How to prove that $\int\limits_0^{\pi} e^{\sin^2(x)}dx > {3\over2}\pi$?

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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

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