How prove this $\int_{0}^{2\pi}e^{\sin{x}}dx<2\pi e^{\frac{1}{4}}$

Question

Show that this intergral inequality $$\int_{0}^{2\pi}e^{\sin{x}}dx<2\pi e^{\frac{1}{4}}$$

I know this use Taylor's formula.But I think is very ugly,maybe this problem have simple methods.Thank you

Answer 1

We have $$e^{\sin(x)} = \sum_{k=0}^{\infty} \dfrac{\sin^k(x)}{k!}$$ Hence, $$\int_0^{2 \pi} e^{\sin(x)} dx = \sum_{k=0}^{\infty} \dfrac1{(2k)!} \int_0^{2\pi} \sin^{2k}(x)dx = \underbrace{\sum_{k=0}^{\infty} \dfrac1{k!k!} \cdot \dfrac{2\pi}{2^{2k}} < 2\pi\left(\sum_{k=0}^{\infty} \dfrac1{k!} \cdot \dfrac1{4^k} \right)}_{\text{Since $k! \geq 1$}} = 2\pi e^{1/4}$$ where $\displaystyle \int_0^{\pi/2} \sin^{2k}(t) dt = \dfrac{\pi}{2^{2k+1}} \dbinom{2k}k$.

Answer 2

Use that $\cosh(x) \leq (\cosh(1)-1) x^2 + 1$ for all $x\in[-1,1]$ to obtain the sharper estimate $$ \int_0^{2\pi}e^{\sin(x)}dx=2 \int_{-\pi/2}^{\pi/2}\cosh(\sin(x))dx\leq \pi\left(1+\cosh(1)\right).$$ Now the question remains to give a simple proof of the inequality $$1+\cosh(1) < 2e^{1/4}.$$