# Calculating $\int^1_0\frac{x \log x}{1 + x^2} dx$

Trying to reinvent the spirit of Calculus, I was trying to Understand the Leibniz Rule of Differentiation under Integration sign. To check my understanding, I was considering this integral: $$I(b) = \int^1_0 \frac{x \log(b + x)}{1 + x^2} dx$$ After some simplification, I came upon the Integral in question: $$2I(1) = \frac{1}{4} \log^2{2} + \frac{\pi^2}{16} + 2\int^1_0\frac{b \log b}{1 + b^2} db$$

Now I am a bit confused about how to progress. Maxima shows the answer to be $-\frac{\pi^2}{48}$.

This is not a homework problem.

\begin{align} \int_0^1\frac{x\log(x)}{1+x^2}\,\mathrm{d}x &=\sum_{k=0}^\infty\int_0^1(-1)^kx^{2k+1}\log(x)\,\mathrm{d}x\tag1\\ &=\sum_{k=0}^\infty\frac{(-1)^{k+1}}{(2k+2)^2}\tag2\\ &=\frac14\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\tag3\\ &=-\frac14\left(\sum_{k=1}^\infty\frac1{k^2}-2\sum_{k=1}^\infty\frac1{(2k)^2}\right)\tag4\\ &=-\frac{\pi^2}{48}\tag5 \end{align} Explanation:
$(1)$: expand $\frac{x}{1+x^2}$ into a geometric series
$(2)$: integrate by parts: $u=\log(x)$, $\mathrm{d}v=x^{2k+1}\,\mathrm{d}x$
$(3)$: substitute $k\mapsto k-1$
$(4)$: an alternating sum is all the terms minus twice the even terms
$(5)$: $\zeta(2)=\frac{\pi^2}6$

• The fourth line seems particularly tricky to me. But the derivation is concise enough. – sarker306 Mar 1 '18 at 11:48
• @sarker306: I've added some explanation to hopefully clarify the steps. – robjohn Mar 1 '18 at 15:21
• @sarker306: I could have skipped step $(4)$ and just used $\eta(2)=\frac{\pi^2}{12}$, where $\eta$ is the Dirichlet Eta Function. – robjohn Mar 1 '18 at 15:26
• Thank you. This solution is worth revisiting. – sarker306 Mar 1 '18 at 16:08
• after getting myself familiar with using infinite series (expansions of elementary functions), this solution looks like the easiest one to apply. – sarker306 Jul 10 at 7:43

This calculation may contains some convergence problem, but it'll be resolved easily.

Using integration by parts, we have $$I(0)=\int_{0}^{1}\frac{x\log x}{1+x^{2}}dx = \left[\frac{1}{2}\log(1+x^{2})\log x\right]_{0}^{1} - \int_{0}^{1} \frac{1}{2}\log(1+x^{2})\frac{dx}{x}$$ Since $$\lim_{x\to 0} \log(1+x^{2})\log x = \lim_{x\to 0} \frac{\log(1+x^{2})}{x^{2}}\cdot x\cdot x\log x = 0,$$ we have $$I(0) = -\frac{1}{2}\int_{0}^{1} \frac{\log(1+x^{2})}{x}dx.$$ Now use substitution $x^{2} = t$, then $$I(0) = -\frac{1}{4} \int_{0}^{1}\frac{\log(1+t)}{t}dt$$ Now use the Taylor series of $\log(1+t)$, $$I(0) = -\frac{1}{4} \int_{0}^{1} \frac{1}{t}\left(t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-\cdots\right)dt = -\frac{1}{4}\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}} = -\frac{1}{4}\left(\sum_{n=1}^{\infty}\frac{1}{n^{2}}-2\sum_{n=1}^{\infty} \frac{1}{(2n)^{2}}\right) = -\frac{1}{4}\cdot \frac{1}{2}\zeta(2) = -\frac{\pi^{2}}{48}.$$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{x\ln\pars{x} \over 1 + x^{2}}\,\dd x & \,\,\,\stackrel{x\ \to\ x^{\large 1/2}}{=}\,\,\, {1 \over 4}\int_{0}^{1}{\ln\pars{x} \over 1 + x}\,\dd x \,\,\,\stackrel{x\ \to\ -x}{=}\,\,\, -\,{1 \over 4}\int_{0}^{-1}{\ln\pars{-x} \over 1 - x}\,\dd x \\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\, {1 \over 4}\int_{0}^{-1}{-\ln\pars{1 - x} \over x}\,\dd x = {1 \over 4}\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\,\dd x = {1 \over 4}\,\mrm{Li}_{2}\pars{-1} \\[5mm] & = {1 \over 4}\,\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2}} = {1 \over 4}\,\sum_{n = 1}^{\infty}\braces{% {1 \over \pars{2n}^{2}} - \bracks{{1 \over n^{2}} - {1 \over \pars{2n}^{2}}}} \\[5mm] & = -\,{1 \over 8}\ \underbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}} _{\ds{\pi^{2} \over 6}}\ =\ \bbx{-\,{\pi^{2} \over 48}} \end{align}

• You don't need series if your purpose is to use $\zeta(1)=\frac{\pi^2}{6}$. Change of variable tool is enough. – FDP Mar 1 '18 at 22:33
• @FDP $\zeta\left(\color{#f00}{2}\right) = \pi^{2}/6$. I was late. So, I was trying to build something different. Sorry for the $overkill$ !!!. – Felix Marin Mar 1 '18 at 22:43
• You are right, my mistake ! It's weird for me to use integral definition of dilog and use series for $\zeta(2)$ in the same computation. – FDP Mar 1 '18 at 22:58