How to compute $\int_0^1\frac{t\ln t}{1+t^2}$ ?

Question

How to compute the integral $$\int_0^1\frac{t\ln t}{1+t^2}\ ?$$

So Wolfram alpha says it is exactly $-\dfrac{\pi^2}{48}$ .

I tried many substitutions without success, and partial integration as well.

Any help is welcome.

Answer 1

Integrating by parts $$ -\int_0^1\frac{t\ln t}{1+t^2}\,dt=\frac12\int_0^1\bigl(\log(1+t^2)\bigr)'\log t\,dt=\frac12\int_0^1\frac{\log(1+t^2)}{t}\,dt. $$ Expad the integrand in a power series to get $$ \int_0^1\frac{\log(1+t^2)}{t}\,dt=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\int_0^1t^{2n-1}\,dt=\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^2}. $$ This is a well known sum.

Answer 2

I would expand $(1+t^2)^{-1}$ around $t=0$ and find out what $$\int_0^1 t^k \log t dt$$ is for each $k$.

ADD The sequence of functions $\displaystyle f_n=-t\log t\sum_{k=0}^n (-1)^k t^{2k}$ is dominated by $-t\log t$ over $[0,1]$ and converges to $f=-t\log t(1+t^2)^{-1}$ so Lebesgue's DCT ensures $$\int_0^1 \frac{t\log t}{1+t^2}dt= \sum_{k\geqslant 1}(-1)^k\int_0^1 t^{2k+1}\log t dt$$

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \color{#66f}{\large\int_{0}^{1}{t\ln\pars{t} \over 1 + t^{2}}\,\dd t}&= \Re\int_{0}^{1}{\ln\pars{t} \over \ic + t}\,\dd t =-\Re\int_{0}^{1}{\ln\pars{-\ic\bracks{\ic t}} \over 1 - \pars{\ic t}} \,\pars{\ic\,\dd t} =-\,\Re\int_{0}^{\ic}{\ln\pars{-\ic t} \over 1 - t}\,\dd t \\[3mm]&=-\,\Re\int_{0}^{\ic}\ln\pars{1 - t}\,{1 \over -\ic t}\,\pars{-\ic}\,\dd t =\Re\int_{0}^{\ic}{{\rm Li}_{1}\pars{t} \over t}\,\dd t \end{align} where $\ds{{\rm Li_{s}}\pars{z}}$ is a PolyLogarithm Function and $\ds{{\rm Li}_{1}\pars{z} = -\ln\pars{1 - z}}$.

With the PolyLogaritm Recursive Property and $\ds{{\rm Li}_{2}\pars{0} = 0}$: \begin{align} \color{#66f}{\large\int_{0}^{1}{t\ln\pars{t} \over 1 + t^{2}}\,\dd t}&= \Re\int_{0}^{\ic}{\rm Li}_{2}'\pars{t}\,\dd t=\Re{\rm Li}_{2}\pars{\ic} =\color{#66f}{\large -\,{\pi^{2} \over 48}} \end{align}

Note that \begin{align}&\color{#c00000}{\Re{\rm Li}_{2}\pars{\ic}} =\Re\sum_{n = 1}^{\infty}{\ic^{n} \over n^{2}} =\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over \pars{2n}^{2}} ={1 \over 4}\bracks{\sum_{n = 1}^{\infty}{1 \over \pars{2n}^{2}} -\sum_{n = 1}^{\infty}{1 \over \pars{2n - 1}^{2}}} \\[3mm]&={1 \over 4}\bracks{{1 \over 4}\sum_{n = 1}^{\infty}{1 \over n^{2}} -\sum_{n = 1}^{\infty}{1 \over n^{2}} + \sum_{n = 1}^{\infty}{1 \over \pars{2n}^{2}}} =-\,{1 \over 8}\sum_{n = 1}^{\infty}{1 \over n^{2}} =-\,{1 \over 8}\,{\pi^{2} \over 6}=\color{#c00000}{-\,{\pi^{2} \over 48}} \end{align}

Also, $\ds{\Re{\rm Li}_{2}\pars{\ic}}$ can be calculated by means of ${\sf\mbox{Jonquiere Inversion Formula}}$ as shown in the above cited link.

A 'straightforward' calculation can be performed by expanding $\ds{\ln\pars{1 - t}}$, in powers of $\ds{t}$, in the expression $\ds{-\,\Re\int_{0}^{\ic}{\ln\pars{1 - t} \over t}\,\dd t}$.