How would I go about evaluating this integral? $$\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx.$$ What I've tried so far: I tried a semicircular integral in the positive imaginary part of the complex plane, excluding the negative real axis, but had trouble calculating the residue at $z=i$ (perhaps there is a way of doing this that I don't know of). After that didn't work, I tried a rectangular box integral from $\epsilon$ to $R$, from $R$ to $R+i/2$, from $R+i/2$ to $-S+i/2$, from $-S+i/2$ to $-S+i\epsilon$, from $-S+i\epsilon$ to $-\epsilon+i\epsilon$ and finally a semicircle around the origin, radius $\epsilon$.

Any help would be appreciated.

  • $\begingroup$ Numerically it seems to be $\pi \ln 2$. Now, the contour method is rather usual, you should get $\pi \mathrm{Res}_i(f)$, but you'll have a problem with residue at $i$. $\endgroup$ Mar 25, 2013 at 12:27
  • $\begingroup$ @arbautjc Yes, we can find math.stackexchange.com/questions/177160/… some methods (but not complex, using contour integration). $\endgroup$
    – Cortizol
    Mar 25, 2013 at 12:53
  • $\begingroup$ Oh, my. I have found $\int_0^{\pi/2} \log{(1+\tan^2 x)}\mathrm{d}x$, but I just forgot that $1+\tan^2 x = \frac{1}{\cos^2 x}$. Thanks! $\endgroup$ Mar 25, 2013 at 13:03
  • $\begingroup$ Note that, you have two branch points. $\endgroup$ Mar 25, 2013 at 14:01
  • 3
    $\begingroup$ Integrate $f(z) = \frac{\ln(z+i)}{z^{2}+1}$ around a large closed half-circle in the upper half plane. $\endgroup$ Apr 11, 2013 at 15:12

7 Answers 7


Use $x\to \tan\theta$ and $dx=\sec^2\theta\ d\theta$. The integral becomes $$\int_0^{\pi/2} 2\ln (\sec \theta) d\theta$$

Which is $$-2\int_0^{\pi/2} \ln (\cos\theta) d\theta$$ And can be solved.

Let $$I= \displaystyle \int_0^{\pi/2} \ln \left( \sin \theta \right) \cdot d\theta$$ $$I= \displaystyle \int_0^{\pi/2} \ln \left( \cos \theta \right) \cdot d\theta$$ Adding both. $$2I=\displaystyle \int_0^{\pi/2} \ln \left( \sin \theta \times \cos \theta\right) \cdot d\theta$$ $$2I= \displaystyle \int_0^{\pi/2} \ln \left(2 \sin \theta\times \cos \theta \right) -\ln2 \cdot d\theta$$ $$2I=\int_0^{\pi/2}\ln(\sin{2\theta})-\ln2 \cdot d\theta$$ $$\int_0^{\pi/2}\ln(\sin{2\theta})\cdot d\theta=I$$ So, $$I=-\int_0^{\pi/2}\ln2\cdot d\theta$$ $$I=-\dfrac{\pi\ln2}2$$

And your integral comes out to be

$${\pi \ln2}$$

  • 3
    $\begingroup$ This is tagged as complex analysis, and I believe OP wants to use contour integration. (Nice approach,though, so +1). $\endgroup$
    – Aryabhata
    Apr 11, 2013 at 14:42
  • 10
    $\begingroup$ It is not a perception. This is based on the tags (which now include contour-integration tag) and the question description. $\endgroup$
    – Aryabhata
    Apr 11, 2013 at 14:56
  • $\begingroup$ This answer is incorrect. $\endgroup$ Apr 11, 2013 at 16:26
  • 2
    $\begingroup$ There's a mistake but I think it is a rather minor one and the approach indeed is nice:$$\int\limits_0^{\pi/2}\log\sin 2\theta\,d\theta=\frac{1}{2}\int\limits_0^\pi\log\sin u\,du=\int\limits_0^{\pi/2}\log\sin u\,du\implies$$in the second line from the bottom in the blue zone of the answer it must be $$I=-\int\limits_0^{\pi/2}\log 2\,d\theta\ldots$$ $\endgroup$
    – DonAntonio
    Apr 11, 2013 at 18:34
  • 1
    $\begingroup$ @DonAntonio Thanx for pointing out. $\endgroup$
    – ABC
    Apr 12, 2013 at 2:49

Maybe this is going to seem a lot more involved than it needs to be, but it is likely that complex methods are not the best way to attack an integral like this. Nonetheless, it is possible.

We consider the integral in the complex plane

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

where $C$ is some contour to be determined. Our first instinct is to make $C$ a simple semicircle in the upper half plane. The problem is that the branch point singularity at $z=i$ is extremely problematic, as it coincides with an ostensible pole. Nonetheless, the corresponding integral over the real line is finite (and twice the originally specified integral), so there must be a way to treat this.

The way to go with branch points like this is to avoid them. We thus have to draw $C$ so as to do that, and then use Cauchy's theorem to state that the above complex integral about $C$ is zero. Such a contour $C$ is illustrated below.


The contour integral is then taken along six different segments. I will state without proof that the integral about the two outer arcs vanishes as the radius of those arcs $R \to \infty$. We are then left with four integrals:

$$\int_{-R}^R dx \frac{\log{(1+x^2)}}{1+x^2} + \left [\int_{C_-}+\int_{C_+}+\int_{C_{\epsilon}} \right ] dz \frac{\log{(1+z^2)}}{1+z^2} = 0$$

$C_-$ is the segment to the right of the imaginary axis, down from the arc to the branch point, $C_+$ is the segment to the left of the imaginary axis, up from the branch point to the arc, and $C_{\epsilon}$ is the circle about the branch point of radius $\epsilon$.

It is crucial that we get the arguments of the log correct along each path. I note that the segment $C_-$ is "below" the imaginary axis and I assign the phase of this segment to be $2 \pi$, while I assign the phase of the segment $C_+$ to be $0$.

For the segment $C_-$, set $z=i(1+y e^{i 2 \pi})$:

$$\int_{C_-} dz \frac{\log{(1+z^2)}}{1+z^2} = i\int_R^{\epsilon} dy \frac{\log{[-y (2+y)]}+ i 2 \pi}{-y (2+y)} $$

For the segment $C_+$, set $z=i(1+y)$:

$$\int_{C_-} dz \frac{\log{(1+z^2)}}{1+z^2} = i\int_{\epsilon}^R dy \frac{\log{[-y (2+y)]}}{-y (2+y)} $$

I note that the sum of the integrals along $C_-$ and $C_+$ is

$$-2 \pi \int_{\epsilon}^R \frac{dy}{y (2+y)} = -\pi \left [ \log{R} - \log{(2 + R)} - \log{\epsilon} + \log{(2 + \epsilon)}\right]$$

For the segment $C_{\epsilon}$, set $z=i (1+\epsilon e^{-i \phi})$. The integral along this segment is

$$\begin{align}\int_{C_{\epsilon}} dz \frac{\log{(1+z^2)}}{1+z^2} &= \epsilon \int_{-2 \pi}^0 d\phi e^{-i \phi} \frac{\log{\left [ -2 \epsilon e^{-i \phi} \right]}}{-2 \epsilon e^{-i \phi}}\end{align}$$

Here we use $\log{(-1)}=-i \pi$ and the above integral becomes

$$\begin{align}\int_{C_{\epsilon}} dz \frac{\log{(1+z^2)}}{1+z^2} &= -\frac12 (-i \pi)(2 \pi) - \frac12 \log{2} (2 \pi) - \frac12 \log{\epsilon} (2 \pi) -\frac12 (-i) \frac12 (0-4 \pi^2) \\ &= -\pi \log{2} - \pi \log{\epsilon} \end{align}$$

Adding the above integrals, we have

$$\begin{align}\int_{-R}^R dx \frac{\log{(1+x^2)}}{1+x^2} -\pi \log{R} + \pi \log{(2 + R)} + \pi \log{\epsilon} - \pi \log{(2 + \epsilon)} -\pi \log{2} - \pi \log{\epsilon} &= 0\\ \implies \int_{-R}^R dx \frac{\log{(1+x^2)}}{1+x^2} -\pi \log{R} + \pi \log{(2 + R)} - \pi \log{(2 + \epsilon)} -\pi \log{2} &=0\end{align}$$

Now we take the limit as $R \to \infty$ and $\epsilon \to 0$ and we get

$$\int_{-\infty}^{\infty} dx \frac{\log{(1+x^2)}}{1+x^2} -2 \pi \log{2} = 0$$


$$\int_{0}^{\infty} dx \frac{\log{(1+x^2)}}{1+x^2} = \pi \log{2}$$

  • $\begingroup$ @EricNaslund: many thanks. $\endgroup$
    – Ron Gordon
    Apr 24, 2013 at 14:42
  • 2
    $\begingroup$ (+1) Finally solution with contuor integration, like OP wanted (I suppose). $\endgroup$
    – Cortizol
    Apr 27, 2013 at 8:26
  • $\begingroup$ @Ron You directed me to your post, so I hope you're willing to answer some of my questions. I don't think the branch of $\log(1+z^{2})$ you used coincides with that cut. I'm pretty sure that cut is the result of using the principal branch of $\log(1+z^{2})$. And when I parametrized the contour around the branch point at $z=i$, I assumed I could replace $(i+\epsilon e^{it})^{2}$ with $-1 + 2i\epsilon e^{it}$ since we're only interested in small values of $\epsilon$. You seem to have done something similar but without stating what you did. Could you elaborate? $\endgroup$ Aug 27, 2013 at 15:05
  • $\begingroup$ The branch of the log doesn't coincide with what cut? And I did exactly what you said, see the integral about $C_{\epsilon}$. It is certainly there - we are just making sure we are picking up the contribution from the branch point. $\endgroup$
    – Ron Gordon
    Aug 27, 2013 at 15:13
  • $\begingroup$ I assume we want the branch cuts on the imaginary axis from $i$ to $i \infty$ and from $-i$ to $-i \infty$. Along those cuts $1+z^{2}$ is negative and real. So those cuts come about by choosing the principal branch of $\log(1+z^{2})$. So the argument of $\ln(1+z^{2})$ just to the right of the cut on $[i, i \infty)$ is $\pi$, while just to the left of the cut the argument is $- \pi$. $\endgroup$ Aug 27, 2013 at 15:27

One way to solve this problem is to use parametric integrals. Let $$ I(\alpha)=\int_{0}^{\infty}\frac{\ln(\alpha x^2+1)}{x^2+1}dx. $$ Then \begin{eqnarray*} I'(\alpha)&=&\int_{0}^{\infty}\frac{x^2}{(\alpha x^2+1)(x^2+1)}dx\\ &=&\int_{0}^{\infty}\left(-\frac{1}{\alpha-1}\frac{1}{\alpha x^2+1}+\frac{1}{\alpha-1}\frac{1}{x^2+1}\right)dx\\ &=&-\frac{1}{\alpha-1}\frac{1}{\sqrt{\alpha}}\frac{\pi}{2}+\frac{1}{\alpha-1}\frac{\pi}{2}\\ &=&\frac{\pi}{2}\frac{\sqrt{\alpha}-1}{\sqrt{\alpha}(\alpha-1)}\\ &=&\frac{\pi}{2}\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\sqrt{\alpha}+1}\right). \end{eqnarray*} Thus $$ I(\alpha)=\pi\ln(\sqrt{\alpha}+1)+C. $$ But $I(0)=0$ implies $C=0$. So $I(1)=\pi\ln 2$.

  • $\begingroup$ I'd appreciate if you could address the following doubts. Thanks. (1) How do you relate your $\,I(\alpha)\,$ with the original integral? That has logarithm in the numerator, yours doesn't ; (2) you're apparently differentiating under the integral sign, but if so how do get what you write in $\,I'(\alpha)\,$? $\endgroup$
    – DonAntonio
    Apr 12, 2013 at 3:01
  • $\begingroup$ There should be a log applied to the numerator, then it works fine. $\endgroup$ Apr 12, 2013 at 9:38
  • $\begingroup$ It looks like a neat approach, but something is wrong. Want to fix it? $\endgroup$ Apr 12, 2013 at 10:33
  • $\begingroup$ @nbubis, tell me which place is wrong so that I can fix. $\endgroup$
    – xpaul
    Apr 12, 2013 at 12:56
  • 1
    $\begingroup$ It's a bit easier if we use $\log(\alpha +x^2)$ instead of $\log(1+\alpha x^2)$. See THIS ANSWER. ;-)) $\endgroup$
    – Mark Viola
    Aug 23, 2016 at 23:47

Hints As RandomVariable suggested, use $\log(x^2+1)=\log(x+i)+\log(x-i)$, choosing the branches of the logarithm carefully. It's generally best to isolate unpleasant singular things.

Then write $$\int_0^\infty=\frac{1}{2}\int_{-\infty}^{\infty}$$ and make use of the above splitting to integrate the two parts on different contours, each time avoiding enclosing the singularity of the logarithm. The part on the semicircle vanishes.

Answer The UHP pole gives $\log(i+i)/2i\times2\pi i$. The LHP pole gives $\log(-i-i)/(-2i)\times-2\pi i$. Summing and halving gives the answer $$\frac{\pi}{2}\left(\log(2i)+\log(-2i)\right)$$ so all that remains is choosing the right logarithm. This is easy enough, actually, and the answer is what you expect: $$\pi\ln 2$$

  • $\begingroup$ No, I'm sorry, but you're still wrong! It is $\pi \ln 2$ just like I said, and just like I verified by checking it in Mathematica. $\endgroup$ Apr 11, 2013 at 17:07
  • $\begingroup$ Mathematica gives: Integrate[Log[x^2 + 1]/(1 + x^2), {x, 0, Infinity}] = [Pi] Log[2] $\endgroup$ Apr 11, 2013 at 17:10
  • $\begingroup$ -2 Integrate[Log[Cos[t]], {t, 0, Pi/2}] = [Pi] Log[2] also holds. $\endgroup$ Apr 11, 2013 at 17:11
  • $\begingroup$ If I may ask, what attracted the downvote, whoever did so? If this answer is wrong somehow I and the OP would probably like to know! If you just prefer the other answer, upvote that, don't downvote me... $\endgroup$ Apr 16, 2013 at 16:37
  • $\begingroup$ BTW I did not give you the downvote. $\endgroup$
    – Ron Gordon
    Apr 24, 2013 at 12:58

We can also use contour integration to show that

$$I(a,b,\theta) = \int_{-\infty}^{\infty} \frac{\log \left(a^{2}+2ax \cos \theta + x^{2}\right)}{x^{2}+b^{2}} \, dx = \frac{\pi}{b} \, \log \left(a^{2}+2ab \sin \theta +b^{2} \right), $$ where $a, b >0$ and $0 < \theta < \pi$.

Your integral is $\frac{1}{2} \, I\left(1,1,\frac{\pi}{2} \right)$.

Consider the complex function $$f(z) = \frac{\log \left(z+ae^{i \theta}\right)}{z^{2}+b^{2}},$$

which has a branch point in the lower half-plane at $z=-ae^{i \theta}$.

If we use the principal branch of the logarithm, then $f(z)$ is meromorphic in the upper half-plane.

So by integrating around a closed semicircular contour in the upper half-plane (and using the fact that $ \lim_{|z| \to \infty} z f(z) =0$ ), we get

$$ \int_{-\infty}^{\infty} \frac{\log \left(x+ae^{i \theta} \right)}{x^{2}+b^{2}} \, dx = 2 \pi i \ \text{Res}[f(z),ib] = \frac{\pi}{b} \, \log \left( ib +ae^{i \theta} \right).$$

And equating the real parts on both sides of the equation, we get

$$ \int_{-\infty}^{\infty} \frac{\log(a^{2}+2ax \cos \theta +x^2)}{x^{2}+b^{2}} \, dx = \frac{\pi}{b} \, \log \left(a^{2}+2ab \sin \theta +b^{2} \right).$$


Here is a way by using contour integration and easy to understand:


The branch of the logarithm is principal branch cut.

The argument is using under principal value.

$$\int_{0}^{\infty} \frac{\log(1+x^2)}{1+x^2}~dx = \\ \int_{0}^{\infty} \frac{\log(x^2+1)}{1+x^2}~dx = \\ \int_{0}^{\infty} \frac{\log((x+i)(x-i))}{1+x^2}~dx = \\\int_{0}^{\infty} \frac{\log(|x+i|) + Arg(x+i) + log(|x-i|)+ Arg(x-i) }{1+x^2}~dx = \\ \int_{0}^{\infty} \frac{\log(|x+i|) + log(|x-i|)+ Arg(2x) }{1+x^2}~dx = \\$$$$\int_{0}^{\infty} \frac{\log(|x+i|) + log(|x-i|)}{1+x^2}~dx \tag{1} $$

Doing contour integration with respect to function: $f(x) = log(x+i)/(x^2+1)$

Define contour:

$\Gamma_1:= x \text{ from } 0 \text{ to } \infty$

$\Gamma_2:= x \text{ from } \infty \text{ to } -\infty \text{ along the upper semicircle}$

$\Gamma_3:= x \text{ from } -\infty \text{ to } 0$

Easy to see: $\int_{\Gamma_2} |f(x)| \leq 2 \pi \max_{\Gamma_2} {\frac{|\log(x+i)|}{|1+x^2|}} \leq \frac{log(\sqrt{x^2+1})}{x^2+1}+\frac{\pi^2}{x^2+1} = 0$

So $\int_{\Gamma_2} f(x) = 0$

The residue for $f(x)$ at $x = i$ is $\frac{\log(2)+ \frac{\pi}{2}}{2i}$

By residue theorem: $\int_{\Gamma_1} f(x) + \int_{\Gamma_3} f(x) = 2\pi Res(f,i) = \pi \log(2) + \frac{\pi^2 i}{2} $

$$\int_0^\infty \frac{log(x+i)}{(x^2+1)} + \int_{-\infty}^0 \frac{log(x+i)}{(x^2+1)} = \pi \log(2) + \frac{\pi^2 i}{2} \implies\\\int_0^\infty \frac{log|x+i|}{(x^2+1)} + \int_{-\infty}^0 \frac{log(|x+i|)+ \pi i}{(x^2+1)} = \pi \log(2) + \frac{\pi^2 i}{2} \implies\\\int_0^\infty \frac{log|x+i|}{(x^2+1)} + \int_{-\infty}^0 \frac{log(|x+i|)}{(x^2+1)}+ \int_{-\infty}^0 \frac{ \pi i}{(x^2+1)} = \pi \log(2) + \frac{\pi^2 i}{2} \\$$

Recall: $\int \frac{1}{x^2+1} = arctan(x)+c$

$$ \implies \int_0^\infty \frac{log|x+i|}{(x^2+1)} + \int_{0}^{\infty} \frac{log(|x-i|)}{(x^2+1)}+ \frac{i\pi^2}{2} = \pi \log(2) + \frac{i\pi^2}{2}\\\\\implies \int_0^\infty \frac{log|x+i| + log|x-i|}{(x^2+1)} = \pi \log(2) $$

Substitute into $(1)$,$\int_{0}^{\infty} \frac{\log(|x+i|) + log(|x-i|)}{1+x^2}~dx = \pi \log(2)$


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln\pars{x^{2} + 1} \over x^{2} + 1}\,\dd x} = {1 \over 2}\int_{-\infty}^{\infty}{\ln\pars{x^{2} + 1} \over x^{2} + 1} \,\dd x \\[5mm] = &\ \Re\int_{-\infty}^{\infty}{\ln\pars{1 + \ic x} \over x^{2} + 1}\,\dd x \\[5mm] \stackrel{{\Large 1\ +\ \ic x\ =\ s} \atop {\Large x\ =\ \ic\pars{1 - s}}}{=}\,\,\,& \Re\int_{1 - \infty\ic}^{1 + \infty\ic}{\ln\pars{s} \over -\pars{1 - s}^{2} + 1}\,\pars{-\ic}\dd s \\[5mm] & =\, -\,\Im\int_{1 - \infty\ic}^{1 + \infty\ic}{\ln\pars{s} \over s\pars{s - 2}}\,\dd s \\[5mm] = &\ -\,\Im\bracks{\color{red}{-}2\pi\ic\,{\ln\pars{2} \over 2}} \label{1}\tag{1} \\[5mm]= &\ \bbx{\pi\ln\pars{2}} \approx 2.1776 \\ & \end{align} In (\ref{1}), I "closed" the integration path with a infinite radius semicircle on the "right" ( note the $\ds{\color{red}{-}\mbox{sign}}$ because the contour integration is clockwise ). The magnitude of the contribution along the arc $\ds{< \pi\root{\ln^{2}\pars{R} + \pi^{2}}/R}$ as $\ds{R \to \infty}$ where $\ds{R}$ is the semicircle radius. $\ds{\ln}$ is the logarithm principal branch.


You must log in to answer this question.