$$ I:=\int_0^{\infty} \frac{\ln \cos^2 x}{x^2}dx=-\pi. $$ Using $2\cos^2 x=1+\cos 2x$ failed me because I ran into two divergent integrals after using $\ln(ab)=\ln a + \ln b$ since I obtained $\int_0^\infty x^{-2}dx$ and $\int_0^\infty (1+\cos^2 x)dx $ which both diverge. Perhaps we should try a complex analysis approach? I also tried writing $$ I(\alpha)=\int_0^\infty \frac{\ln \cos^2 \alpha \,x}{x^2}dx $$ and obtained $$ -\frac{dI(\alpha)}{d\alpha}=2\int_0^\infty \frac{\tan \alpha x}{x}dx=\int_{-\infty}^\infty\frac{\tan \alpha x}{x}dx. $$ Taking a second derivative $$ I''(\alpha)=\int_{-\infty}^\infty {\sec^2 (\alpha x)}\, dx $$ Random Variable pointed out how to continue from the integral after the 1st derivative, but is it possible to work with this integral $\sec^2 \alpha x$? Thanks

  • $\begingroup$ Where did you get the $-\pi$ from? $\endgroup$ – Jika Apr 30 '14 at 15:10
  • $\begingroup$ @Jika THat is the closed form result..... $\endgroup$ – Jeff Faraci Apr 30 '14 at 15:11
  • 3
    $\begingroup$ Extend the interval of integration to the entire real line and then consider the Cauchy principal value of the integral. Then since $\text{PV} \int_{-\infty}^{\infty} \frac{\tan ax}{x} \ dx= \pi \ (a >0)$, your second approach gives you the answer fairly quickly. $\endgroup$ – Random Variable Apr 30 '14 at 17:27
  • 1
    $\begingroup$ For what it's worth, I get the answer is $\pi(1-\sqrt{2})$. I'll check again my calculation, but not now. I have to sleep since it's already morning here. :) $\endgroup$ – Tunk-Fey Apr 30 '14 at 21:42
  • 1
    $\begingroup$ @Tunk-Fey Thank you. Goodnight $\endgroup$ – Jeff Faraci Apr 30 '14 at 21:45

Let the desired integral be denoted by $I$. Note that $$\eqalign{ 2I&=\int_{-\infty}^\infty\frac{\ln(\cos^2x)}{x^2}dx= \sum_{n=-\infty}^{+\infty}\left(\int_{n\pi}^{(n+1)\pi}\frac{\ln(\cos^2x)}{x^2}dx\right)\cr &=\sum_{n=-\infty}^{+\infty}\left(\int_{0}^{\pi}\frac{\ln(\cos^2x)}{(x+n\pi)^2}dx\right) \cr &=\int_{0}^{\pi}\left(\sum_{n=-\infty}^{+\infty} \frac{1}{(x+n\pi)^2}\right)\ln(\cos^2x)dx \cr &=\int_{0}^{\pi}\frac{\ln(\cos^2x)}{\sin^2x}dx \cr } $$ where the interchange of the signs of integration and summation is justified by the fact that the integrands are all negative, and we used the well-known expansion: $$ \sum_{n=-\infty}^{+\infty} \frac{1}{(x+n\pi)^2}=\frac{1}{\sin^2x}.\tag{1} $$ Now using the symmetry of the integrand arround the line $x=\pi/2$, we conclude that $$\eqalign{ I&=\int_{0}^{\pi/2}\frac{\ln(\cos^2x)}{\sin^2x}dx\cr &=\Big[-\cot(x)\ln(\cos^2x)\Big]_{0}^{\pi/2}+\int_0^{\pi/2}\cot(x)\frac{-2\cos x\sin x}{\cos^2x}dx\cr &=0-2\int_0^{\pi/2}dx=-\pi. } $$ and the announced conclusion follows.$\qquad\square$

Remark: Here is a proof of $(1)$ that does not use residue theorem. Consider $\alpha\in(0,1)$, and let $f_\alpha$ be the $2\pi$-periodic function that coincides with $x\mapsto e^{i\alpha x}$ on the interval $(-\pi,\pi)$. It is easy to check that the exponential Fourier coefficients of $f_\alpha$ are given by $$ C_n(f_\alpha)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f_\alpha(x)e^{-inx}dx=\sin(\alpha\pi)\frac{(-1)^n}{\alpha \pi-n\pi} $$ So, by Parseval's formula we have $$ \sum_{n\in\Bbb{Z}}\vert C_n(f_\alpha)\vert^2=\frac{1}{2\pi}\int_{-\pi}^\pi\vert f_\alpha(x)\vert^2dx $$ That is $$ \sin^2(\pi\alpha) \sum_{n\in\Bbb{Z}}\frac{1}{(\alpha\pi-n\pi)^2}=1 $$ and we get $(1)$ by setting $x=\alpha\pi\in(0,\pi)$.

  • 1
    $\begingroup$ Could you elaborate a bit more on that series expansion for $\dfrac1{\sin^2x}$ ? Are you aware of any ways to prove it which do not require residues or contour integrals? $\endgroup$ – Lucian Apr 30 '14 at 17:10
  • $\begingroup$ @Omran yes do you have proof of this series? Thank you for your answer, I would appreciate if you could add detail on this matter however. $\endgroup$ – Jeff Faraci Apr 30 '14 at 17:11
  • $\begingroup$ I will try, to elaborate. $\endgroup$ – Omran Kouba Apr 30 '14 at 17:35
  • $\begingroup$ @OmranKouba THank you for the update. IT is very clear now. Thanks +1 $\endgroup$ – Jeff Faraci Apr 30 '14 at 21:45

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.