Evaluate an improper integral involving log

Question

Studying for complex analysis, I stumbled upon this problem. Evaluate $$\int_{0}^{\infty} \frac{\log(1+x^2)}{1+x^2}~dx.$$ So the integrand suggests that I need to use the function $f(z)=\frac{\log 1+z^2}{1+z^2},$ which has two branch points and two simple poles at $z=i$ and $z=-i,$ which lead me to use a semi-circle contour in the right-halfplane having branch cuts at $i$ and $-i.$ But then the process become too lengthy. Is there any other method to go by doing the problem. I was thinking of making a substitution and reducing this to a form to use Gamma function, but wasn't successful. Any help is appreciated.

Answer 1

Here is a straightforward methodology that relies on "Feynman's Trick" for differentiating under the integral sign. We proceed as follows.

First, let $I(a)$ be the integral defined by

$$\bbox[5px,border:2px solid #C0A000]{I(a)=\int_0^\infty \frac{\log(a+x^2)}{1+x^2}\,dx} \tag 1$$

where $I(1)=\int_0^\infty \frac{\log(1+x^2)}{1+x^2}\,dx$ is the term of interest.

Second, differentiating under the integral sign yields

$$\begin{align} I'(a)&=\int_0^\infty \frac{1}{(a+x^2)(1+x^2)}\,dx\\\\ &=\frac{1}{1-a}\int_0^\infty \left(\frac{1}{a+x^2}-\frac{1}{1+x^2}\right)\,dx\\\\ &=\frac{\pi}{2}\frac{1}{\sqrt{a}(\sqrt{a}+1)} \tag 2 \end{align}$$

Third, integrating $(2)$ reveals that $I(a)=\pi \,\log(1+\sqrt{a})+C$. Then, since $I(0)=0$, which can be seen easily by enforcing the substitution $x\to 1/x$ in $(1)$, $C=0$ and $I(a) =\pi\,\log(1+\sqrt{a})$.

Finally, we find that $I(1)=\pi \,\log(2)$ and hence

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{\log(1+x^2)}{1+x^2}\,dx=\pi \, \log(2)}$$

in agreement with the result reported by @jackd'aurizio.

Answer 2

Your intuition is correct: it is not difficult to compute such integral through a real-analytic method that involves differentiation under the integral sign and Euler's beta function (so $\Gamma$ function, too).

$$ \int_{0}^{+\infty}\frac{\log(1+t^2)}{1+t^2}\,dt = \left.\frac{d}{d\alpha}\int_{0}^{+\infty}(1+t^2)^{\alpha-1}\,dt\right|_{\alpha=0^+} \tag{1}$$ and by setting $t=\tan\theta$ we have $$ I(\alpha) = \int_{0}^{+\infty}(1+t^2)^{\alpha-1}\,dt = \int_{0}^{\pi/2}\frac{d\theta}{\cos^{2\alpha}(\theta)}=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(\frac{1}{2}-\alpha\right)}{\Gamma\left(1-\alpha\right)}.\tag{2}$$ In order to compute $I'(\alpha)$, we may exploit $$ I'(\alpha)=I(\alpha)\cdot\frac{d}{d\alpha}\log I(\alpha),\qquad \frac{d}{dx}\log\Gamma(x)=\psi(x)\tag{3} $$ leading to: $$ I'(\alpha) = -I(\alpha)\cdot\left[\psi\left(\frac{1}{2}-\alpha\right)-\psi(1-\alpha)\right],\qquad I'(0)=-\frac{\pi}{2}\cdot\left[\psi\left(\frac{1}{2}\right)-\psi(1)\right]\tag{4}$$ At least, the identity $$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{5}$$ proves that $\psi\left(\frac{1}{2}\right)-\psi(1)$ is related with a simple series whose value is $-2\log 2$.
Putting all together, $$ \int_{0}^{+\infty}\frac{\log(1+t^2)}{1+t^2}\,dt = \color{red}{\pi\log 2}.\tag{6} $$

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This is clearly not the simplest or fastest method, I went through it just to demonstrate its power and flexibility. The fastest method here is probably to directly set $t=\tan\theta$ and deduce $(6)$ from the symmetry of the integrand function or from a well known Fourier series. That technique was showed on MSE many times.

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\mathscr{I}} & = \int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over 1 + x^{2}}\,\dd x \,\,\,\stackrel{x\ =\ \tan\pars{\theta}}{=}\,\,\,\ \overbrace{-2\int_{0}^{\pi/2}\ln\pars{\cos\pars{\theta}}\,\dd\theta}^{\ds{=\ \color{#f00}{\mathscr{I}}}}\ =\ \overbrace{-2\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta} ^{\ds{=\ \color{#f00}{\mathscr{J}}}} \end{align}

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\begin{align} \color{#f00}{\mathscr{I}} & = {\color{#f00}{\mathscr{I}} + \color{#f00}{\mathscr{I}} \over 2} = -\int_{0}^{\pi/2}\bracks{\ln\pars{\cos\pars{\theta}} + \ln\pars{\sin\pars{\theta}}}\,\dd\theta = -\int_{0}^{\pi/2}\ln\pars{\sin\pars{2\theta} \over 2}\,\dd\theta \\[5mm] & = \half\,\pi\ln\pars{2} - \half\int_{0}^{\pi}\ln\pars{\sin\pars{\theta}}\,\dd\theta \\[5mm] & = \half\,\pi\ln\pars{2} - \half\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta - \half\int_{\pi/2}^{\pi}\ln\pars{\sin\pars{\theta}}\,\dd\theta \\[5mm] & = \half\,\pi\ln\pars{2} - \half\int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta - \half\int_{-\pi/2}^{0}\ln\pars{-\sin\pars{\theta}}\,\dd\theta \\[5mm] & = \half\,\pi\ln\pars{2} - \int_{0}^{\pi/2}\ln\pars{\sin\pars{\theta}}\,\dd\theta = \half\,\pi\ln\pars{2} + \half\,\color{#f00}{\mathscr{I}} \\[5mm] & \imp\quad \color{#f00}{\mathscr{I}} = \int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over 1 + x^{2}}\,\dd x = \color{#f00}{\pi\ln\pars{2}} \end{align}