Integrals

[Tolaso J Kos]

Prove that:

\( \rm{i.} \) \( \displaystyle \int_{0}^{1}\frac{\ln x}{x-1}\, dx=\sum_{n=1}^{\infty }\frac{1}{n^2}=\zeta (2)=\frac{\pi^2}{6} \).

\( \rm{ii.} \) \( \displaystyle \int_{0}^{1}\frac{x-1}{\ln x}\, dx=\ln 2 \)

[Grigorios Kostakos]

\(\rm{i.}\) \begin{align*}
\displaystyle \int_{0}^{1}{\frac{\log x}{x-1}\, dx}&=\int_{0}^{1}{\log x\biggl({-\mathop{\sum}\limits_{n=0}^{\infty}x^n}\biggr) dx}\\
&=\mathop{\sum}\limits_{n=0}^{\infty}\biggl({\int_{0}^{1}{-x^n\log{x} \, dx}}\biggr)\\
&=\mathop{\sum}\limits_{n=0}^{\infty}{\frac{1}{(n+1)^2}}\\
&=\mathop{\sum}\limits_{n=1}^{\infty}{\frac{1}{n^2}}=\zeta(2)=\frac{\pi^2}{6}\,.
\end{align*}


\(\rm{ii.}\) \begin{align*}
\displaystyle \int_{0}^{1}{\frac{x-1}{\ln x}\, dx}&\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,-\log{x}}\\
{-{\rm{e}}^{-t}dt\,=\,dx}\\
\end{subarray}}\, -\int_{+\infty}^{0}{\frac{{\rm{e}}^{-t}-1}{-t}\,{\rm{e}}^{-t}dt}\\
&= \int_{0}^{+\infty}{\frac{{\rm{e}}^{-t}-{\rm{e}}^{-2t}}{t}\,dt}\\
&=\int_{0}^{+\infty}{\biggl({\int_{1}^{2}{{\rm{e}}^{-st}ds}}\biggr)\,dt}\\
&=\int_{1}^{2}{\biggl({\int_{0}^{+\infty}{{\rm{e}}^{-st}dt}}\biggr)\,ds}\\
&=\int_{1}^{2}{\frac{1}{s}ds}\\
&=\log 2-\log1\\
&=\log 2\,.
\end{align*}

[Tolaso J Kos]

With the risk of flogging a dead horse, another solution for the second one.

\( \rm{ii.} \) Consider the integral: \( \displaystyle I(a)=\int_{0}^{1}\frac{x^a-1}{\ln x}\, dx \).
We differentiate with respect to \( a \) to get: \( \displaystyle \frac{\partial }{\partial a}I(a)=\int_{0}^{1}\frac{\partial }{\partial a}\frac{x^a-1}{\ln x}\, dx=\int_{0}^{1}x^a\, dx=\frac{1}{a+1} \) thus \( I(a)=\ln \left ( a+1 \right )+c \). However \(I(0)=0 \implies c=0 \) thus \( I(1)=\ln 2\).