I would like to evaluate the following integral
$\int_{-\infty}^{\infty}(\sin(ax)-1)^ne^{-\frac{x^2}{4b}}dx$.
Is there an analytical expression for this integral (maybe in terms of special functions)? I suppose that there probably is no general expression for arbitrary $n$. Are there any approximations to be made, that are valid for large $a$, $b$ and $n$? I already tried expanding $(\sin(ax)-1)^n$ as a Taylor series around $0$ up to second order but as expected this is not a good approximation.
Best Answer
The integral can be expressed as a summation of an analytical series. I hope it come out handy enough to work with.
First, $$ \int_{-\infty}^{\infty}(\sin(ax)-1)^ne^{-\frac{x^2}{4b}}dx {= 2\sqrt{b}\int_{-\infty}^{\infty}(\sin(2a\sqrt{b}x)-1)^ne^{-x^2}dx \\= 2\sqrt{b}\int_{-\infty}^{\infty}\left[\frac{e^{i2a\sqrt{b}x}-e^{-i2a\sqrt{b}x}-2i}{2i}\right]^ne^{-x^2}dx \\= \frac{2\sqrt{b}}{(2i)^n}\int_{-\infty}^{\infty}\left[{e^{ia\sqrt{b}x}-ie^{-ia\sqrt{b}x}}\right]^{2n}e^{-x^2}dx \\= \frac{2\sqrt{b}}{(2i)^n}\int_{-\infty}^{\infty}\sum_{k=0}^{2n}\binom{2n}{k}{e^{i2a\sqrt{b}x(k-n)}}(-i)^{2n-k}e^{-x^2}dx \\= \frac{2\sqrt{b}}{(2i)^n}\sum_{k=0}^{2n}\binom{2n}{k}(-i)^{2n-k}\int_{-\infty}^{\infty}{e^{i2a\sqrt{b}x(k-n)}}e^{-x^2}dx }. $$ Second, by using the Fourier transform, we achieve $$ \sqrt{\pi}\exp(-k^2)=\int_{-\infty}^\infty \exp(-x^2)\exp(-i2kx)dx. $$ Finally, $$ \int_{-\infty}^{\infty}(\sin(ax)-1)^ne^{-\frac{x^2}{4b}}dx{= 2\sqrt{b\pi}\left(\frac{i}{2}\right)^n\sum_{k=0}^{2n}\binom{2n}{k}i^{k} e^{-a^2b(k-n)^2} } . $$
Update
A simpler series is as follows: $$ \int_{-\infty}^{\infty}(\sin(ax)-1)^ne^{-\frac{x^2}{4b}}dx{= 2\sqrt{b\pi}\left(\frac{i}{2}\right)^n\sum_{k=0}^{2n}\binom{2n}{k}i^{k} e^{-a^2b(k-n)^2} \\= 2\sqrt{b\pi}\left(\frac{i}{2}\right)^n\Big[ \binom{2n}{n}i^{n} + \sum_{k=0}^{n-1}\Bigg\{\binom{2n}{k}i^{k} e^{-a^2b(k-n)^2} \\+ \binom{2n}{2n-k}i^{2n-k} e^{-a^2b(2n-k-n)^2}\Bigg\}\Big] \\= 2\sqrt{b\pi}\left(\frac{i}{2}\right)^n\Big[ \binom{2n}{n}i^{n} + \sum_{k=0}^{n-1}\binom{2n}{k}\left(i^{k}+i^{2n-k}\right) e^{-a^2b(k-n)^2} \Big] \\= \binom{2n}{n}2\sqrt{b\pi}\left(-\frac{1}{2}\right)^n + 2\sqrt{b\pi}\left(\frac{i}{2}\right)^n \sum_{k=0}^{n-1}\binom{2n}{k}\left(i^{k}+i^{2n-k}\right) e^{-a^2b(k-n)^2} \\= \binom{2n}{n}2\sqrt{b\pi}\left(-\frac{1}{2}\right)^n + 2\sqrt{b\pi}\left(-\frac{1}{2}\right)^n \sum_{k=1}^{n}\binom{2n}{n-k}\left(i^{-k}+i^{k}\right) e^{-a^2bk^2} \\= 2\sqrt{b\pi}\left(-\frac{1}{2}\right)^n\cdot\left[ \binom{2n}{n} + 2 \sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{2n}{n-2k}(-1)^k e^{-4a^2bk^2} \right] } . $$