I am trying to solve the following by elementary methods: $$\int_0^{\pi} \frac{x}{x^2+\ln^2(2\sin x)}\,dx$$

I wrote the integral as: $$\Re\int_0^{\pi} \frac{dx}{x-i\ln(2\sin x)}$$ But I don't find this easier than the original integral. I have seen solutions which make use of complex analysis but I am interested in elementary approaches.

Any help is appreciated. Thanks!

  • $\begingroup$ Why the downvote? $\endgroup$ – Pranav Arora Jul 26 '14 at 12:04
  • $\begingroup$ I don't know, even using non elementary methods but the beautiful thing is that your last integral is a real ! +1 for this interesting question. Cheers. $\endgroup$ – Claude Leibovici Jul 26 '14 at 12:05
  • $\begingroup$ Beats me why there's a downvote, but I upvoted because I've been wondering this too and I want a darn answer. Is the integral really exactly $2$?? $\endgroup$ – David H Jul 26 '14 at 12:09
  • $\begingroup$ Here math.stackexchange.com/questions/625456/… $\endgroup$ – Shine Jul 26 '14 at 14:15

Here is an approach without using contour integration (Cauchy's theorem).

I've found the following result.

Theorem. Let $a$ be any real number.

Then $$ \begin{align} \displaystyle \int_{0}^{\pi} \frac{x}{x^2+\ln^2(2 e^{a} \sin x)}\mathrm{d}x & = \, \frac{2 \pi^2}{\pi^2+4a^2},\tag1\\\\ \int_{0}^{\pi} \frac{\ln(2 e^{a} \sin x)}{x^2+\ln^2(2 e^{a} \sin x)}\mathrm{d}x & = \frac{4\pi a}{\pi^2+4a^2}.\tag2 \end{align} $$

from which, by putting $a=0$, you deduce the evaluation of the desired integral:

Example $1 (a)$. $$ \begin{align} \int_{0}^{\pi} \frac{x}{x^2+\ln^2(2 \sin x)}\:\mathrm{d}x & = 2 \tag3 \end{align} $$


Example $1 (b)$. $$ \begin{align} \int_{0}^{\pi} \frac{\ln(2\sin x)}{x^2+\ln^2(2\sin x)}\mathrm{d}x& = 0. \\ \tag4 \end{align} $$

Proof of the Theorem. We consider throughout the principal value of $\log(z)$ defined for $z\neq 0$ by $$ \log(z)= \ln |z| + i \arg z, \, - \pi < \arg z \leq \pi, $$ where $\ln$ is the base-$e$ logarithm $\ln e = 1$.

Let $a$ be any real number. Since $\displaystyle z \mapsto i \pi/2+a+\log(1+z) $ is analytic on $|z| < 1$ non-vanishing at the point $z=0$, one may obtain the following power series expansion $$ \begin{align} \displaystyle \frac{1}{i \pi/2+a+\log \left(1+z\right)} - \frac{1}{i \pi/2+a} = \sum_{n=1}^{\infty} \displaystyle \frac{a_n(\alpha)}{n!} z^n,\tag5 \end{align} $$ with $$ \displaystyle a_1(\alpha) = -\alpha^2, \quad a_2(\alpha) = \alpha^3+\alpha^2/2, \, ... , $$ $\displaystyle {a}_n(\cdot)$ being a polynomial of degree $n+1$ in $\displaystyle \alpha$, where we have set $\displaystyle \alpha:=1/(i \pi/2+a)$. Plugging $z=-e^{2ix}$ in $(3)$, for $0<x<\pi$, noticing $$ i x +\ln(2e^{a} \sin x) = \log\left(ie^{a}\left(1-e^{2ix} \right) \right), $$ separating real and imaginary parts, gives the Fourier series expansions: $$ \begin{align} \displaystyle \frac{x}{x^2 +\ln^2(2e^{a}\sin x)} & = \frac{2\pi}{\pi^2+4a^2}+ \sum_{n=1}^{\infty}(-1)^n \left( \frac{a_n^{-}(\alpha)}{n!} \cos (2nx)+\frac{a_n^{+}(\alpha)}{n!} \sin (2nx)\right) \\ \frac{\ln(2e^{a}\sin x)}{x^2 +\ln^2(2e^{a}\sin x)} & = \frac{4a}{\pi^2+4a^2}+ \sum_{n=1}^{\infty}(-1)^n \left( \frac{a_n^{+}(\alpha)}{n!} \cos (2nx)-\frac{a_n^{-}(\alpha)}{n!} \sin (2nx)\right) \end{align} $$ with $\displaystyle a_n^{+}(\alpha):= {\mathfrak{R}}a_n(\alpha)$ and $\displaystyle a_n^{-}(\alpha):= {\mathfrak{I}}a_n(\alpha)$. Now, a termwise integration with respect to $x$ from 0 to $\pi$, justified by the convergence of the series $\sum_{n=1}^{\infty} \displaystyle \frac{a_n(\alpha)}{n!} $, leads to the Theorem due to $$ \begin{align} \displaystyle \int_{0}^{\pi} \cos(2nx)\:\mathrm{d}x = \int_{0}^{\pi} \sin(2nx)\:\mathrm{d}x = 0. \end{align} $$

One may observe that, by uniqueness of the Fourier coefficients, you have the following closed forms.

Example $2 (a)$. $$ \begin{align} \int_{0}^{\pi} \frac{x}{x^2+\ln^2(2 \sin x)}\cos^{2} x \:\mathrm{d}x & = 1, \tag6 \\ \int_{0}^{\pi} \frac{x}{x^2+\ln^2(2 \sin x)}\sin^{2} x\:\mathrm{d}x & = 1. \tag7 \end{align} $$

$$ $$

Example $2 (b)$. $$ \begin{align} \int_{0}^{\pi} \frac{\ln(2 \sin x)}{x^2+\ln^2(2 \sin x)} \cos^{2} x \:\mathrm{d}x & = -\frac{1}{\pi}, \tag8 \\ \int_{0}^{\pi} \frac{\ln(2 \sin x)}{x^2+\ln^2(2 \sin x)}\sin^{2} x\:\mathrm{d}x & = \frac{1}{\pi}. \tag9 \end{align} $$

A general result does exist.

  • $\begingroup$ Thank you Oliver Oloa for the response but I cannot understand most of it. :( What does the statement "Since $z↦\log(1+z)$ is analytic on $|z|<1$, one may obtain the following power series expansion" mean? And then, why did you introduce the subsequent series expansion? Can you please share some motivation behind introducing $e^a$? Sorry for too many dumb questions. Thank you again. $\endgroup$ – Pranav Arora Jul 26 '14 at 12:42
  • 1
    $\begingroup$ @PranavArora. There is no dumb question ! I am fascinated by this answer but I am almost sure I just understand a few percents of it. Cheers :) $\endgroup$ – Claude Leibovici Jul 26 '14 at 12:50
  • $\begingroup$ @Oliver Oloa. Do you have many things like that ? By the way, where are you in Paris ? Cheers from Pau :) $\endgroup$ – Claude Leibovici Jul 26 '14 at 12:52
  • $\begingroup$ @Pranav Arora When an analytic function $h$ say near $z_0=0$ is such that $h(0) \neq 0$, then the reciprocal $z \mapsto 1/h(z)$ is also analytic near $0$. We are in this case here. Thank you. $\endgroup$ – Olivier Oloa Jul 26 '14 at 12:58
  • $\begingroup$ I do not know what an analytic function is. I think I will have to learn about that before trying to understanding your answer. Thank you for your time, I accept your answer. :) $\endgroup$ – Pranav Arora Jul 26 '14 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.