\text dy}{1-y^2},\quad I_{01}=2\int\limits_0^\infty \ln\,\dfrac{5y^4-2y^2+1}{(y^2+1)^2}\,\dfrac{\text dy}{1-y^2},\tag3$$ By the substitution $\,y=\dfrac1z\,$ easily to get $$\int\limits_0^\infty \ln\,\dfrac{5y^4-2y^2+1}{(y^2+1)^2}\,\dfrac{\text dy}{1-y^2}=-\int\limits_0^\infty \ln\,\dfrac{z^4-2z^2+5} {(z^2+1)^2}\,\dfrac{\text dz}{1-z^2},$$ $$I_{01}=-I_{00},\quad I=2I_{00},\tag4$$

$$I=2I_{00}=4\int\limits_0^\infty \ln\,\dfrac{y^4-2y^2+5}{(y^2+1)^2}\,\dfrac{\text dy}{1-y^2}$$ $$=4\int\limits_0^\infty \dfrac{\ln(y^2-1-2i)+\ln(y^2-1+2i)-2\ln(y^2+1)}{1-y^2}\,\text dy$$ $$I=4(J(-1-2i)+J(-1+2i)-2J(1)),\tag5$$ where $$J(p)=\int\limits_0^\infty \dfrac{\ln(y^2+p)}{1-y^2}\,\text dy\tag6,$$ $$\dfrac{\,\text dJ(p)}{\,\text dp} =\int\limits_0^\infty \dfrac{\,\text dy}{(y^2+p)(1-y^2)} =\dfrac{F_1+F_2(p)}{1+p},$$ $$F_1=\int\limits_0^\infty \dfrac1{1-y^2}\,\text dy,\quad F_2(p)=\int\limits_0^\infty \dfrac1{y^2+p}\,\text dy.$$ By the substitution $\,y=\dfrac1t\,$ easily to get $$\int\limits_0^1 \dfrac1{1-y^2}\,\text dy =-\int\limits_1^\infty \dfrac1{1-t^2}\,\text dt,\quad F_1=0.$$ At the same time, $$F_2(p)=\dfrac1{\sqrt p}\,\arctan\dfrac{y}{\sqrt p}\bigg|_0^\infty =\dfrac \pi{2\sqrt p}.$$ Therefore, $$\dfrac{\,\text dJ(p)}{\,\text dp}=\dfrac\pi{2(1+p)\sqrt p},$$ $$J(p)=\int\dfrac\pi{1+p}\,\dfrac{\,\text dp}{2\sqrt p}=\pi \arctan \sqrt p + \text C.$$ From $(5)$ should $$\begin{align} &I=4\pi\left(\arctan\sqrt{-1-2i}+\arctan\sqrt{-1+2i} -2\arctan1\right)\\[4pt] &=4\pi\left(\arctan\dfrac{\sqrt{-1-2i}+\sqrt{-1+2i}}{1-\sqrt5} -\dfrac\pi2\right) =4\pi\left(\arctan\dfrac{\sqrt{2(\sqrt5-1)}}{1-\sqrt5} -\dfrac\pi2\right)\\[4pt] &=4\pi\left(\arctan\left(-\sqrt{\dfrac{\sqrt5+1}2}\right) -\dfrac\pi2\right) =4\pi\left(\pi-\arctan\sqrt{\dfrac{\sqrt5+1}2}-\dfrac\pi2\right),\\[4pt] &\color{green}{\mathbf{I=4\pi\operatorname{arccot}\sqrt{\varphi}\approx 8.37221162660127566162574712109841263808172805388220741371709,}} \end{align}$$ where $\;\varphi=\dfrac{\sqrt5+1}2\;$ is the golden ratio.

  • 2
    $\begingroup$ Nice solution. Although are you sure the wolfram alpha result is correct? For positive real a, it seems the integral should return a real value (assuming the Cauchy principal value), which means the imaginary term probably shouldn't be there. Wolfram also says the answer is for Re(a)>0 which is not that case for a = -1 +/- 2i. Maybe you could try Feynman integration if you want to do it explicitly. Or just look at the explicit antiderivative from Wolfram and treat it carefully. $\endgroup$
    – Andrew
    Commented Jun 23, 2023 at 7:48
  • $\begingroup$ @Andrew Thank you for the deep comment! Honestly, I was not sure in the correctness of this integral too. Undoubtedly, if 𝑎 is real, then its negative value leads to the singularity of ln(z^2+𝑎) in the domain of the integration. At the same time, this problem does not exist in the task conditions. $\endgroup$ Commented Jun 25, 2023 at 9:16
  • $\begingroup$ @Andrew New version (by your notes) is ready. $\endgroup$ Commented Sep 17, 2023 at 15:32

Eight years later.

Starting from @Ron Gordon's substitution $$8 \int_0^{\infty} \frac{(u^2-1)(u^4-6 u^2+1)}{u^8+4 u^6+70 u^4+4 u^2+1} \log(u)\,du$$ since the roots of the polynomials in $\color{red}{ u^2}$ are simple, we can use partial fraction decomposition (I shall not type the formulae) and we face four integrals $$I=\int \frac \alpha {\beta x^2+\gamma}\log(x)\,dx$$ where all coefficients are complex numbers. Then $$I=\frac{i \alpha \left(\text{Li}_2\left(\frac{i x \sqrt{\beta }}{\sqrt{\gamma }}\right)-\text{Li}_2\left(-\frac{i x \sqrt{\beta }}{\sqrt{\gamma }}\right)+\log (x) \left(\log \left(1-\frac{i \sqrt{\beta } x}{\sqrt{\gamma }}\right)-\log \left(1+\frac{i \sqrt{\beta } x}{\sqrt{\gamma }}\right)\right)\right)}{2 \sqrt{\beta\gamma }}$$ which make $$J=\int_0^\infty \frac \alpha {\beta x^2+\gamma}\log(x)\,dx=\frac{i \alpha \left(\log ^2\left(\frac{i \sqrt{\beta }}{\sqrt{\gamma }}\right)-\log ^2\left(-\frac{i \sqrt{\beta }}{\sqrt{\gamma }}\right)\right)}{4 \sqrt{\beta\gamma }}=-\frac{\pi \alpha \log \left(\frac{\beta }{\gamma }\right)}{4 \sqrt{\beta\gamma }}$$ This gives as a result $$8 \int_0^{\infty} \frac{(u^2-1)(u^4-6 u^2+1)}{u^8+4 u^6+70 u^4+4 u^2+1} \log(u)\,du=\pi \left(\pi -\cot ^{-1}\left(\frac{1}{4} \sqrt{22+17 \sqrt{5}}\right)\right)$$ I have not been able to simplify further.


If you look at this question of mine, @Jyrki Lahtonen made the simplification I was not able to do.


I have a relatively straightforward elementary approach.

First you use the substitution $x \rightarrow -x$ and add the integrals to obtain $$I=\int_{-1}^1\frac1x\frac{1}{\sqrt{1-x^2}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx.$$

A further substitution, $x\rightarrow\frac{2x}{1+x^2}$ yields

$$I=\int_{-1}^1\frac1x\ln\left(\frac{4 x^2+(x+1)^4}{4 x^2+(x-1)^4}\right)\ \mathrm dx=2\int_{-1}^1\frac1x\ln\left(4 x^2+(x+1)^4\right)\ \mathrm dx.$$

Define the function J by $$J(\lambda)=\int_{-1}^{1}\frac{\ln\left((x+1)^2+2\, i \,\lambda\, x\right)}{x} dx.$$ Then $$I=2(J(1)+J(-1)).$$

Differentiating under the integral sign gives $$J^{'}(\lambda)=2i\int_{-1}^{1}\frac{1}{\left((x+1)^2+2\, i \,\lambda\, x\right)} dx=\frac{i \pi}{\sqrt{\lambda^2-2i\lambda}}=\frac{i \pi}{\sqrt{1-(1+i\lambda)^2}}.$$

Hence $$J(\lambda)=\pi\arcsin(1+i\lambda)+C.$$ But $$J(0)=2\int_{-1}^{1}\frac{\ln\left(x+1\right)}{x} dx=\pi^2/2$$ which implies $C=0$, hence $$I=2\pi(\arcsin(1+I)+\arcsin(1-I))=4\pi\arcsin(\frac{1}{\phi})=4\,\pi\operatorname{arccot}\sqrt\phi$$ using standard identities.


Since this question received a lot of answers recently, I thought I might partake as well. My solution to this historical question (on this site, at least) requires a lemma, namely: $$ \int_0^{\infty}{\frac{dx^2+e}{ax^4+bx^2+c}}\mathrm{d}x=\frac{ \pi}{2\sqrt{2\sqrt{ac}+b}}\left[ \frac{d}{\sqrt{a}}+\frac{e}{\sqrt{c}} \right] $$ This works for $a,b,c>0, d,e\in \mathbb{R}$. (the variable range can be larger, but this is enough for this question.) I shall provide a proof, too. The first time I had derived this was through complex analysis, but here I will do it more elementarily.
To start with, assume $f$ is a even real function, then, \begin{align*} \int_{-\infty}^{\infty}{f\left( \frac{x}{p^2}-\frac{q^2}{x} \right) \mathrm{d}x}=&\underset{x=pqe^t}{\underbrace{\int_0^{\infty}{f\left( \frac{x}{p^2}-\frac{q^2}{x} \right) \mathrm{d}x}}}+\underset{x=pqe^{-t}}{\underbrace{\int_{-\infty}^0{f\left( \frac{x}{p^2}-\frac{q^2}{x} \right) \mathrm{d}x}}} \\ =&\int_{-\infty}^{\infty}{f\left( \frac{q}{p}\left( e^t-e^{-t} \right) \right) pqe^t\mathrm{d}t}+\int_{-\infty}^{\infty}{f\left( \frac{q}{p}\left( e^{-t}-e^t \right) \right) pqe^{-t}\mathrm{d}t} \\ =&\int_{-\infty}^{\infty}{f\left( 2\frac{q}{p}\sinh \left( t \right) \right) 2pq\cosh \left( t \right) \mathrm{d}t} \stackrel{2\frac{q}{p}\sinh \left( t \right) =x}{=}p^2\int_{-\infty}^{\infty}{f\left( x \right) \mathrm{d}x} \end{align*} So we have \begin{align*} &\int_0^{\infty}{\frac{dx^2+e}{ax^4+bx^2+c}}\mathrm{d}x=\frac{1}{2}\int_{-\infty}^{\infty}{\frac{dx^2+e}{ax^4+bx^2+c}}\mathrm{d}x \\ =&\frac{d}{2}\int_{-\infty}^{\infty}{\frac{x^2}{ax^4+bx^2+c}}\mathrm{d}x+\frac{e}{2}\int_{-\infty}^{\infty}{\frac{1}{ax^4+bx^2+c}}\mathrm{d}x \\ =&\frac{d}{2}\int_{-\infty}^{\infty}{\frac{x^2}{cx^4+bx^2+a}}\mathrm{d}x+\frac{e}{2}\int_{-\infty}^{\infty}{\frac{x^2}{ax^4+bx^2+c}}\mathrm{d}x \\ =&\frac{d}{2}\int_{-\infty}^{\infty}{\frac{1}{cx^2+b+\frac{a}{x^2}}}\mathrm{d}x+\frac{e}{2}\int_{-\infty}^{\infty}{\frac{1}{ax^2+b+\frac{c}{x^2}}}\mathrm{d}x \\ =&\frac{d}{2}\int_{-\infty}^{\infty}{\frac{1}{\left( \sqrt{c}x-\frac{\sqrt{a}}{x} \right) ^2+2\sqrt{ac}+b}}\mathrm{d}x+\frac{e}{2}\int_{-\infty}^{\infty}{\frac{1}{\left( \sqrt{a}x-\frac{\sqrt{c}}{x} \right) ^2+2\sqrt{ac}+b}}\mathrm{d}x \\ =&\frac{d}{2\sqrt{c}}\int_{-\infty}^{\infty}{\frac{1}{x^2+2\sqrt{ac}+b}}\mathrm{d}x+\frac{e}{2\sqrt{a}}\int_{-\infty}^{\infty}{\frac{1}{x^2+2\sqrt{ac}+b}}\mathrm{d}x \\ =&\frac{\pi}{2\sqrt{2\sqrt{ac}+b}}\left[ \frac{d}{\sqrt{a}}+\frac{e}{\sqrt{c}} \right] \end{align*}

Now, let's prove the integral at hand! Actually, let's prove a broader case. Given $r>\left| s \right|$, $$ \int_{-1}^1{\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\ln \left( \frac{\left( 2r-1 \right) x^2+2sx+1}{\left( 2r-1 \right) x^2-2sx+1} \right)}\mathrm{d}x=4\pi \mathrm{sgn} \left( s \right) \mathrm{arcsin} \left( \sqrt{r-\sqrt{r^2-s^2}} \right) $$ notice that $I\left( r,0 \right) =0$ and $I\left( r,-s \right) =-I\left( r,s \right) $, therefore we can only consider the case where $s>0$ \begin{align*} I\left( r,s \right) =&\int_{-1}^1{\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\ln \left[ \frac{\left( 2r-1 \right) x^2+2sx+1}{\left( 2r-1 \right) x^2-2sx+1} \right]}\mathrm{d}x \\ =&2\int_0^1{\frac{1}{x\sqrt{1-x^2}}\ln \left[ \frac{\left( 2r-1 \right) x^2+2sx+1}{\left( 2r-1 \right) x^2-2sx+1} \right]}\mathrm{d}x \\ \stackrel{(1)}{=}&2\int_0^{\frac{\pi}{2}}{\frac{1}{\cos \left( \theta \right)}\ln \left[ \frac{\left( 2r-1 \right) \cos ^2\left( \theta \right) +2s\cos \left( \theta \right) +1}{\left( 2r-1 \right) \cos ^2\left( \theta \right) -2s\cos \left( \theta \right) +1} \right]}\mathrm{d}\theta \\ \stackrel{(2)}{=}&4\int_0^1{\frac{1}{1-t^2}\ln \left[ \frac{\left( 2r-1 \right) \left( 1-t^2 \right) ^2+2s\left( 1-t^2 \right) \left( 1+t^2 \right) +\left( 1+t^2 \right) ^2}{\left( 2r-1 \right) \left( 1-t^2 \right) ^2-2s\left( 1-t^2 \right) \left( 1+t^2 \right) +\left( 1+t^2 \right) ^2} \right]}\mathrm{d}t \end{align*} Then differentiate \begin{align*} \partial_s I\left( r,s \right) =&4\int_0^1{\left[ \frac{1+t^2}{\left( r-s \right) t^4+\left( 2-2r \right) t^2+\left( r+s \right)}+\frac{1+t^2}{\left( r+s \right) t^4+\left( 2-2r \right) t^2+\left( r-s \right)} \right]}\mathrm{d}t \\ \stackrel{(3)}{=}&4\int_1^{\infty}{\left[ \frac{1+t^2}{\left( r+s \right) t^4+\left( 2-2r \right) t^2+\left( r-s \right)}+\frac{1+t^2}{\left( r-s \right) t^4+\left( 2-2r \right) t^2+\left( r+s \right)} \right]}\mathrm{d}t \\ =&2\int_0^{\infty}{\left[ \frac{1+t^2}{\left( r+s \right) t^4+\left( 2-2r \right) t^2+\left( r-s \right)}+\frac{1+t^2}{\left( r-s \right) t^4+\left( 2-2r \right) t^2+\left( r+s \right)} \right]}\mathrm{d}t \\ \stackrel{(4)}{=}&\frac{2\pi \left( \sqrt{r-s}+\sqrt{r+s} \right)}{\sqrt{r^2-s^2}\sqrt{2\sqrt{r^2-s^2}+2-2r}} \end{align*} Integrate again \begin{align*} I\left( r,\lambda \right) =\int_0^{\lambda}{\frac{\partial}{\partial s}I\left( r,s \right) \mathrm{d}s}=&\sqrt{2}\pi \int_0^{\lambda}{\frac{\sqrt{r-s}+\sqrt{r+s}}{\sqrt{r^2-s^2}\sqrt{\sqrt{r^2-s^2}+1-r}}\mathrm{d}s} \\ \stackrel{(5)}{=}&4\pi \sqrt{2r}\int_0^{X}{\frac{1}{\left( t^2+1 \right) \sqrt{\left( 1-2r \right) t^2+1}}\mathrm{d}t} \\ \stackrel{(6)}{=}&4\pi \int_0^{\tan \left( X \right)}{\frac{\sqrt{2r}\cos \left( \theta \right)}{\sqrt{1-2r\sin ^2\left( \theta \right)}}\mathrm{d}\theta} \\ =&4\pi \mathrm{arcsin} \left( \sqrt{2r}\sin \left( \tan \left( X \right) \right) \right) \\ =&4\pi \mathrm{arcsin} \left( \sqrt{r-\sqrt{r^2-\lambda ^2}} \right) \end{align*} (1) $x=\cos \left( \theta \right)$
(2) $\tan \left( \frac{\theta}{2} \right) =t$
(3) $t\rightsquigarrow \frac{1}{t}$
(4) Use the lemma
(5) $s=\frac{2rt}{1+t^2}$
(6) $t=\tan \left( \theta \right)$
(7) $X=\frac{r-\sqrt{r^2-\lambda ^2}}{\lambda}$

So, for this specific question, we have $r=1.5, s=1$. After plugging these values in, we got $$ \int_{-1}^1{\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\ln \left( \frac{2x^2+2x+1}{2x^2-2x+1} \right)}\mathrm{d}x=4\pi \mathrm{arcsin} \left( \sqrt{\frac{3-\sqrt{5}}{2}} \right) =4\pi \mathrm{arccot} \left( \sqrt{\frac{1+\sqrt{5}}{2}} \right) $$

  • $\begingroup$ The formula $$\int_{-\infty}^\infty f\left(\frac x{p^2}-\frac{q^x}x\right)dx=p^2\int_{-\infty}^\infty f(x)dx$$ seems like something worthy of being published, no? $\endgroup$ Commented Oct 5, 2023 at 18:53
  • $\begingroup$ @KamalSaleh It's a modified version of Glasser's master theorem, not that big of a deal. :) $\endgroup$
    – oO_ƲRF_Oo
    Commented Oct 5, 2023 at 18:56
  • $\begingroup$ Oh, I just noticed. Here $\frac1{p^2}=|a|$ and the "sum" is the second part of the input. Makes sense now. $\endgroup$ Commented Oct 5, 2023 at 19:02
  • $\begingroup$ Also it should be $q^2$ and not $q^x$ $\endgroup$ Commented Oct 5, 2023 at 19:02

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