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 Post subject: $\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}$Posted: Tue Jun 13, 2017 3:57 pm
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
And the cause of this:

Evaluate $$\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}\,,\quad {n}\in\mathbb{N}\,,\, n\geqslant3\,.$$

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Grigorios Kostakos

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 Posted: Wed Jun 14, 2017 2:16 am

Joined: Tue May 10, 2016 3:56 pm
Posts: 33
Let the integral be $I_n$. We show that
$$I_n = \left\{\begin{array}{ll} \frac{1}{2k} \left(\frac{1}{2k-1} - \frac{1}{2k-2} + \cdots + 1\right), & \mbox{if n = 2k+1, k \geq 1}\\ \frac{2\ln 2}{2k-1} + \frac{1}{2k-1}\left(\frac{1}{2k-2} - \frac{1}{2k-3} + \cdots - 1\right), & \mbox{if n = 2k, k \geq 2.}\end{array}\right.$$
To this end, we show the case when $n = 2k+1$ only. The case when $n=2k$ is similar. Integrating by parts and partial fractions yields
\begin{eqnarray*}
I_{2k+1} & = & \int_1^\infty\ln(1+x)d\left(-\frac{1}{2k}\,x^{-2k}\right)\\
& = & \frac{1}{2k}\,\ln 2 + \frac{1}{2k}\,\int_1^\infty\frac{dx}{x^{2k}(1 + x)}\\
& = & \frac{1}{2k}\,\ln 2 + \frac{1}{2k}\,\int_1^\infty\left(\frac{1}{x^{2k}} - \frac{1}{x^{2k-1}} + \cdots + \frac{1}{x^2} - \frac{1}{x} +\frac{1}{1 + x}\right)\,dx\\
& = & \frac{1}{2k}\,\ln 2 + \frac{1}{2k}
\left(\frac{1}{2k-1} - \frac{1}{2k-2} + \cdots + 1 -\ln 2\right)\\
& = & \frac{1}{2k}
\left(\frac{1}{2k-1} - \frac{1}{2k-2} + \cdots + 1\right).
\end{eqnarray*}

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 Posted: Wed Jun 14, 2017 5:42 am
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
Nice solution mathofusva! Here is my attempt for $n>2$.
\begin{align*}
\int_{1}^{+\infty}{\frac{\log(x+1)}{x^n}\,dx} &\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,\frac{1}{x}}\\
{dx\,=\,-\frac{1}{t^2}dt}\\
\end{subarray}}\,-\int_{1}^{0}{t^n\log\big(\tfrac{1}{t}+1\big)\frac{1}{t^2}\,dt}\\
&=\int_{0}^{1}{t^{n-2}\log\big(\tfrac{1}{t}+1\big)\,dt}\\
&=\int_{0}^{1}{\Big(\frac{t^{n-1}}{n-1}\Big)'\log\big(\tfrac{1}{t}+1\big)\,dt}\\
&=\Big[\frac{t^{n-1}}{n-1}\log\big(\tfrac{1}{t}+1\big)\Big]_{0}^{1}+\frac{1}{n-1}\int_{0}^{1}{\frac{t^{n-2}}{t+1}\,dt}\\
&=\frac{\log2}{n-1}-\frac{1}{n-1}\,\cancelto{0}{\mathop{\lim}\limits_{t\to0^{+}}t^{n-1}\log\big(\tfrac{1}{t}+1\big)}\;+\frac{1}{n-1}\int_{0}^{1}{t^{n-2}\mathop{\sum}\limits_{m=0}^{+\infty}(-1)^mt^m\,dt}\\
&=\frac{\log2}{n-1}+\frac{1}{n-1}\mathop{\sum}\limits_{m=0}^{+\infty}(-1)^m\int_{0}^{1}{t^{m+n-2}\,dt}\\
&=\frac{\log2}{n-1}+\frac{1}{n-1}\mathop{\sum}\limits_{m=0}^{+\infty}\frac{(-1)^m}{m+n-1}\\
&=\frac{\log2}{n-1}+\frac{1}{2\,(n-1)}\Big(\psi\big(\tfrac{n}{2}\big)-\psi\big(\tfrac{n-1}{2}\big)\Big)\,,
\end{align*}
from which this question came up.

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Grigorios Kostakos

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