How to prove that $\int\limits_0^{\pi} e^{\sin^2(x)}\ dx > {3 \over 2}\pi$?


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  • $\begingroup$ Could you please edit the question, the math notations are not showing. $\endgroup$ – John_Wick Nov 29 '18 at 21:53
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    $\begingroup$ I don't understand why anyone would down-vote this question. It's perfect fine and compiles with the requirements of the page. $\endgroup$ – user150203 Nov 30 '18 at 3:48
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    $\begingroup$ @DavidG Hover your cursor over the voting arrows... "This question (shows / does not show any) research effort". $\endgroup$ – epimorphic Dec 4 '18 at 17:52
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    $\begingroup$ @DavidG Go to How to ask a good question and report back to us in which ways this question satisfies any of the criteria. This is a problem statement question, essentially expecting answerers to do this users work for them. Hence the downvote, hence the close votes. This site is not a "do my work/proof for me" service. $\endgroup$ – Namaste Dec 4 '18 at 22:42
  • $\begingroup$ @amWhy - will do. Thanks for the link. $\endgroup$ – user150203 Dec 4 '18 at 22:44

Use the inequality $e^t\ge 1+t$, valid for all $t$, to get: $$ \int_0^\pi e^{\sin^2x}dx\ge\int_0^\pi(1+\sin^2x)dx=\int_0^\pi\left(\frac32 + \frac12\cos 2x\right)\,dx $$ You should be able to take it from here. The inequality is in fact strict, because the difference $e^{\sin^2x}-(1+\sin^2x)$ is continuous, non-negative and not identically zero.

  • $\begingroup$ I've taken and it equals to $int_0^\pi(1+\sin^2x)dx = {3\over2}\pi$ But is it really right to use $e^t\ge 1+t$? $\endgroup$ – TBox Nov 29 '18 at 22:20
  • $\begingroup$ $$e^t > 1+t$$ is true for all nonzero real numbers, hence it is not necessary to use the second term of Taylor expansion. $\endgroup$ – Crostul Nov 29 '18 at 22:22
  • $\begingroup$ got it, thank you! $\endgroup$ – TBox Nov 29 '18 at 22:28
  • $\begingroup$ @ArseniyBakaev The inequality $e^t\ge 1+t$ is easily proven: See math.stackexchange.com/q/1330815/215011 $\endgroup$ – grand_chat Nov 29 '18 at 22:28
  • $\begingroup$ @Crostul Right, and the justification that $\int_0^\pi \sin^4 (x) dx>0$ will equally apply to $\int_0^\pi e^{\sin^2x}dx > \int_0^\pi (1+\sin^2x)dx$. $\endgroup$ – grand_chat Nov 29 '18 at 22:55

We can also use Taylor expansion: $e^{\sin^2(x)}=1+\sin^2(x)+\sin^4(x)/2+...$ and integrate it from $0$ to $\pi$

$\int_0^\pi(1+\sin^2(x)+(\sin^4(x))/2)dx=(27\pi)/16 ≈ 5.3014 > 3\pi/2$

  • $\begingroup$ Why post this, which copies the idea of a previous answer, only more clumsily? $\endgroup$ – Did Dec 5 '18 at 0:06

The integral is a Bessel function:

$$\sqrt{e} \pi I_0\left(\frac{1}{2}\right)$$

which has numerical value $5.50843$, which is indeed less than $3 \pi/2$. This, it would seem, answers the question completely in a principled, and correct manner. One might try any number of other techniques, but I cannot see how any could be better than solving the integral exactly.

  • $\begingroup$ Is it possible to compare without Bessel function? $\endgroup$ – TBox Nov 29 '18 at 21:57
  • $\begingroup$ The only other way to compare would be a numerical integration. $\endgroup$ – David G. Stork Nov 29 '18 at 21:58
  • $\begingroup$ @DavidG.Stork Bessel function does the trick, but in order to just find a soft result, you can go by the ML inequality as elaborated below. $\endgroup$ – Rebellos Nov 29 '18 at 22:03
  • $\begingroup$ "The only other way to compare would be a numerical integration" Not true. $\endgroup$ – Did Dec 5 '18 at 0:05

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