Integral with logarithm

Let $n \in \mathbb{N}, \; n >2$  then prove that:

$$\int_{0}^{\infty}\frac{\log \left(\frac{1}{x}\right)}{\left ( 1+x \right )^n}\, {\rm d}x = \frac{1}{n-1}\sum_{k=1}^{n-2}\frac{1}{k}$$

Solution



We present two solutions. However in the link below the interested reader can find three more.

1st:

Let $\displaystyle I_n = \int_{0}^{\infty}\frac{\log x}{\left ( 1+x \right )^n}\, {\rm d}x $. Then we can easily see that $I_2=0$. Hence:

$$\begin{aligned}I_n &=\int_{0}^{\infty}\frac{\log x}{\left ( 1+x \right )^n}\, {\rm d}x \\&=\int_{0}^{\infty}\frac{(1+x)\log x}{\left ( 1+x \right )^{n+1}}\, {\rm d}x \\
 &= \int_{0}^{\infty}\frac{\log x }{\left ( 1+x \right )^{n+1}}\, {\rm d}x + \int_{0}^{\infty}\frac{x \log x}{\left ( 1+x \right )^{n+1}}\, {\rm d}x\\
&= I_{n+1}- \frac{1}{n}\int_{0}^{\infty}x \log x \left ( \frac{1}{(1+x)^n} \right )' \, {\rm d}x \\
&=I_{n+1}- \frac{1}{n}\left [ \frac{x\log x}{(1+x)^n} \right ]_0^{\infty}+ \frac{1}{n}\int_{0}^{\infty}\frac{\log x +1}{(1+x)^n}\, {\rm d}x \\
 &=I_{n+1}+ \frac{1}{n}I_n + \frac{1}{n}\int_{0}^{\infty}\frac{{\rm d}x}{\left ( 1+x \right )^n}
\end{aligned}$$

Therefore:

$$I_n = I_{n+1}+ \frac{1}{n}I_n + \frac{1}{n(n-1)}$$

From the recurrence relation we deduce that:

$$\left.\begin{matrix}
(n-1)I_{n} - (n-2)I_{n-1}& = &-\dfrac{1}{n-2}\\
(n-2)I_{n-1}- (n-3)I_{n-2}& = &-\dfrac{1}{n-3}\\
\cdots \cdots \cdots \cdots& \cdots &\cdots \cdots \\
2I_3 -I_2& = & -1
\end{matrix}\right\} \overset{(+)}{\Rightarrow }(n-1)I_n -I_2 $$

$$=-\left ( 1+ \frac{1}{2}+ \frac{1}{3}+\cdots + \frac{1}{n-2} \right )=- \sum_{k=1}^{n-2}\frac{1}{k}$$

and the result follows.


2nd:

We are invoking the following Laplace transformations:

${\color{gray} \blacksquare} \;\; \displaystyle \int_{0}^{\infty}\log x e^{-xy}\, {\rm d }x = - \frac{\gamma+ \log y}{y}$

${\color{gray} \blacksquare} \;\; \displaystyle \int_{0}^{\infty}f(x)e^{-xy}\, {\rm d}x = F(y) \Longrightarrow \int_{0}^{\infty}x^n f(x)e^{-xy}\,{\rm d}x = (-1)^n F^{(n)}(y)$

${\color{gray} \blacksquare} \;\; \displaystyle  \frac{\mathrm{d}^n }{\mathrm{d} x^n} \left [ \frac{1}{x} \right ]= \frac{(-1)^n n!}{x^{n+1}} , \;\;\;\;\; \frac{\mathrm{d}^n }{\mathrm{d} x^n} \left [ \frac{\log x}{x} \right ]= \frac{(-1)^{n-1} n! \mathcal{H}_n}{x^{n+1}}+ \frac{(-1)^n n! \log x}{x^{n+1}}$

where $\mathcal{H}_n$ denotes the $n$-th harmonic number. The third formulae used are proved by induction.

Successively, then, we get:

 $$\begin{aligned}\int_{0}^{\infty}\frac{\log x}{\left ( 1+x \right )^n}\, {\rm d}x &=\frac{1}{(n-1)!}\int_{0}^{\infty}\log x \left( \int_{0}^{\infty}y^{n-1}e^{-(1+x)y}\, {\rm d}y \right)\, {\rm d}x\\
&=\frac{1}{(n-1)!}\int_{0}^{\infty}y^{n-1}e^{-y}\left ( \int_{0}^{\infty}\log x e^{-xy}\, {\rm d}x \right )\, {\rm d}y \\
&\overset{(1)}{=}- \frac{1}{(n-1)!}\int_{0}^{\infty}y^{n-2}e^{-y}\left ( \gamma +\log y \right )\, {\rm d}y \\
&= -\frac{\gamma}{(n-1)!}\int_{0}^{\infty}y^{n-2}e^{-y}\, {\rm d}y -\\
& \quad \quad \quad -\frac{1}{(n-1)!}\int_{0}^{\infty}y^{n-2}e^{-y}\log y \, {\rm d}y\\
&\overset{(2)}{=} -\frac{\gamma}{n-1}+ \frac{(-1)^{n-2}}{(n-1)!}\left ( \frac{\gamma+ \log x}{x} \right )^{(n-2)} \bigg|_{x=1} \\
&=- \frac{\gamma}{n-1}+ \frac{\gamma (-1)^{n-2}}{(n-1)!}\left ( \frac{1}{x} \right )^{(n-2)}\bigg|_{x=1}+ \\
& \quad \quad \quad +\frac{(-1)^{n-2}}{(n-1)!}\left ( \frac{\log x}{x} \right )^{(n-2)} \bigg|_{x=1} \\
&\overset{(3)}{=}- \frac{\gamma}{n-1}+ \frac{\gamma (-1)^{n-2}}{(n-1)!}\cdot \frac{(-1)^{n-2}(n-2)!}{x^{n-1}} \bigg|_{x=1}+ \\
& \quad \quad \quad + \frac{(-1)^{n-2}}{(n-1)!}\frac{(-1)^{n-3}(n-2)!\mathcal{H}_{n-2}}{x^{n-1}} \bigg|_{x=1} \\
&= \cancel{\frac{-\gamma}{n+1}} + \cancel{\frac{\gamma}{n+1}} - \frac{1}{n-1}\mathcal{H}_{n-2}\end{aligned}$$

leading us to the wanted result once again.

The exercise can also be found in mathematica.gr