Computing $\int_{-1}^{1} \frac{\ln (1+y)}{y} d y$ with simple method.

Question

Using the series $$ \ln (1+y)=\sum_{n=0}^{\infty} \frac{(-1)^{n} y^{n+1}}{n+1} \text { for }|y|<1 $$ to convert the integral into $$ \begin{aligned} \int_{-1}^{1} \frac{\ln (1+y)}{y} d y &=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} \int_{-1}^{1} y^{n} d y \\ &=\sum_{n=0}^{\infty} \frac{1-(-1)^{n+1}}{(n+1)^{2}} \\ &=\zeta(2)+\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n+1)^{2}} \\ &=\zeta(2)+\frac{1}{2}\zeta(2)\\&=\frac{\pi^{2}}{4} \end{aligned} $$

Question: Is there any other simple solution? Your suggestion and solution is warmly welcome.

Answer 1

An elementary solution \begin{align} \int_{-1}^{1} \frac{\ln (1+y)}{y} &{d y} \overset{y\to -y} = \frac12 \int_{-1}^{1} \frac{\ln \frac{1+y}{1-y}}{y} d y \overset{y\to \frac{1+y}{1-y}}=\int_0^\infty \frac{\ln y}{y^2-1}dy\\ =& \int_0^\infty \int_0^1 \frac{x}{1+(y^2-1)x^2}dx\>dy =\frac\pi2 \int_0^1 \frac1{\sqrt{1-x^2}}dx=\frac{\pi^2}4 \end{align}

Answer 2

$\newcommand{\d}{\,\mathrm{d}}\newcommand{\li}{\operatorname{Li}_2}$Using the logarithm series is definitely the simplest method. I will instead, for the sake of variety as you request, use a series expansion in the $1/y$ term, which is slightly longer and requires knowledge that $\li(1/2)=\frac{\pi^2}{12}-\frac{1}{2}\ln^22$, which I in all honesty had to look up :) but actually then this integral provides a nice proof of that result, by comparing the two methods.

Using $y\mapsto y-1$ and then $y\mapsto e^{-y}$ we obtain: $$\int_{-1}^1\frac{\ln(1+y)}{y}\d y=\int_0^2\frac{\ln(y)}{y-1}\d y=-\int_{-\ln2}^\infty\frac{y}{e^{-y}-1}\cdot e^{-y}\d y$$

Now we split into two integrals. Using the standard Gamma function transformation $y\mapsto ny$: $$\int_0^\infty\frac{y}{e^{-y}-1}e^{-y}\d y=-\sum_{n\ge1}\int_0^\infty y\cdot e^{-ny}\d y=-\zeta(2)=-\frac{\pi^2}{6}$$

Using the same geometric series idea, and the fact that $(-\ln2,0)$ is a negative interval: $$\int_{-\ln2}^0\frac{y}{e^{-y}-1}e^{-y}\d y=\int_{-\ln2}^0\frac{y}{1-e^y}\d y=\sum_{n\ge0}\int_{-\ln2}^0y\cdot e^{ny}\d y$$

Integration by parts for $n\ge1$: $$\begin{align}\int_{-\ln2}^0y\cdot e^{ny}\d y&=\frac{1}{n}y\cdot e^{ny}\Big|_{-\ln2}^0-\frac{1}{n}\int_{-\ln2}^0e^{ny}\d y\\&=\frac{\ln2}{n}2^{-n}-\frac{1}{n^2}(1-2^{-n})\end{align}$$And of course for $n=0$ the integral straightforwardly equals $-\frac{1}{2}\ln^22$. Using the (di)logarithm series: $$\sum_{n\ge0}\int_{-\ln2}^0y\cdot e^{ny}\d y=-\frac{1}{2}\ln^22-\ln2\cdot\ln(1-1/2)-\zeta(2)+\li(1/2)=\frac{\pi^2}{12}-\frac{\pi^2}{6}$$

So finally we have: $$\int_{-1}^1\frac{\ln(1+y)}{y}\d y=-\left[-\frac{\pi^2}{6}+\frac{\pi^2}{12}-\frac{\pi^2}{6}\right]=\frac{\pi^2}{4}\quad\blacksquare$$

To get: $$\li(1/2)=\frac{\pi^2}{12}-\frac{1}{2}\cdot\ln^22$$By other means, plug $x=1/2$ into formula $7$ of this derivation by RobJohn.

Answer 3

The function $\log(1+z)/z$ is holomorphic in the region ${\rm Re} \,z>-1,$ ${\rm Im}\, z\ge 0.$ Thus (actually the limit procedure should be applied to justify what follows) $$\int\limits_{-1}^1 {\log(1+y)\over y}\,dy= -\int\limits_{\gamma} {\log(1+z)\over z}\,dz$$ where $\gamma$ is the half-circle parametrized in the positive direction, i.e. $z=e^{it},$ $0\le t\le \pi.$ Hence $$\int\limits_{-1}^1 {\log(1+y)\over y}\,dy={\rm Re}\left \{ -i\int\limits_{0}^\pi \left [\log\left (2\cos{t\over 2}\right)+i{t\over 2}\right ]\,dt\right \}={1\over 2}\int\limits_0^\pi t\,dt={\pi^2\over 4}$$