Integral of $\sin\left( \frac{x}{a^x} \right)$ How would I find the (definite) integral of a function looking something like this,
$\sin\left( \frac{x}{a^x} \right)$
...where $a$ is a constant. I thought it would have something to do with making the function fit the form of $\sin(A+B)$, which I can then easily integrate, but I am unable to do this.
 A: not a solution.  Just some observations.
I used the case $a=e$, so I wanted the integral
$$
A = \int \sin\left(\frac{x}{e^x}\right)\;dx
$$
Substitute $y=x/e^x$ to get
$$
A = \int\frac{\sin(y) W(y)}{y(1+W(y))}\;dy
$$
where $W$ is the Lambert $W$ function.  Integrate by parts to get
$$
A = \cos(y) W(y) - \int \cos(y)W(y)\;dy .
$$
For this integral we might substitute $z=\sin(y)$ to get
$$
\int \cos(y) W(y)\;dy = \int W(\arcsin z)\;dz .
$$

Of course, none of these integrals seems to have a "closed form" in terms of known functions.
A: Recall the Maclaurin series for the sine:
$$
\sin(y) = \sum_{n=0}^{\infty} \frac{y^{2n+1}}{(2n+1)!}
$$
Then, setting $y=xe^{-x}$:
$$
\sin(xe^{-x}) = \sum_{n=0}^{\infty} \frac{x^{2n+1} e^{-(2n+1)x}}{(2n+1)!}
$$
Letting $s = (2n+1)$:
$$
\sin(xe^{-x}) = \sum_{n=0}^{\infty} \frac{x^{s} e^{-sx}}{s!}
$$
Since, according to WolframAlpha:
$$
\int x^a exp(-a x) dx = -a^{-a-1} \, \Gamma (a + 1, a x) + C
$$
And noting the property for integer $k$ reference here:
$$
\Gamma(k,z) = (k-1)! e^{-z} \sum^{k-1}_{i=0} \frac{z^i}{k!}
$$
Then the previous integral yields:
$$
\int x^a e^{-a x} dx = -a^{-a-1} \, (a!) e^{-ax} \sum^{a}_{i=0} \frac{(ax)^i}{i!} + C
$$
And conveniently:
$$
\int \frac{x^a e^{-a x}}{a!} dx = -a^{-a-1} \, e^{-ax} \sum^{a}_{i=0} \frac{(ax)^i}{i!} + C
$$
Switching back $a=2n+1$
$$
\int \frac{x^{2n+1} e^{-(2n+1)x}}{(2n+1)!} dx = -(2n+1)^{-2n-2} \, e^{-(2n+1)x} \sum^{(2n+1)}_{i=0} \frac{((2n+1)x)^i}{i!} + C
$$
Adding the summation for $n$ varying between zero and infinity, we get:
$$
\int \sin(xe^{-x}) dx = \sum_{n=0}^{\infty}\left[ -(2n+1)^{-2n-2} \, e^{-(2n+1)x} \sum^{(2n+1)}_{i=0} \frac{((2n+1)x)^i}{i!} \right] +C
$$
The bad new is that, even if the integral I'm presenting is not wrong due to some mistake or typo, I have no promising ideas on how to simplify it. Good news is, if that you need is a numerical method, then this looks promising enough.
A: In the same spirit as @Mefitico, using
$$\sin(y) = \sum_{n=0}^{\infty} (-1)^n\frac{y^{2n+1}}{(2n+1)!}$$ which is valid for all $y$, let $y=x a^{-x}$ and you face integrals
$$I_n=\int \left(x a^{-x}\right)^{2 n+1}\,dx$$
Using $x=\frac{t}{\log (a)}$, then
$$I_n=\log ^{-2 (n+1)}(a)\int \left(e^{-t} t\right)^{2 n+1}\,dt$$ and
$$J_n=\int \left(e^{-t} t\right)^{2 n+1}\,dt=-t^{2(n+1)} E_{-(2 n+1)}((2 n+1) t)$$ where appear (as one could expect) the exponential integral function.
