$\int_{0}^{+\infty}{\frac{\log({\frac{1}{x}})}{(1+x)^n}dx}$

[Grigorios Kostakos]

For \(n\in{\mathbb{N}}\,,\;n>2\,,\) prove that \[\displaystyle \int_{0}^{+\infty}{\frac{\log\bigl({\tfrac{1}{x}}\bigr)}{(1+x)^n}\, dx}=\frac{1}{n-1}\mathop{\sum}\limits_{k=1}^{n-2}{\frac{1}{k}}\,.\]

[Tolaso J Kos]

Good morning Grigoris.

First I will rewrite the given integral as: \( \displaystyle -\int_{0}^{\infty }\frac{\ln x}{\left ( 1+x \right )^n}\, {\rm d}x \).
Now let me consider the integral \( \displaystyle \int_{0}^{\infty }\frac{x^{m-1}}{\left ( 1+x \right )^n}\, {\rm d}x \) whereas \( m\neq n \).
Thus applying \( \Gamma \) to the previous integral we have \( \displaystyle \int_{0}^{\infty }\frac{x^{m-1}}{\left ( 1+x \right )^n}\, {\rm d}x=\frac{\Gamma (m)\Gamma (n-m)}{\Gamma (n)} \). Differentiating both parts with respect to \( m \) we get:

$$-\int_{0}^{\infty }\frac{x^{m-1}\ln x}{\left ( 1+x \right )^n}dx=-\frac{\Gamma (m)\Gamma (n-m)}{\Gamma (n)}\left ( \psi (m)-\psi (n-m) \right ) $$

Now if we let \( m=1 \) we get:

$$\int_{0}^{\infty }\frac{\ln (\frac{1}{x})}{\left ( 1+x \right )^n}\, dx=-\frac{1}{n-1}\left ( - \gamma -\psi (n-1)\right )=\frac{1}{n-1}\left ( \gamma +\psi (n-1) \right )=\frac{1}{n-1}\sum_{k=1}^{n-2}\frac{1}{k} $$

since \( \displaystyle \gamma +\psi (n+1)=\sum_{k=1}^{n}\frac{1}{k} \).

[Grigorios Kostakos]

A 2nd solution(?) :

\begin{align*}
\int_{0}^{+\infty}{\frac{\log\bigl({\tfrac{1}{x}}\bigr)}{(1+x)^n}\, dx}&=\int_{0}^{1}{\frac{\log\bigl({\tfrac{1}{x}}\bigr)}{(1+x)^n}\, dx}+\int_{1}^{+\infty}{\frac{\log\bigl({\tfrac{1}{x}}\bigr)}{(1+x)^n}\, dx}\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,\frac{1}{x}}\\
{-\frac{1}{t^2}\,dt\,=\,dx}\\
\end{subarray}}\,\int_{0}^{1}{\frac{\log{x}}{(1+x)^n}\, dx}-\int_{0}^{1}{\frac{t^{n-2}\log{t}}{(1+t)^n}\, dt}\\
&=\int_{0}^{1}{\frac{\log{x}}{(1+x)^n}\, dx}-\int_{0}^{1}{\frac{x^{n-2}\log{x}}{(1+x)^n}\, dx}\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{x\,=\,{\mathrm{e}}^{-t}}\\
{dx\,=\,-{\mathrm{e}}^{-t}dt} \\
\end{subarray}}\,\int_{0}^{+\infty}{\frac{t\,{\rm{e}}^{-t}}{(1+{\rm{e}}^{-t})^n}dt}-\int_{0}^{+\infty}{\frac{t\,{\rm{e}}^{-(n-1)t}}{(1+{\rm{e}}^{-t})^n} dt}\\
&=\int_{0}^{+\infty}{t\biggl({\mathop{\sum}\limits_{k=1}^{+\infty}\frac{(-1)^{k+1}(n+k-2)!}{(k-1)!(n-1)!}\,{\mathrm{e}}^{-kt}}\biggr)\, dt}\,-\\
&\hspace{2.0cm}\int_{0}^{+\infty}{t\biggl({\mathop{\sum}\limits_{k=1}^{+\infty}\frac{(-1)^{k+1}(n+k-2)!}{(k-1)!(n-1)!}\,{\mathrm{e}}^{-(n+k-2)t}}\biggr)\, dt}\\
&=\mathop{\sum}\limits_{k=1}^{+\infty}\biggl({\frac{(-1)^{k+1}(n+k-2)!}{(k-1)!(n-1)!}\int_{0}^{+\infty}{t\,{\mathrm{e}}^{-kt}\, dt}}\biggr)\,-\\
&\hspace{2.0cm}\mathop{\sum}\limits_{k=1}^{+\infty}\biggl({\frac{(-1)^{k+1}(n+k-2)!}{(k-1)!(n-1)!}\int_{0}^{+\infty}{t\,{\mathrm{e}}^{-(n+k-2)t}\, dt}}\biggr)\\
&=\mathop{\sum}\limits_{k=1}^{+\infty}\biggl({\frac{(-1)^{k+1}(n+k-2)!}{(k-1)!(n-1)!}\,\frac{1}{k^2}}\biggr)\,-\\
&\hspace{2.0cm}\mathop{\sum}\limits_{k=1}^{+\infty}\biggl({\frac{(-1)^{k+1}(n+k-2)!}{(k-1)!(n-1)!}\,\frac{1}{(n+k-2)^2}}\biggr)\\
&\stackrel{(*)}{=}\mathop{\sum}\limits_{k=1}^{+\infty}\biggl({\frac{(-1)^{k+1}(n+k-2)!}{(k-1)!(n-1)!}\,\frac{1}{k^2}}\biggr)\,+\\
&\hspace{1.0cm}\frac{1}{n-1}\Bigl({\psi(n)+\gamma-\frac{1}{n-1}}\Bigr)-\mathop{\sum}\limits_{k=1}^{+\infty}\biggl({\frac{(-1)^{k+1}(n+k-2)!}{(k-1)!(n-1)!}\,\frac{1}{k^2}}\biggr)\\
&=\frac{1}{n-1}\Bigl({H_{n-1}-\frac{1}{n-1}}\Bigr)\\
&=\frac{1}{n-1}\,H_{n-2}\,.
\end{align*}

[akotronis]

Another solution:

We set \(\displaystyle{a_n:=-\int_{0}^{+\infty}\frac{\ln x}{(1+x)^n}\,dx}\), multiply with \(y^n\) and sum for \(n\geq2\).

For \(y\in(-1,0)\cup(0,1)\) we will have:

\[\begin{aligned}\sum_{n\geq2}a_ny^n&=-\sum_{n\geq2}\left(\int_{0}^{+\infty}\frac{\ln x}{(1+x)^n}\,dx\right)y^n
\stackrel{*}{=}-y^2\int_{0}^{+\infty}\frac{\ln x}{(1+x)^2}\frac{1}{1-\frac{y}{1+x}}\,dx\\
&=-y^2\int_{0}^{+\infty}\frac{\ln x}{x^2+x(2-y)+1-y}\,dx\stackrel{1-y=:a\in(0,1)\cup(1,2)}{\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}-y^2\int_{0}^{+\infty}\frac{\ln x}{x^2+x(1+a)+a}\,dx\\&=(a-1)\left(\int\frac{\ln x}{x+a}\,dx-\int\frac{\ln x}{x+1}\,dx\right)\Bigg|_{0}^{+\infty}\hspace{20ex}(1)\end{aligned}\]

Now,

\[\begin{aligned}\int\frac{\ln x}{x+k}\,dx&=\ln x\ln(x+k)-\int\frac{\ln(x+k)}{x}\,dx\stackrel{x=-y}{=}\ln x\ln(x+k)-\int\frac{\ln(k-y)}{y}\,dy\\&=\ln x\ln(k+x)-\ln k\int\frac{1}{y}\,dy-\int\frac{\ln(1-y/k)}{y}\,dy\\&\stackrel{y/a=z}{\!=\!=\!=\!=\!=\!=}\ln x\ln\left(\frac{k+x}{k}\right)-\int\frac{\ln(1-z)}{z}\,dz=\ln x\ln\left(\frac{k+x}{k}\right)+\mathrm{Li}_2(-x/k)\end{aligned}\]

where \(\mathrm{Li}_2(z)=-\int_{0}^{z}\frac{\ln(1-x)}{x}\,dx\) is the dilogarithmic function, so \((1)\) for \(k=a,\;1\) will give

\[\displaystyle{\sum_{n\geq2}a_ny^n=(a-1)\left(\ln x\ln\left(\frac{a+x}{a(1+x)}\right)+\mathrm{Li}_2(-x/a)-\mathrm{Li}_2(-x)\right)\Bigg|_{0}^{+\infty}\hspace{20ex}(2)}\].

it is straightforward that \(\displaystyle{\mathrm{Li}_2(x)= o\left(1\right) \qquad x\to0\quad(**)}\) and furthermore

one can show, by the definition of the dilogarithmic function, that

\[\displaystyle{\mathrm{Li}_2(-x)+\mathrm{Li}_2(-1/x)=-\pi^2/6-\frac{\ln^2x}{2}}\] from where it is clear that

\[\displaystyle{\mathrm{Li}_2(-x)=-\pi^2/6-\frac{\ln^2x}{2}+o(1)\qquad x\to+\infty\quad(***)}\]

plugging ** and *** in \((2)\) and going back from \(a\) to \(y\) we get that

\[\displaystyle{\sum_{n\geq2}a_ny^n=\frac{y\ln^2(1-y)}{2}}\]


But

\[\begin{aligned}\left[y^{n}\right]\frac{y\ln^2(1-y)}{2}&=\frac{1}{2}[y^{n}]y\left(\sum_{n\geq0}\left(\sum_{k=0}^{n}\frac{1}{k+1}\cdot\frac{1}{n-k+1}\right)y^n\right)\\&=\frac{1}{2}[y^{n-1}]\left(\sum_{n\geq0}\left(\sum_{k=0}^{n}\frac{1}{k+1}\cdot\frac{1}{n-k+1}\right)y^n\right)\\&=\frac{1}{2}\sum_{k=0}^{n-1}\frac{1}{k+1}\cdot\frac{1}{n-k}=\frac{1}{2(n+1)}\sum_{k=0}^{n-1}\left(\frac{1}{k+1}+\frac{1}{n-k}\right)=\frac{\mathrm{H}_n}{n+1}\end{aligned}\]

and we 're done.
__________________________________________________________________________
* An argument is need here to justify the change of the summation-integration order. I leave it for now.