It is currently Tue Jun 18, 2019 1:42 am

 All times are UTC [ DST ]

 Print view Previous topic | Next topic
Author Message
 Post subject: Logrithmic Integral Posted: Sat May 12, 2018 7:35 am

Joined: Thu Nov 12, 2015 5:26 pm
Posts: 102
$$\int^{\pi}_{0}x^2\ln(\sin x)dx$$

Top   Post subject: Re: Logrithmic Integral Posted: Sat May 12, 2018 1:43 pm
 Team Member Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
Using that the Fourier series of $\log(\sin{x})$ on $(0,\pi)$ is(*) \begin{align}
\log(\sin{x})=-\log2-\mathop{\sum}\limits_{n=1}^{\infty}\frac{\cos(2nx)}{n}
\end{align} then \begin{align*}
\int_{0}^{\pi}x^2\log(\sin{x})\,dx&\stackrel{(1)}{=}\int_{0}^{\pi}\Big(-x^2\log2-\mathop{\sum}\limits_{n=1}^{\infty}\frac{x^2\cos(2nx)}{n}\Big)\,dx\\
&=-\log2\int_{0}^{\pi}x^2\,dx-\int_{0}^{\pi}\mathop{\sum}\limits_{n=1}^{\infty}\frac{x^2\cos(2nx)}{n}\,dx\\
&=-\frac{\pi^3\log2}{3}-\mathop{\sum}\limits_{n=1}^{\infty}\Big(\frac{1}{n}\int_{0}^{\pi}x^2\cos(2nx)\,dx\Big)\\
&=-\frac{\pi^3\log2}{3}-\mathop{\sum}\limits_{n=1}^{\infty}\Big(\frac{1}{n}\frac{\pi}{2n^2}\Big)\\
&=-\frac{\pi^3\log2}{3}-\frac{\pi}{2}\mathop{\sum}\limits_{n=1}^{\infty}\frac{1}{n^3}\\
&=-\frac{\pi^3\log2}{3}-\frac{\pi}{2}\,\zeta(3)\,.\end{align*}

(*) which is not easy to prove! So, I hope that someone else will give a simpler solution.

_________________
Grigorios Kostakos

Top   Post subject: Re: Logrithmic Integral Posted: Sat May 12, 2018 4:31 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
The Fourier series is the way to go here. Recall that

$$\cos(2kx) = \dfrac{e^{i2kx} + e^{-i2kx}}{2}$$

Hence,

\begin{align*}
\sum_{k=1}^{\infty} \dfrac{\cos(2kx)}k
&= \sum_{k=1}^{\infty} \dfrac{e^{i2kx} + e^{-i2kx}}{2k}
\\&= \dfrac12 \big(-\log (1-e^{i2x} )-\log (1-e^{-i2x} ) \big)
\\&= - \dfrac12 \log \big(2 - 2\cos(2x) \big)
\\&= - \dfrac12 \log\big(4 \sin^2(x)\big)
\\&= - \log 2 - \log\big(\sin(x)\big)
\end{align*}

and the result follows. The rest is history.

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

Top   Post subject: Re: Logrithmic Integral Posted: Thu May 24, 2018 6:44 am

Joined: Thu Nov 12, 2015 5:26 pm
Posts: 102
Thanks kostakos and Riemann

Top   Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending

 All times are UTC [ DST ]

Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum

Search for:
 Jump to:  Select a forum ------------------ Algebra    Linear Algebra    Algebraic Structures    Homological Algebra Analysis    Real Analysis    Complex Analysis    Calculus    Multivariate Calculus    Functional Analysis    Measure and Integration Theory Geometry    Euclidean Geometry    Analytic Geometry    Projective Geometry, Solid Geometry    Differential Geometry Topology    General Topology    Algebraic Topology Category theory Algebraic Geometry Number theory Differential Equations    ODE    PDE Probability & Statistics Combinatorics General Mathematics Foundation Competitions Archives LaTeX    LaTeX & Mathjax    LaTeX code testings Meta 