I recently learned $\int^{\infty}_{-\infty}\sin(e^x)dx$ can be solved like this:





It inspired me to come up with this integral which wolfram alpha approximates to $-0.98771815$.

I employed a similar strategy on my integral:


I don't know how to solve this so I tried a different substitution:


I feel like I am close here but I don't know how to progress.

  • 3
    $\begingroup$ What makes you think there is an explicit antiderivative? $\endgroup$ Feb 13 at 23:44
  • $\begingroup$ @TedShifrin Nothing, I was just curious about this integral $\endgroup$ Feb 14 at 0:00
  • 1
    $\begingroup$ This is equivalent to finding $\frac i 2\int_0^\infty x^{-i x-1}-x^{i x-1}dx$ $\endgroup$ Feb 14 at 1:18
  • $\begingroup$ Additionally, $\int_0^1\frac{\sin(x\ln(x))}xdx=\sum\limits_{n=0}^\infty (-1)^{n+1}(2n+1)^{-2 (n+1)}$ and the integral over other bounds requires summing over special functions. How do you know the improper integral converges? $\endgroup$ Feb 14 at 1:51
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    $\begingroup$ For $x > 0$, change variable to $y = xe^x$. Since $\frac{dx}{dy}$ is positive, decreases to $0$ as $x \to \infty$, $\int_0^\infty \sin(y) \frac{dx}{dy} dy$ converges by Dirichlet's test. $\endgroup$ Feb 16 at 19:34

3 Answers 3


Here is the graph of $\sin(xe^x)$
Of course the integral $\int_{-\infty}^0$ converves. But how do we know that $\int_0^\infty$ converges?


$$\int_{-\infty}^\infty\sin(xe^x)dx=\underbrace{\int_{-\infty}^0\sin(xe^x)dx}_A+ \underbrace{\int_0^\infty\sin(xe^x)dx}_B=\underbrace{\frac i2\int_0^1x^{-i x-1}-x^{ix-1}}_Adx+ \underbrace{\frac i2\int_1^\infty x^{-i x-1}-x^{ix-1}}_B$$

Integral $A$:

$A$ is reminiscent of the sophomore’s dream integrals. We expand using sine’s Maclaurin series: $$\int_{-\infty}^0\sin(xe^x)dx=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\int_{-\infty}^0 x^{2n+1} e^{(2n+1)x}dx$$

Finally substituting $(2n+1)x=-t$ and using the factorial’s integral representation gives:


Integral $B$:

$B$ is like the “natural” sophomore’s dream $\int_0^\infty x^{-x}dx$. Taking a limit and applying the same procedure, we get the lower $P(a,x)$ regularized gamma functions:

$$B=\int_0^\infty\sin(xe^x)=\lim_{b\to\infty}\int_0^b\sin(xe^x)dx=\lim_{b\to\infty}\sum_{n=1}^\infty\frac{(-1)^n P(2n,(1-2n)b)}{(2n-1)^{2n}}$$

Truncating the sum at $n=N$, the result numerically matches if $b\ll N$ as shown here. For larger $N$, there looks to be computation errors, but you can still find expected results like here, after clicking “More digits”.

  • $\begingroup$ Is there any other interesting form for the integral, which is about $-0.528$? $\endgroup$ Feb 22 at 13:51

This answer is only to prove that the integral converges.

\begin{align} \int_0^{A} \sin \left(x e^x\right) \mathrm d x &= \int_1^{e^A} \frac{\sin(u\ln u)}{u}\mathrm du\\ &= \int_1^{e^A} \frac1{u\left(1+\ln u\right)} (1+\ln u) \sin(u\ln u)\mathrm du\\ &= \left[-\frac{1}{u(1+\ln u)} \cos(u\ln u)\right]_1^{e^A} - \int_1^{e^A} \frac{2 + \ln u}{u^2 (1+\ln u)^2} \cos(u\ln u)\mathrm d u\\ &= -\frac1{e^A(1+A)}\cos\left(Ae^{A}\right) + 1 - \int_{1}^{e^A} \frac{\cos(u\ln u)}{u^2(1+\ln u)}\mathrm d u - \int_{1}^{e^A} \frac{\cos(u\ln u)}{u^2(1+\ln u)^2}\mathrm d u. \end{align}

It is clear that $\lim\limits_{A\to \infty} \frac1{e^A(1+A)}\cos\left(Ae^{A}\right) = 0$ since $$\left|\frac{\cos(u\ln u)}{u^2 (1 + \ln(u))}\right| \le \frac1{u^2},$$ $$\left|\frac{\cos(u\ln u)}{u^2 (1 + \ln(u))^2}\right| \le \frac1{u^2}$$ and $\int_1^{\infty} \frac1{u^2}\mathrm du$ converges, then the integral converges.

  • 1
    $\begingroup$ Why the downvote? $\endgroup$
    – Kroki
    Feb 16 at 16:11
  • $\begingroup$ Thanks for the help! $\endgroup$ Feb 16 at 21:25
  • 2
    $\begingroup$ This proves only the convergence. Are you looking for the exact value ? $\endgroup$
    – Kroki
    Feb 16 at 22:55
  • 1
    $\begingroup$ upvoted ! I see no reasons to downvote ! $\endgroup$
    – mick
    Feb 17 at 12:21
  • $\begingroup$ Me too and I don't know why it was downvoted $\endgroup$
    – Kroki
    Feb 17 at 15:48

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