4
$\begingroup$

$$\lim_{n\to \infty}\int_0^{\pi\over 2}\cos x\cos\left({x\over n } \right) \log(\log\csc (x)) \, dx$$

surprisingly give a result $-0.577156\ldots$, which went we check up it, it is Euler's constant value!!!.

Is this true or Wolfram alpha integration has a bug in its software?

I don't think using basic Integration by part or substitution will work to solve this problem. This is pass my knowledge. If this integral is correct, then it Just by luck we found on wolfram integral.

$\endgroup$
  • $\begingroup$ Here is the link wolframalpha.com/input/… $\endgroup$ – user549999 Apr 7 '18 at 21:40
  • $\begingroup$ If you replace $\cos(x/n)$ with $1$, Wolfram Alpha still gives the -0.577216, which is correct for all the first digits. So I’d try to get that integral (but WA find no closed form for that integral, either.) $\endgroup$ – Thomas Andrews Apr 7 '18 at 21:50
  • $\begingroup$ yeah you are right, well I didn't know that. $\endgroup$ – user549999 Apr 7 '18 at 21:53
6
$\begingroup$

Yes, that's true!

First, observe that you can replace the $\cos(\frac{x}{n})$ term with $1$ by the Monotone convergence theorem.

Then, do the substitution $t=\sin(x)$, so $dt=\cos(x)dx$ and $t$ takes values from 0 to 1. We get

$\int_0^1 \log(\log(\frac{1}{t}))dt = \int_0^1 \log(-\log(t))dt$

The last integral is indeed $\gamma$, as one can see in Wikipedia.

Alternatively, one can make another substitution to get $\int_0^{\infty} e^{-t}\log(t)dt$ and look at this question to see a proof the value is $\gamma$.

$\endgroup$

Your Answer

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