Integral of $\sin\left( \frac{x}{a^x} \right)$

Question

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.

Answer 1

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 $$ \int \cos(y) W(y)\;dy = \int W(\arcsin z)\;dz . $$

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Of course, none of these integrals seems to have a "closed form" in terms of known functions.

Answer 2

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.

Answer 3

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.