Integrating this natural log using complex analysis

Question

Question: Integrate$$\mathscr{I}=\int\limits_{-\infty}^{\infty}dx\,\frac {\log(1+x^2)}{1+x^2}$$

I tried using a semicircle with radius $R$ with a smaller semicircle detour inside So we have that$$\oint\limits_{C}dz\, f(z)=\int\limits_r^Rdx\, f(x)+\int\limits_{\gamma_R}dz\, f(z)+\int\limits_{-R}^{-r}dx\,\frac {\log|1+x^2|+i\arg(1+x^2)}{1+x^2}+\int\limits_{\gamma_r}dz\, f(z)$$where $f(z)$ represents the integrand. However, I'm having trouble computing the residue at $z=i$ because the limit leads to infinity.

Are branch cuts required? If so, is it possible for you to walk me through it? I'm not very familiar with them...

Answer 1

Yes we require a branch cut, recall that $$\log(z)=\log(|z|)+i\arg(z)$$

In general with these sort of things we want to pick our branch cut so that our contour doesn't go through it, let's pick the branch cut where we insist $\arg(z)\in[-\pi/2,3\pi/2)$, i.e. the branch cut is down the negative imaginary axis, note that for real $x$, $$\log(1+x^2)=\log(1+ix)+\log(1-ix)$$

So let $f(z)=\frac{\log(1-iz)}{1+z^2}$ which is suitably nice on the upper semicircle from $-R$ to $R$ just having the one pole at $i$ with $\text{Res}(f,i)=\frac{\log(2)}{2i}$.

We then have by residue theorem that $$\int_{C_R} f(z) dz +\int_{-R}^{0}f(z) dz+ \int_{0}^{R}f(z) dz = \pi\log(2)$$

Where $C_R$ is the bit of the contour not on the real line, by the usual sort of reasoning the first integral tends to zero as $R$ tends to infinity. While after making the change of variables $z\rightarrow -z$ in the $-R$ to $0$ integral we get $$\int_0^{\infty}\frac{\log(1+iz)+\log(1-iz)}{1+z^2} dz =\int_0^{\infty}\frac{\log(1+z^2)}{1+z^2}dz=\pi\log(2)$$

And since your integrand is even...

Answer 2

It is well known that a good alternative to contour integration is differentiation under the integral sign: it perfectly applies here, so I am going to outline an alternative approach. For any $a>\frac{1}{2}$, by exploiting symmetry and by enforcing the substitution $\frac{1}{1+x^2}=u$, we have $$ \int_{-\infty}^{+\infty}\frac{dx}{(1+x^2)^a} = \frac{\sqrt{\pi}\,\Gamma\left(a-\tfrac{1}{2}\right)}{\Gamma(a)}$$ by Euler's Beta function. By differentiating both sides with respect to $a$, and exploiting $\frac{d}{da}\,f=f\cdot\frac{d}{da}\log f$, we get: $$ \int_{-\infty}^{+\infty}\frac{-\log(1+x^2)}{(1+x^2)^a}\,dx = \frac{\sqrt{\pi}\,\Gamma\left(a-\tfrac{1}{2}\right)}{\Gamma(a)}\left[\psi\left(a-\tfrac{1}{2}\right)-\psi(a)\right]$$ and by evaluating both sides at $a=1$: $$ \int_{-\infty}^{+\infty}\frac{\log(1+x^2)}{(1+x^2)}\,dx = \pi\left[\psi(1)-\psi\left(\tfrac{1}{2}\right)\right]=\color{red}{2\pi\log 2}.$$ The equivalent claim $$ \int_{0}^{\pi/2}\log\sin\theta\,d\theta = -\frac{\pi}{2}\log 2$$ can also be proved through $\prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)=\frac{2n}{2^n}$ and Riemann sums, or just by symmetry tricks.

Answer 3

$\newcommand{\Log}{\operatorname{Log}}\newcommand{\Arg}{\operatorname{Arg}}$EDIT I+II. First there was a silly mistake, but now it is fixed. On the second version, it was just too sinuous. Now it is made straight to the point and short.

These kind of integrals are tricky. For getting started see $\ln(1+x^2)=\text{Re}(2\Log(i+x))$ where $\Log(\cdot) $ is the principal value of the logarithm. So it is reasonable to consider integrating something with $\Log(i+z)$ instead of something with $\Log(1+z^2)$. Let's integrate the following: \begin{align} \oint_C \frac{\Log(i+z)}{1+z^2}\,dz =\int_{C_R}\frac{\Log(i+z)}{1+z^2}\,dz+ \int^R_{-R}\frac{\Log(i+z)}{1+z^2}\,dz \end{align} where $C$ is a semi circle in the upper half plane; $C_R$ is the circle part and the other one is obvious. The branch cut is on $z=-i-\lambda$ with $\lambda\in (0,\infty) $, so we are (luckily) far away from the branch cut. The only residue inclosed is $z=i$ and therefore the equality: \begin{align} \int_{C_R}\frac{\Log(i+z)}{1+z^2}\,dz+ \int^R_{-R}\frac{\Log(i+z)}{1+z^2}\,dz=2\pi i \text{Res}_{z=i} \frac{\Log(i+z)}{1+z^2} \end{align} by The Residue Theorem. The integral on $C_R$ will have no contribution when $R\to\infty$. Now lets have a look at the real part of the one on $[-R,R]$: \begin{align} \Re\left(\int^R_{-R}\frac{\Log(i+z)}{1+z^2}\,dz\right) &= \frac{1}{2}\int^R_{-R} \frac{\ln(1+x^2)}{1+x^2}\,dx \end{align} That is what we have observed in the beginning. Let $R\to \infty$ and we get: \begin{align}\tag{1} \frac{1}{2}\int^\infty_{-\infty} \frac{\ln(1+x^2)}{1+x^2}\,dx = \Re\left(2\pi i \text{Res}_{z=i} \frac{\Log(i+z)}{1+z^2}\right) \end{align} Doing the residue times $2\pi i$: \begin{align} 2\pi i\text{Res}_{z=i} \frac{\Log(i+z)}{1+z^2} = 2\pi i\frac{\Log(2i)}{2i}= 2\pi i\frac{\ln(2)+i\frac{\pi}{2}}{2i}=\pi\ln(2)+\frac{i\pi^2}{2} \end{align} The real parts in both sides of $(1)$ must be equal and mulitply both sides with $2$ and get the final result:

\begin{align} \int^\infty_{-\infty}\frac{\ln(1+x^2)}{1+x^2}\,dx = 2\pi\ln(2) \end{align}

Answer 4

The function $ln(1+x^2)$ has two branch points at $\pm i$, then we cannot choose the contour you propose, as noted in the comment by @Sangchul Lee. Also, the point $x=0$ is not special for the function, the contribution of $\gamma_r$ is useless too. Instead, we can consider a branch cut on the vertical segment between $i$ and $-i$. It is convenient to decompose $$ f(x)=\frac{1}{2}\frac{\ln(1+x^2)}{1-ix}+\frac{1}{2}\frac{\ln(1+x^2)}{1+ix}$$ By enforcing the substitution $y=-x$ in the second integral, one has \begin{align} I&=\frac{1}{2}\int_{-\infty}^\infty\frac{\ln(1+x^2)}{1-ix}\,dx+\frac{1}{2}\int_{-\infty}^\infty\frac{\ln(1+x^2)}{1+ix}\,dx\\ &=\int_{-\infty}^\infty\frac{\ln(1+x^2)}{1-ix}\,dx \end{align}

We define a contour along the real axis which avoids the branch cut by turning around $x=i$ and which is closed by the large upper semi-circle $C_R$. From the residue theorem, it holds \begin{equation} 0=\left[\int_{C_R}+\int_{-\infty}^{-\varepsilon}+J+\int_{\varepsilon}^{\infty}\right] f^-(x)\,dx \end{equation} where $f^-(x)=\left( 1-ix \right)^{-1}\ln\left( 1+x^2 \right)$ and $J$ is the contribution of the path which turns around the branch cut. It can be shown that the large semi-circle contribution vanishes as $R\to\infty$, then \begin{equation} I=\int_{-\infty}^\infty f^-(x)\,dx=-J \end{equation} and

$$J=\left[\int_{-\varepsilon}^{i-\varepsilon} +\int_{\gamma_i} +\int^{\varepsilon}_{i+\varepsilon} \right] f^-(x)\,dx$$ $\int_{\gamma_i}$ is the contribution of a small semi-circle around $x=i$, it can be shown to vanish when $\varepsilon\to 0$. On the left and right sides of the vertical contribution, with $x=-\varepsilon+ iy$, and passing to the limit $\varepsilon\to 0$ \begin{align} \int_{-\varepsilon}^{i-\varepsilon}f^-(x)\,dx&=i\int_0^1\frac{\ln(1-y^2)+2i\pi}{1+y}\,dy\\ \int^{\varepsilon}_{i+\varepsilon}f^-(x)\,dx&=i\int_1^0\frac{\ln(1-y^2)}{1+y}\,dy \end{align} Adding both terms, the logarithmic contributions cancel each other and we obtain \begin{equation} J=-2\pi\ln2 \end{equation} Finally, \begin{equation} I=2\pi\ln2 \end{equation} In this method, like in the other given answers, a decomposition is used in order to separate the poles and the branch points near the contour.