# Prove integral inequality: $\int_{0}^{\frac{\pi}{2}}e^{\sin x}\,dx\geq\frac{\pi}{2}(e-1)$

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?

• 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. Jul 15 '20 at 1:48
• 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 Jul 15 '20 at 1:55
• $sinx$ 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 Jul 15 '20 at 2:03
• I tried making sinx to equal to -1, it didn't work Jul 15 '20 at 2:14

## 2 Answers

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

• 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. Jul 15 '20 at 18:09
• @NicholasRoberts You are right, that was a typo. Now I fix it. Thank you. Jul 15 '20 at 22:12

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)

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