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What it says in the title. If $$I=\int_{2/\pi}^{+\infty }\ln\left({\cos\left({\frac{1}{x} }\right) }\right) \, \mathrm dx,$$ how could I proceed in order to find the value of $I$?

Healg
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  • Do you know there is a closed form? – Simon S Jun 01 '15 at 17:42
  • I am pretty sure there is. – Healg Jun 01 '15 at 17:45
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    First step is to put $u=1/x$, I'd say. – mickep Jun 01 '15 at 17:46
  • Is the question just "does it converge"? or do you actually need to find the value? – TravisJ Jun 01 '15 at 17:50
  • Indeed, then I have $I=\int_{0}^{\frac{\pi}{2} }\frac{\ln(\cos(u))}{u^{2}}du$ But what can I do with it? – Healg Jun 01 '15 at 17:52
  • I need the value. – Healg Jun 01 '15 at 17:52
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    wolframalpha times out computing the exact value, but if you replace $\infty$ with 100000000 it seems to have converged to about $-1.03294$. So it looks like it probably actually converges... considering that wolframalpha fails to find a closed form, I suspect it will be very hard to do so. – TravisJ Jun 01 '15 at 18:01
  • If you substitute $y = 1/x$ and then $z = \cos y$ then you should get the integral $\int_0^1 \frac{\ln z dz}{(\cos^{-1} z)^2 \sqrt{1-z^2}}.$ Wolfram Alpha numerically integrates this to -1.03294 but doesn't find a closed form solution. I can't tell you if there is one or not, but it looks like it converges. – Bridgeburners Jun 01 '15 at 18:03
  • @Travis: Wolfram doesn't find a close form of $\int_0^{\pi/2}\ln(\cos(x))dx$ and yet it's very easy to compute this integral and find $\int_0^{\pi/2}\ln(\cos(x))dx=-\frac{\pi\ln(2)}{2}$. Therefore your argument is not valide ! :-) – Surb Jun 01 '15 at 18:05
  • @Surb, my argument is more of a heuristic... plus, I don't think I see the easy way to compute the integral you describe (not immediately at least). – TravisJ Jun 01 '15 at 18:07
  • The little astuce of the integral I proposed is simply to show that $\int_0^{\pi/2}\ln(\cos(t))dt=\int_0^{\pi/2}\ln(\sin(t))dt$ and thus that $$\int_0^{\pi/2}\ln(\cos(t))dt=\frac{1}{2}\left(\int_0^{\pi/2} \ln(\cos(t))dt+ \int_0^{\pi/2}\ln(\sin(t))dt\right)=...$$ It's of course not obvious, but when you know the trick it's not complicate :-) What I wanted to say is that wolfram is not enough to justify that there is not any close form. I know that it's not what you wanted to say directly, but it looks like you thought about that :-) Good evening, – Surb Jun 01 '15 at 18:17
  • @Surb The integral of interest is not $\int_0^{\pi/2}\log \cos x dx$. Rather, it is $\int_0^{\pi/2} \frac{\log \cos x}{x^2} dx$ – Mark Viola Jun 01 '15 at 19:48
  • @alkabary Please avoid edits worsening the readability of the question: using \frac{2}{\pi} instead of 2/\pi, and \large, were two bad ideas. – Did Jun 01 '15 at 19:58

1 Answers1

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Using a change of variable and a well-known fact: $$I=\int_{0}^{\pi/2}\frac{\log\cos x}{x^2}\,dx=2\sum_{k\geq 1}\frac{(-1)^k}{k}\int_{0}^{\pi/2}\left(\frac{\sin(kx)}{x}\right)^2\,dx\tag{1}$$ but since: $$ \cos x = \prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right)\tag{2}$$ we have: $$ \log \cos x = -\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m(2n+1)^{2m}\pi^{2m}}=-\sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)}{m\,\pi^{2m}}x^{2m}\tag{3}$$ so the RHS of $(1)$ also equals: $$ -\frac{2}{\pi} \sum_{m\geq 1}\frac{(1-4^{-m})\zeta(2m)}{ m(2m-1)}\tag{4}$$ that is a fast convergent series.

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Jack D'Aurizio
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