Is the integral $\int_0^{\infty} \sin(e^x) \, dx$ convergent? Can anyone help me with this problem? I've tried substitution method but I do not know how to continue. Let $u = e^x$ and $du = u\,dx$, so $dx = du/u$. So we have, $$ \int \sin(u) \frac{1}{u} du = \int \frac{\sin(u)}{u} du.$$
 A: Integrate by parts, using $u=e^{-x}$ and $dv=e^x \sin(e^x)\,dx$.
Then $du=-e^{-x}\,dx$ and we can take $v$ to be $-\cos(e^x)$. The rest is easy, the integral we end up with converges.
Added: The integral from $0$ to $B$ is
$$\left. -e^{-x}\cos(e^x)\right|_0^B -\int_0^B e^{-x}\cos(e^x)\,dx.$$
There is no problem as $B\to\infty$, since $|\cos(e^x)|$  is bounded.
A: We have
$$\int_0^{\infty} \sin(e^x) \, dx = \int_1^{\infty} \sin(t) \dfrac{dt}t = \int_0^{\infty} \sin(t) \dfrac{dt}t - \int_0^1 \sin(t) \dfrac{dt}t = \dfrac{\pi}2 - \text{Si}(1)$$
where the last integral can be obtained from here. Also the integral $ \displaystyle \int_0^1 \dfrac{\sin(t)}t dt$ is clearly bounded since $\sin(t) \in \left( \dfrac2{\pi}t, t\right)$ for $t \in (0,\pi/2)$.
A: Here's an alternative (and elementary) way to solve the problem. Consider 
$$
f(x) = \int_x^{x+1} \sin e^t \,dt.
$$
Making the substitution $v = e^t,$ and integrating by parts, one obtains
$$
e^{x} f(x) = \cos e^x - e^{-1}\cos e^{x+1} + r(x),
$$
where $|r(x)|< e^{-x}.$ 
Accordingly, the integral in question, is equivalent to calculating $f(0) + f(1) + \cdots, $ which is now a telescoping sum, with remainder term going to $0$ as $x \to \infty.$ 
Remark: This is essentially Andre Nicolas's solution written slightly differently. 
