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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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Do you have any reason to believe there is a closed form for that horrid-looking thing? –  dfeuer Nov 11 '13 at 17:12
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. –  Ron Gordon Nov 12 '13 at 0:21
A related problem. –  Mhenni Benghorbal Nov 19 '13 at 18:46
@MhenniBenghorbal: Related how? –  Ron Gordon Nov 19 '13 at 21:02
Out of curiosity, is there an application for this integral? –  linuxfreebird Nov 16 at 0:35

5 Answers 5

up vote 364 down vote accepted

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:

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

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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Amazing solution! –  Liu Jin Tsai Nov 13 '13 at 17:54
He's unstoppable! One for the "best of" - for sure. –  Bennett Gardiner Nov 14 '13 at 6:48
UpVote $0$ k. It's quite nice. A tour of force. –  Felix Marin Nov 16 '13 at 9:36
+1. $\text{arccot}(\sqrt{\phi})$-definitely one of the most weirdest closed form solutions to have ever been obtained! –  user17762 Dec 2 '13 at 6:31
@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 $\phi$. –  Ron Gordon Mar 14 at 17:08

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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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. –  Ron Gordon Nov 17 '13 at 8:45
@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. –  sos440 Nov 17 '13 at 15:44
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. –  Ron Gordon Nov 17 '13 at 15:53
@RonGordon, I'm happy to hear that my solution was helpful to you! Thanks. –  sos440 Nov 17 '13 at 17:16


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This should really be a comment. An answer would contain a proof. –  LTS Nov 11 '13 at 21:47
@OliverBel IMHO, if the question is not asking explicitly for a proof, and there is no obvious conjectured closed form, I think it's OK to post the result first and add a proof later, when time permits. The result itself can be useful for the person asking. It may be downvoted if not useful, but it still qualifies as an answer. The guideline for comments say "Avoid answering questions in comments", and I see no reason to withhold the result until the full proof is ready to be posted. –  Vladimir Reshetnikov Nov 11 '13 at 22:20
@VladimirReshetnikov: I defer to Hamming: "The purpose of computing is insight, not numbers." Unless the result itself is particularly illuminating, I do not agree that it is an answer. There is a user who continues to insist on putting Maple output as an answer, and I think that degrades the site. I'd prefer if Cleo gave some insight as to where this answer came from. –  Ron Gordon Nov 11 '13 at 22:53
@RonGordon Of course, I would prefer to see the proof too. I hope that Cleo posts it eventually, or somebody else does it, inspired by the fact the closed form exists (and maybe also guided by its shape). My point is that writing a proof clearly and typesetting it can take hours (at least for me) and I can imagine that not everybody can allocate the required time promptly. I would prefer to at least see the result posted promptly in such cases. (And from my experience being a PhD student in theoretical physics I can say that sometimes the result is the only thing a person cares of :) –  Vladimir Reshetnikov Nov 11 '13 at 23:17
This style of answer is complete disrespect. This situation seems for me like this: Cleo found interesting problem, and solved it. He is lazy to write the solution but want to show how clever he is, so decided to post only the final result. The reference to the definition of golden ratio made me laugh. If OP asks question of such level he definitely familiar with this constant. Note that this is not a single example. ALL Cleo's answer are of this style, and even after polite ephasis that these answers is not what OP's wanted he continues to post only final results! –  no identity Nov 17 '13 at 9:14

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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@RonGordon Thanks for checking my calculations. I numbered the integrals. Could you tell me the number of integral which is not finite? sos440 says that the denominator is strictly positive, so (1),(2),(3) are finite (hopefully). Introducing new variables do not change the finiteness, so (4),(5),(6),(7),(8) are finite. Do you think (9) is not correct? I'm very curious to find the mistake, logical or miscount. –  vesszabo Feb 17 at 20:41
@RonGordon I see. It was a typo, corrected. –  vesszabo Feb 20 at 11:22
Oh, you did it. Congratulation! (+1 upvote) –  sos440 Mar 9 at 18:48
@sos440 Thanks. It was not a trivial task. How could you discover your general Proposition? –  vesszabo Mar 9 at 19:28
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. –  sos440 Mar 9 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 $$\tan^{-1}(z) = \frac{i}{2}\log\left(\frac{1-iz}{1+iz}\right),$$ 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 can also make the change of variable $y=\text{sech}^{-1}x$ to obtain

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

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