$ \int_{0}^{\infty} e^{-\sin x}~dx $ How can I integrate this ?
$$ \int_{0}^{\infty} e^{-\sin{x}}~dx $$
I have tried using dominating convergence theorem to evaluate this to $0$ . Is there any other way to evaluate this integral ?
Thank you. 
 A: This integral doesn't converge. Since the integrand is periodic with period $2\pi$, we have that
$$\int_0^{2\pi n}e^{-\sin(x)} dx = n\int_0^{2\pi}e^{-\sin(x)}dx := n\cdot I$$
where $n$ is a natural number. Since the function $e^{-\sin(x)}$ is positive everywhere, $I$ is positive, and so as $n\to\infty$ the above goes to infinity, and so the integral does not converge. 
A: One may observe that
$$
e^{-\sin{x}} \ge e^{-1},\qquad x \in \mathbb{R},
$$ yielding, by comparison, the divergence of the given integral:
$$
\int_0^M e^{-\sin{x}}\:dx \ge \int_0^M e^{-1}\:dx.
$$
A: Note that both sine and its exponential are periodic and since exponential is non-negative everywhere, the integral diverges.
A: Hint : $$\forall x\in[0,\infty),|e^{-\sin{x}}| \geq|e^{-1}|$$
A: Because $e^{-\sin(x)} \geq e^{-1}$, one has
$$
\liminf_{a \to +\infty} \int_0^a e^{-\sin(x)} dx\geq \liminf_{a \to +\infty} \int_0^a e^{-1}dx = \liminf_{a \to +\infty} ae^{-1} = +\infty
$$
whereby
$$
\int_0^{+\infty}e^{-\sin(x)}dx = \infty .
$$
