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, $$ \int \sin(u) \frac{1}{u} du = \int \frac{\sin(u)}{u} du.$$
[Math] Is the integral $\int_0^{\infty} \sin(e^x) \, dx$ convergent
calculus
Best Answer
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.