$\displaystyle \int_0^{+\infty} \dfrac{\ln(x+1)}{x^2+1} \ \mathrm dx$

Question

Find the value of:

$$\int_0^{+\infty} \dfrac{\ln(x+1)}{x^2+1} \ \mathrm dx$$

I tried using substitution but it doesn't work. Is there any way to solve it ?

Answer 1

Hint. One may set $$ I(t):=\int_0^{+\infty} \dfrac{\ln(1+tx)}{x^2+1} \ \mathrm dx,\qquad t>0. $$ Then one may obtain $$ \begin{align} I'(t)=&\int_0^{+\infty} \dfrac{x}{(x^2+1)(1+tx)} \ \mathrm dx \\\\&=\int_0^{+\infty} \left(\dfrac{x}{(x^2+1)(1+tx)}-\frac{t}{\left(t^2+1\right) (t x+1)} \right) \mathrm dx \\\\&=\left[\frac{\ln \left(x^2+1\right)}{2 t^2+2}+\frac{t \arctan x-\ln (t x+1)}{t^2+1} \right]_0^{+\infty} \\\\&=\frac{\pi t}{2 t^2+2}-\frac{2 \ln t}{2 t^2+2},\qquad t>0, \end{align} $$ giving, with $I(0)=0$,

$$ \int_0^{+\infty} \dfrac{\ln(1+x)}{x^2+1}=I(1)=I(1)-I(0)=\int_0^1I'(t)\:dt=C+\frac{\pi}{4} \ln 2 $$

where $C$ is the Catalan constant.

Answer 2

I tried using substitution but it doesn't work.

A trig substitution will do. Letting $x = \tan \theta$, the integral becomes $\int_0^{\pi/2} \ln(\tan \theta + 1)\, d\theta$, which is the same as

$$\int_0^{\pi/2} \ln(\sin \theta + \cos \theta)\, d\theta - \int_0^{\pi/2}\ln(\cos \theta)\, d\theta$$

Since $\sin(\theta + \cos \theta) = \sqrt{2}\sin(\theta + \pi/4)$,

\begin{align}\int_0^{\pi/2}\ln(\sin \theta + \cos \theta)\, d\theta &= \int_0^{\pi/2} \ln\sqrt{2}\, d\theta + \int_0^{\pi/2} \ln\sin(\theta + \pi/4)\, d\theta\\ &= \frac{\pi}{4}\ln2 + \int_0^{\pi/2}\ln\sin(\theta + \pi/4)\, d\theta \end{align}

Note $\int_0^{\pi/4}\ln \sin(\theta + \pi/4)\, d\theta = \int_{\pi/4}^{\pi/2} \ln \sin\theta\,d\theta$ and $$\int_{\pi/4}^{\pi/2} \ln\sin(\theta + \pi/4)\, d\theta = \int_{\pi/4}^{\pi/2}\ln \cos(\theta - \pi/4)\, d\theta = \int_0^{\pi/4}\ln\cos \theta\, d\theta$$ The total integral is then $$\frac{\pi}{4}\ln 2 + \int_{\pi/4}^{\pi/2} \ln \sin \theta\, d\theta + \int_{0}^{\pi/4} \ln\cos \theta\, d\theta - \int_0^{\pi/2}\ln \cos \theta\, d\theta$$ or $$\frac{\pi}{4}\ln 2 + \int_{\pi/4}^{\pi/2}\ln \tan\theta\, d\theta$$ The latter integral equals $\int_0^{\pi/4}\tan(\pi/2 - \theta)\, d\theta = \int_0^{\pi/4}\ln \cot \theta = C$, Catalan's constant. Hence, the integral evaluates to

$$\frac{\pi}{4}\ln 2 + C$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\ln\pars{x + 1} \over x^{2} + 1}\,\dd x & = \int_{0}^{1}{\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x + \bracks{\int_{1}^{\infty}{\ln\pars{1/x + 1} \over x^{2} + 1}\,\dd x + \int_{1}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x} \end{align}

Note that the last integral is the Catalan Constant $C$. Namely, $\ds{\int_{1}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x = C}$.

Lets $\ds{x\ \mapsto\ 1/x}$ in the second integral: \begin{align} \int_{0}^{\infty}{\ln\pars{x + 1} \over x^{2} + 1}\,\dd x - C & = \int_{0}^{1}{\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x + \int_{1}^{0}{\ln\pars{x + 1} \over 1/x^{2} + 1}\pars{-\,{1 \over x^{2}}}\,\dd x \\[5mm] & = 2\int_{0}^{1}{\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x \,\,\,\stackrel{x\ =\ \tan\pars{\theta}}{=}\,\,\, 2\int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta \\[5mm] & = \int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta + \int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{{\pi \over 4} - \theta}}\,\dd\theta \\[5mm] & = \int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta + \int_{0}^{\pi/4} \ln\pars{1 + {1 - \tan\pars{\theta} \over 1 + \tan\pars{\theta}}}\,\dd\theta \\[5mm] & = \int_{0}^{\pi/4}\ln\pars{1 + \tan\pars{\theta}}\,\dd\theta + \int_{0}^{\pi/4} \ln\pars{2 \over 1 + \tan\pars{\theta}}\,\dd\theta \\[5mm] & = \int_{0}^{\pi/4}\ln\pars{2}\,\dd\theta = {\pi \over 4}\,\ln\pars{2} \\[5mm] \implies &\ \bbox[15px,#ffe,border:1px dotted navy]{\ds{% \int_{0}^{\infty}{\ln\pars{x + 1} \over x^{2} + 1}\,\dd x = {\pi \over 4}\,\ln\pars{2} + C}} \end{align}