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\(\int_{0}^{\frac{\pi}{2}} \frac{x^{k}}{\tan{x}}\, {\rm d}x\)

Calculus (Integrals, Series)
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Grigorios Kostakos
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\(\int_{0}^{\frac{\pi}{2}} \frac{x^{k}}{\tan{x}}\, {\rm d}x\)

#1

Post by Grigorios Kostakos » Wed Sep 21, 2016 2:09 pm

Evaluate \[\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{x^{k}}{\tan{x}}\, {\rm d}x\,,\] for $k=1,3,5,7$.
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Re: \(\int_{0}^{\frac{\pi}{2}} \frac{x^{k}}{\tan{x}}\, {\rm d}x\)

#2

Post by Riemann » Wed Sep 21, 2016 7:45 pm

Let us answer the cases $k=1, 3$ for the moment and give some extensions.
  • The $k=1$ case: This is the trivial once since:

    \begin{align*}
    \int_{0}^{\pi/2} \frac{x}{\tan x} \, {\rm d}x&=\int_{0}^{\pi/2} x \cot x \, {\rm d}x \\
    &= \int_{0}^{\pi/2} x \left ( \log \sin x \right ) ' \,{\rm d}x\\
    &= \left [ x \log \sin x \right ]_0^{\pi/2} - \int_{0}^{\pi/2} \log \sin x \, {\rm d}x \\
    &= \frac{\pi \log 2}{2}
    \end{align*}
  • The $k=3$ case: We'll nail it using Fourier series:

    \begin{align*}
    \int_{0}^{\pi/2} \frac{x^3}{\tan x} \, {\rm d}x &=\int_{0}^{\pi/2} x^3 \cot x \, {\rm d}x \\
    &=\left [ x^3 \log \sin x \right ]_0^{\pi/2} - \int_{0}^{\pi/2} 3x^2 \log \sin x \, {\rm d}x \\
    &= -3 \int_{0}^{\pi/2} x^2 \left ( -\log 2 - \sum_{n=1}^{\infty} \frac{\cos 2nx}{n} \right ) \, {\rm d}x \\
    &= \frac{\pi^3 \log 2}{8} + 3 \int_{0}^{\pi/2} x^2 \sum_{n=1}^{\infty} \frac{\cos 2nx}{n} \, {\rm d}x \\
    &= \frac{\pi^3 \log 2}{8} + 3 \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\pi/2} x^2 \cos 2nx \, {\rm d}x \\
    &= \frac{\pi^3 \log 2}{8} + \frac{3\pi}{4} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3} \\
    &= \frac{\pi^3 \log 2}{8} - \frac{3\pi}{4} \eta(3) \\
    &= \frac{\pi^3 \log 2}{8} - \frac{9 \pi \zeta(3)}{16}
    \end{align*}
Here is a very interesting result. In what follows $\psi$ is the digamma function.

$$\int_{0}^{\pi/2} x \tan^p x \, {\rm d}x= \frac{\pi}{4 \sin \frac{\pi p}{2}}\left [ \psi^{(0)} \left ( \frac{1}{2} \right ) - \psi^{(0)} \left ( \frac{1-p}{2} \right ) \right ]$$

where $p \in \mathbb{Q} \cup \{-1\} \setminus \mathbb{Z} \cap \left[ (-\infty, 0) \cup (0, 1) \right]$.

Of course $\displaystyle \psi^{(0)}\left ( \frac{1}{2} \right ) = - \gamma -\log 2$ and in general

$$\psi^{(0)} \left ( \frac{p}{q} \right ) = -\gamma - \log 2 q - \frac{\pi}{2} \cot \frac{\pi p}{q} + 2 \sum_{n=1}^{\left \lfloor \frac{q-1}{2} \right \rfloor} \cos \frac{2 \pi n p}{q} \log \sin \frac{\pi n}{q}$$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Re: \(\int_{0}^{\frac{\pi}{2}} \frac{x^{k}}{\tan{x}}\, {\rm d}x\)

#3

Post by Riemann » Fri Sep 23, 2016 8:51 am

The $k=5$ case:

\begin{align*}
\int_{0}^{\pi/2} \frac{x^5}{\tan x}\, {\rm d}x &=\int_{0}^{\pi/2} x^5 \cot x \, {\rm d}x \\
&=-5 \int_{0}^{\pi/2} x^4 \log \sin x \, {\rm d}x \\
&= -5 \int_{0}^{\pi/2} x^4 \left ( -\log 2 - \sum_{n=1}^{\infty} \frac{\cos 2nx}{n} \right ) \, {\rm d}x\\
&= \frac{\pi^5 \log 2}{32} + 5 \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\pi/2} x^4 \cos 2nx \, {\rm d}x\\
&= \frac{\pi^5 \log 2}{32} + 5 \sum_{n=1}^{\infty} \frac{4 \pi (-1)^n\left ( \pi^2 n^2 -6\right )}{32 n^5}\\
&= \frac{\pi^5 \log 2}{32} + \frac{5 \pi}{8} \sum_{n=1}^{\infty} \frac{(-1)^n \left ( \pi^2 n^2 -6 \right )}{n^5} \\
&= \frac{\pi^5 \log 2}{32} + \frac{225 \pi \zeta(5)}{64} - \frac{15 \pi^3 \zeta(3)}{32}
\end{align*}
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Re: \(\int_{0}^{\frac{\pi}{2}} \frac{x^{k}}{\tan{x}}\, {\rm d}x\)

#4

Post by Riemann » Tue Oct 04, 2016 8:17 pm

The $k=7$ case:


\begin{align*}
\int_{0}^{\pi/2} \frac{x^7}{\tan x} \, {\rm d}x &= \int_{0}^{\pi/2} x^7 \cot x \, {\rm d}x \\
&= -7 \int_{0}^{\pi/2} x^6 \log \sin x \, {\rm d}x\\
&= -7 \int_{0}^{\pi/2} x^6 \left ( -\log 2 - \sum_{n=1}^{\infty} \frac{\cos 2nx}{n} \right ) \, {\rm d}x\\
&=\frac{\pi^7 \log 2}{128} + 7 \int_{0}^{\pi/2} x^6 \sum_{n=1}^{\infty} \frac{\cos 2nx}{n} \, {\rm d}x \\
&= \frac{\pi^7 \log 2}{128} + 7 \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\pi/2} x^6 \cos 2nx \, {\rm d}x \\
&= \frac{\pi^7 \log 2}{128} + \frac{21 \pi}{64} \sum_{n=1}^{\infty} \frac{\left ( \pi^4 n^4 -20 \pi^2 n^2 +120 \right ) \cos \pi n}{n^7} \\
&= \frac{\pi^7 \log 2}{128} + \frac{21 \pi}{64} \pi \sum_{n=1}^{\infty} \frac{(-1)^n \left ( \pi^4 n^4 -20 \pi^2 n^2 +120 \right )}{n^7} \\
&= \frac{\pi^7 \log 2}{128} + \frac{1575 \pi^3 \zeta(5)}{256} - \frac{63 \pi^5 \zeta(3)}{256} - \frac{19845 \pi \zeta(7)}{512}
\end{align*}

Throughout the calculations we used the fact that:

$$\eta(s) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^s} = \left ( 1-2^{1-s} \right ) \zeta(s) , \; \mathfrak{Re}(s) \geq 1$$

where $\eta$ is Dirichlet eta function.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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