4
$\begingroup$

I am trying to prove $$\int_{0}^{\frac{\pi}{2}}e^{\sin x}\,dx\geq\frac{\pi}{2}(e-1)$$

I found the Taylor series of $e^{\sin x}$ then approximated $\sin x$ as $\frac{2}{\pi}x$. I have no idea what to do next; any suggestions or different method?

$\endgroup$
4
  • $\begingroup$ Have you looked at the integral as area under the graph? The RHS looks like the area of a rectangle which one might be able to compare with that. $\endgroup$ – Torsten Schoeneberg Jul 15 '20 at 1:48
  • $\begingroup$ I tried to think about that, but I found that the rectangle will have more area in some parts and less in another, I don't have an indea on how should I compare them $\endgroup$ – rashed a564 Jul 15 '20 at 1:55
  • $\begingroup$ $sinx$ is always larger, or same to -1, so $e^{sinx}$ must be larger (or same) than $e^{-1}$. If we integrate here, and change the RHS appropriately, I think the inequality is solved $\endgroup$ – Joshua Woo Jul 15 '20 at 2:03
  • $\begingroup$ I tried making sinx to equal to -1, it didn't work $\endgroup$ – rashed a564 Jul 15 '20 at 2:14
10
$\begingroup$

Note that $\sin x \geq \frac 2 \pi x$ holds for $x \in [0, \frac \pi 2]$. Thus $$\int_{0}^{\pi/2}e^{\sin x}dx \geq \int_{0}^{\pi/2}e^{\frac 2 \pi x}dx =\frac \pi 2 (e-1)$$

$\endgroup$
2
  • $\begingroup$ I'm just a little confused. The estimate that $\sin x \geq \frac 2 \pi x$ holds for $x \in [0, \frac2 \pi]$. But our domain of integration is $[0, \frac{\pi}{2}]$ And $[0, \frac2 \pi]$ is a proper subset of $[0, \frac{\pi}{2}]$. So how can you apply the estimate on the domain of integration when the estimate is not true on the whole domain of integration? I apologize if I'm just not seeing this correctly. $\endgroup$ – Nicholas Roberts Jul 15 '20 at 18:09
  • $\begingroup$ @NicholasRoberts You are right, that was a typo. Now I fix it. Thank you. $\endgroup$ – Luxerhia Jul 15 '20 at 22:12
3
$\begingroup$

Since @sera already provided a good and simple answer, this is a loog comment.

Your idea of using Taylor series was good but it should have been $$e^{\sin(x)}=\sum_{n=0}^\infty \frac {\sin^n(x)} {n!}\implies \int_0^{\frac \pi 2} e^{\sin(x)}\, dx=\sum_{n=0}^\infty \frac {1} {n!}\int_0^{\frac \pi 2}\sin^n(x)\,dx$$ Since $$\int_0^{\frac \pi 2}\sin^n(x)\,dx=\frac{\sqrt{\pi }}2 \frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma\left(\frac{n}{2}+1\right)}$$ So, consider $$S_p=\frac{\sqrt{\pi }}2\sum_{n=0}^p \frac{ \Gamma \left(\frac{n+1}{2}\right)}{n! _, \Gamma\left(\frac{n}{2}+1\right)}$$ and remember that $\frac \pi 2 (e-1) <e $

the partial sums (which are increasing) generate the sequence $$\left\{\frac{\pi }{2},1+\frac{\pi }{2},1+\frac{5 \pi }{8}, \frac{10}{9}+\frac{5 \pi}{8},\frac{10}{9}+\frac{81 \pi }{128}, \frac{251}{225}+\frac{81 \pi}{128},\frac{251}{225}+\frac{2917 \pi }{4608}\right\} $$ and the third term is already larger than the rhs

$\endgroup$

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.