# 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, $$\int \sin(u) \frac{1}{u} du = \int \frac{\sin(u)}{u} du.$$

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This may help in evaluating the second integral: math.stackexchange.com/questions/103030/… –  tylerc0816 Dec 3 '13 at 1:07

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 $t \in (0,\pi/2)$.

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Is the lower limit of integration $1$ in the second integral? –  tylerc0816 Dec 3 '13 at 1:11
@tylerc0816 Yes, thanks. Corrected. –  user17762 Dec 3 '13 at 1:13
Another question dealing with the computation of $\displaystyle\int_0^\infty\frac{\sin x}x\,dx$ is here. I added a real variable argument as an answer. –  Andres Caicedo Dec 3 '13 at 2:17

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.

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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.

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