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

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.

• Shouldn't the integral be negative? – hot_queen Jan 20 '14 at 15:05
• Forgot the minus sign, my bad. – Gabriel Romon Jan 20 '14 at 15:05
• Wolfram shows $Li_2$-terms in the antiderivate, so there is no easy antiderivate, but the terms obviously can be calculated exactly in this case. The other approach could be the residue theorem. – Peter Jan 20 '14 at 15:07

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.

• Your last sentence is one big assumption :). BTW you forgot a minus sign from the beginning. – Gabriel Romon Jan 20 '14 at 21:56

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 $$\int_0^1 \frac{t\log t}{1+t^2}dt= \sum_{k\geqslant 1}(-1)^k\int_0^1 t^{2k+1}\log t dt$$

• Wouldn't there be some issues of uniform convergence? Why can you pass the integral through the sum? – abnry Jan 20 '14 at 15:07
• @nayrb Consider them as Lebesgue integrals. – Pedro Tamaroff Jan 20 '14 at 15:12


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 $\ds{-\,\Re\int_{0}^{\ic}{\ln\pars{1 - t} \over t}\,\dd t}$.