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$$ I:=\int_0^{\infty} \frac{\ln \cos^2 x}{x^2}dx=-\pi. $$ 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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @Tunk-Fey Thank you. Goodnight $\endgroup$ – Jeff Faraci Apr 30 '14 at 21:45
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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 $x=\alpha\pi\in(0,\pi)$.

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    $\begingroup$ 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

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