Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$

We can prove using the Beta-Function identity that

$$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \quad \lambda>\frac{1}{2}$$

Differentiating the above equation with respect to $\lambda$, we obtain an expression involving the Digamma Function $\psi_0(z)$.

$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^\lambda}dx = \sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \left(\psi_0(\lambda)-\psi_0 \left( \lambda-\frac{1}{2}\right) \right)$$

Putting $\lambda=2$, we get

$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx = -\frac{\pi}{4}+\frac{\pi}{2}\log(2)$$


But, does anybody know how to evaluate $\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$?

Mathematica gives the values

  • $\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx = -\frac{G}{6}+\pi \left(-\frac{3}{8}+\frac{1}{8}\log(2)+\frac{1}{3}\log \left(2+\sqrt{3} \right) \right)$

  • $\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx = -\frac{\pi}{2}+\frac{\pi \log \left( 6+4\sqrt{2}\right)}{4}$

Here, $G$ denotes the Catalan's Constant.

Initially, my approach was to find closed forms for

$$\int_0^\infty \frac{1}{(1+x^2)^2(1+x^3)^\lambda}dx \ \ , \int_0^\infty \frac{1}{(1+x^2)^2(1+x^4)^\lambda}dx$$

and then differentiate them with respect to $\lambda$ but it didn't prove to be of any help.

Please help me prove these two results.

  • 1
    $\begingroup$ A trick that should work on both: $f(t)=\displaystyle\int_0^{\infty}\textstyle \frac{\ln (1+tx^4)}{(1+x^2)^2}dx$, then you simply seek $\displaystyle\int_0^1 f'(t)\,dt $ which can be done through partial fractions and a couple simple manipulations (I expect it to be a little tedious though). $\endgroup$ – L. F. Jun 8 '13 at 13:33
  • $\begingroup$ @L.F: Nice idea! I am going to try it. $\endgroup$ – Shobhit Bhatnagar Jun 8 '13 at 13:54

The 2nd evaluation can be obtained from the residue theorem:

  • Using parity, write the integral as $\displaystyle \frac12\int_{-\infty}^{\infty}\frac{\ln(1+x^4)\,dx}{(1+x^2)^2}$.

  • Interpret this as a complex integral and pull the integration contour to, say, $i\infty$. The result will be given by the residue at 2nd order pole at $x=i$ and two integrals of the logarithm jump (equal to $2\pi i$) over the branch cuts emanating from $x=e^{i\pi/4}$ and $x=e^{3i\pi /4}$ in the radial directions.

  • For the first contribution, we have $$2\pi i \cdot\mathrm{res}_{x=i}\frac{\ln(1+x^4)}{(1+x^2)^2}=\frac{\pi}{2}\left(\ln2 -2\right).$$

  • The integral over the branch cut $(e^{i\pi/4},e^{i\pi/4}\infty)$ is $$2\pi i \int_{e^{i\pi/4}}^{e^{i\pi/4}\infty}\frac{dx}{(1+x^2)^2}=\frac{\pi i}{2}\left(\pi-\sqrt{2}-2\arctan e^{i\pi/4}\right),$$ and, similarly, for the second branch cut $(e^{3i\pi/4},e^{3i\pi/4}\infty)$ we find $$2\pi i \int_{e^{3i\pi/4}}^{e^{3i\pi/4}\infty}\frac{dx}{(1+x^2)^2}=\frac{\pi i}{2}\left(\sqrt{2}-\pi-2\arctan e^{3i\pi/4}\right).$$

  • Combining everything, one obtains the answer: \begin{align} \frac12\left\{\frac{\pi}{2}\left(\ln2 -2\right)+\pi\, \mathrm{arccoth}\sqrt{2}\right\}= -\frac{\pi}{2}+\frac{\pi}{4}\ln(6+4\sqrt{2}). \end{align}

  • $\begingroup$ Very cool. The shape of the contour reminds me of a sand dollar for some reason. $\endgroup$ – Random Variable Jun 8 '13 at 17:16
  • $\begingroup$ $\displaystyle\int_{-\infty}^{\infty} \frac{\ln(1+x^{4})}{(1+x^{2})^{2}} \ dx - \pi i \Big(\arctan( e^{\frac{\pi i}{4}}) + \arctan (e^{\frac{3 \pi i}{4}}) \Big) = \frac{\pi}{2} (\ln 2-2)$ And $\displaystyle \arctan( e^{\frac{\pi i}{4}}) + \arctan (e^{\frac{3 \pi i}{4}}) = i \ \text{arccoth}(\sqrt{2})$ So why isn't the $\text{arccoth}$ term negative when brought to the other side of the equation? $\endgroup$ – Random Variable Jun 9 '13 at 19:09
  • $\begingroup$ @RandomVariable Not sure to understand the question. The sum of three pieces is $$\frac{\pi}{2}\left(\ln2-2\right)-i\pi(\arctan e^{i\pi/4}+\arctan e^{3i\pi/4})$$ Then $(-i)\cdot i=+1$. $\endgroup$ – Start wearing purple Jun 9 '13 at 19:22
  • $\begingroup$ But if $C$ is the entire closed contour (which can be divided into ten pieces), $\int_{C} f(z) \ dz = \frac{\pi}{2} (\ln 2-2)$. So the integrals that don't go to zero have to be brought to the other side of the equation so that we can isolate $\int_{-\infty}^{\infty} f(x) \ dx$, no? $\endgroup$ – Random Variable Jun 9 '13 at 19:39
  • $\begingroup$ @RandomVariable I think there is some misunderstanding here. If you want to write something like "integral over real line + integral around the 1st branch cut + integral around the 2nd branch cut = $2\pi i\cdot$ residue at $i$", then integration contours around branch cuts should be oriented clockwise. That would change the sign of both of them, since my convention was to use counterclockwise contours and write rather ""integral over real line = integral around the 1st branch cut + integral around the 2nd branch cut + $2\pi i\cdot$ residue at $i$"". $\endgroup$ – Start wearing purple Jun 9 '13 at 20:06

We can attack the other integral

$$I = \int_0^{\infty} dx \frac{\log{(1+x^3)}}{(1+x^2)^2}$$

in a manner similar to what @O.L. outlined in his answer for the other case, but with a different contour. To wit, consider

$$\oint_C dz \frac{\log{(1+z^3)} \log{z}}{(1+z^2)^2}$$

where $C$ is the following contour


This is a keyhole contour about the positive real axis, but with additional keyholes about the branch points at $z=e^{i \pi/3}$, $z=-1$, and $z=e^{i 5 \pi/3}$. There are poles of order $2$ at $z=\pm i$.

I will outline the procedure for evaluation. The integral about the circular arcs, large and small, go to zero as the radii go to $\infty$ and $0$, respectively. Each of the branch points introduces a jump of $i 2 \pi$ due to the logarithm in the integrand. By the residue theorem, we have

$$-i 2 \pi \int_0^{\infty} dx \frac{\log{(1+x^3)}}{(1+x^2)^2} - i 2 \pi \int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} \\ - i 2 \pi \int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{(1+t^2)^2} - i 2 \pi \int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} = \\ i 2 \pi \sum_{\pm} \frac{d}{dz} \left[\frac{\log{(1+z^3)} \log{z}}{(z\pm i)^2} \right]_{z=\pm i} $$

Without going into too much detail, I will illustrate how the integrals are done by evaluating one of them. Consider

$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{(1+t^2)^2} = -\int_1^{\infty} dy \frac{\log{y}+i \pi}{(1+y^2)^2}$$


$$\int_1^{\infty} \frac{dy}{(1+y^2)^2} = \int_{\pi/4}^{\pi/2} d\theta \cos^2{\theta} = \frac{\pi}{8}-\frac14$$

$$\begin{align}\int_1^{\infty} dy\frac{\log{y}}{(1+y^2)^2} &= -\int_0^1 du \frac{u^2 \log{u}}{(1+u^2)^2}\\ &= -\sum_{k=0}^{\infty} (-1)^k (k+1) \int_0^1 u^{2 k+2} \log{u} \\ &= \sum_{k=0}^{\infty} (-1)^k \frac{k+1}{(2 k+3)^2} \\ &= \frac{G}{2} - \frac{\pi}{8}\end{align}$$

so that

$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{(1+t^2)^2} = - \left ( \frac{G}{2} - \frac{\pi}{8} \right ) - i \pi \left ( \frac{\pi}{8}-\frac14\right ) $$

Along similar lines,

$$\int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} = \frac{G}{3}-\frac{\pi }{8}+\frac{1}{12} \pi \log \left(2+\sqrt{3}\right)+i \left(\frac{1}{4} \log \left(2+\sqrt{3}\right)-\frac{\pi }{6}\right)$$

$$\int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{(1+t^2)^2} = \frac{G}{3}-\frac{\pi }{8}-\frac{5}{12} \pi \log \left(2+\sqrt{3}\right)+i \left(-\frac{5 \pi }{6}+\frac{\pi ^2}{4}-\frac{1}{4} \log \left(2+\sqrt{3}\right)\right)$$

Combining the integrals, I get

$$\frac{G}{6} -\frac{\pi}{8}-\frac{\pi}{3} \log{(2+\sqrt{3})} + i \left [-\frac{3 \pi}{4} + \frac{\pi^2}{8}\right ] $$

The sum of the residues on the RHS is relatively simple to evaluate; I get

$$\sum_{\pm} \frac{d}{dz} \left[\frac{\log{(1+z^3)} \log{z}}{(z\pm i)^2} \right]_{z=\pm i} = \frac{\pi}{2}-\frac{\pi}{8}\log (2)+i \left(\frac{3 \pi }{4}-\frac{\pi ^2}{8}\right)$$

The integral we seek is then the negative of the sum of the combined integrals and the sum of the residues, which gives us

$$\int_0^{\infty} dx \frac{\log{(1+x^3)}}{(1+x^2)^2} = -\frac{G}{6} - \frac{3\pi}{8} + \frac{\pi}{8} \log{2} + \frac{\pi}{3} \log{(2+\sqrt{3})} \approx 0.320555$$

which agrees with Mathematica. Note how the imaginary parts fortuitously canceled.

It should be understood that the above technique may be applied to the other integral. As O.L. has demonstrated, however, one may exploit symmetry and use a less computationally demanding technique for that particular case.

  • 2
    $\begingroup$ I thought the first integral is hopeless because of loss of parity, and here you crack it. GREAT answer!!! $\endgroup$ – Start wearing purple Jul 23 '13 at 8:33
  • 2
    $\begingroup$ @O.L.: this answer does not exist without yours, so thank you. $\endgroup$ – Ron Gordon Jul 23 '13 at 10:54

I hope it is not too late. Define \begin{eqnarray} I(a)=\int_0^\infty\frac{\log(1+ax^4)}{(1+x^2)^2}dx. \end{eqnarray} Then \begin{eqnarray} I'(a)&=&\int_0^\infty \frac{x^4}{(1+ax^4)(1+x^2)^2}dx\\ &=&\frac{1}{(1+a)^2}\int_0^\infty\left(-\frac{2}{1+x^2}+\frac{1+a}{(1+x^2)^2}+\frac{1-a+2ax^2}{1+a x^4}\right)dx\\ &=&\frac{1}{(1+a)^2}\left(-\pi+\frac{1}{4}(1+a)\pi+\frac{(1-a)\pi}{2\sqrt2a^{1/4}}+\frac{\pi a^{1/4}}{\sqrt2}\right)\\ &=&\frac{1}{4(1+a)^2}\left(a-3+\frac{\sqrt2(1-a)}{a^{1/4}}+2\sqrt2 a^{1/4}\right). \end{eqnarray} and hence \begin{eqnarray} I(1)&=&\int_0^1\frac{1}{4(1+a)^2}\left(a-3+\frac{\sqrt2(1-a)}{a^{1/4}}+2\sqrt2 a^{1/4}\right)da\\ &=&-\frac{\pi}{2}+\frac{1}{4}\log(6+4\sqrt2). \end{eqnarray} For the other integral, we can do the same thing to define $$ J(a)=\int_0^\infty\frac{\log(1+ax^3)}{(1+x^2)^2}dx. $$ The calculation is similar and more complicated and here I omit the detail.

  • $\begingroup$ It's never too late to join the party! $\endgroup$ – omegadot Nov 14 '17 at 8:56

Another approach for evaluating the second integral using contour integration that avoids having to deform the contour around branch cuts is to consider $$ \displaystyle f(z) = \frac{\log(z+ e^{i \pi /4})}{(1+z^{2})^{2}}$$ and integrate around a contour that consists of the line segment $[-R,R]$ and the upper half of $|z|=R$.

Then letting $R \to \infty$,

$$ \begin{align} &\int_{-\infty}^{0} \frac{\log(x+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx + \int_{0}^{\infty} \frac{\log(x+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx \\ &= 2 \pi i \ \text{Res}[f(z),i] \\ &= 2 \pi i \lim_{z \to i} \frac{d}{dz} \frac{\log(z+e^{i \pi /4})}{(z+i)^{2}} \\ &=2 \pi i \lim_{z \to i} \left(\frac{1}{(z+e^{i \pi /4})(z+i)^{2}} - \frac{2 \log(z+e^{i \pi /4})}{(z+i)^{3}} \right) \\ &= 2 \pi i \left(- \frac{1}{4} \frac{\sqrt{2}}{1+i(1+\sqrt{2})} + \frac{\log|i+e^{i \pi /4}| + i \arg (i +e^{i \pi/4}) }{4i}\right)\\ &= 2 \pi i \left(\frac{1-\sqrt{2}+i}{8} + \frac{\frac{1}{2} \log (2+\sqrt{2})+ i \frac{3 \pi}{8}}{4i} \right) \\ &= \frac{\pi}{4} \Big(\log(2+\sqrt{2})-1 \Big) + \frac{i\pi}{4} \left(1-\sqrt{2}+\frac{3 \pi}{4} \right) . \end{align}$$

But notice that $$ \begin{align} &\text{Re} \left( \int_{-\infty}^{0} \frac{\log(z+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx + \int_{0}^{\infty} \frac{\log(z+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx \right) \\ &= \text{Re} \left(\int_{0}^{\infty} \frac{\log(-u + e^{i \pi /4})}{(1+u^{2})^{2}} \ du + \int_{0}^{\infty} \frac{\log(z+e^{i \pi /4})}{(1+x^{2})^{2}} \ dx \right) \\ &= \int_{0}^{\infty} \frac{\log|-u + e^{i \pi /4}|}{(1+u^{2})^{2}} \ du + \int_{0}^{\infty} \frac{\log |x+e^{i \pi /4}|}{(1+x^{2})^{2}} \ dx\\ &= \int_{0}^{\infty} \frac{\frac{1}{2} \log(x^{2}-\sqrt{2}x+1) + \frac{1}{2} \log(x^{2}+\sqrt{2}x+1)}{(1+x^{2})^{2}} \ dx \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{\log(1+x^{4})}{(1+x^{2})^{2}} \ dx. \end{align}$$


$$ \begin{align} \int_{0}^{\infty} \frac{\log(1+x^{4})}{(1+x^{2})^{2}} \ dx &= \frac{\pi}{2} \Big(\log(2+\sqrt{2}) -1 \Big) \\ &= \frac{\pi}{2} \Big(\frac{1}{2} \log \big((2+\sqrt{2})^{2} \big) -1 \Big) \\ &= \frac{\pi}{2} \Big(\frac{\log(6+4\sqrt{2})}{2} -1\Big) \\ &= - \frac{\pi}{2} + \frac{\pi \log(6+4\sqrt{2})}{4}. \end{align}$$


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