Is there a closed form of this integral $ \int_0^\infty \sin(xe^{-x})dx\, $? I have tried by subsititution method and it got more  complicate than before. 
Can anyone help me to evaluate this integral.

$$
\int_0^\infty \sin(xe^{-x})dx\,.
$$

 A: I don't know if there exists a closed form in terms of some special known functions. One may recall that the following standard power series
$$
\sin u = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} u^{2n+1}, \quad u \in \mathbb{R},\tag1
$$ has an infinite radius of convergence allowing us to obtain
$$
\int_0^\infty \sin(xe^{-x})\:dx=\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \int_0^\infty x^{2n+1}e^{-(2n+1)x}dx \tag2
$$ the latter integral may be evaluated by using the classic integral representation of the Euler $\Gamma$ function,
$$
\int_0^\infty x^{2n+1}e^{-(2n+1)x}dx=\frac{(2n)!}{(2n+1)^{2n+1}}, \tag3
$$ we then obtain

$$
\int_0^\infty \sin(xe^{-x})\:dx=\sum^{\infty}_{n=0} \frac{(-1)^n\quad}{(2n+1)^{2n+2}}.\tag4
$$ 

The convergence of the preceding series is very fast. One gets, using Mathematica,
NSum[(-1)^n/(2 n + 1)^(2 n + 2), {n, 0, \[Infinity]},  WorkingPrecision -> 50]

0.98771814780760758564289940276054141639545399170883

One gets that $\displaystyle \int_0^\infty \sin(xe^{-x})\:dx>0$, and that the numerical value is not far from $1$ as could be guessed by approximating the initial integral with $\displaystyle \int_0^\infty xe^{-x}\:dx=1.$
