Elementary integral of exponential

Question

I am keen to find the simplest expression for the following integral: $$ \int_{-1}^{1}(x^{2}-1)^{n}e^{i\sigma x}dx $$ where $n$ is a non-negative integer. The best I could come up with involves binomially expanding out the first part and using $$ \int_{-1}^{1}x^{k-1}e^{i\sigma x}dx=\frac{i^{k}}{\sigma^{k}}(\Gamma(k,i\sigma)-\Gamma(k,-i\sigma)) $$ where $\Gamma$ is the incomplete Gamma function. Please help me find a better way to obtain a nicer expression?

Answer 1

There are several direct solutions for $$I_n=\int_{-1}^{1}(x^{2}-1)^{n}\,e^{i\sigma x}\,dx$$

The first one is not very pleasant $$I_n=(-1)^n \sqrt{\pi }\, \Gamma (n+1) \,\, _0\tilde{F}_1\left(;n+\frac{3}{2};-\frac{\sigma^2}{4}\right)$$ where appears the regularized confluent hypergeometric function.

Fortunately, it also write $$I_n=(-1)^n \sqrt{\pi }\, \Gamma (n+1)\, \left(\frac{2}{\sigma }\right)^{n+\frac{1}{2}}\,J_{n+\frac{1}{2} }(\sigma )$$ where appears the Bessel function of the first kind.

This even works for non-integer values of $n$ (leading to comlex values).

If $\sigma$ is small, you have nice expansions $$I_n=(-1)^n \sqrt{\pi }\,\frac{ \Gamma (n+1)}{\Gamma \left(n+\frac{3}{2}\right)}\Bigg[1-\frac{\sigma ^2}{2 (2 n+3)}+\frac{\sigma ^4}{8 (2 n+3) (2 n+5)}-\frac{\sigma ^6}{48 ((2 n+3) (2 n+5) (2 n+7))}+O\left(\sigma ^8\right) \Bigg]$$

Answer 2

Note that $e^{i\sigma x}=\cos(\sigma x)+i\sin(\sigma x)$, and therefore, by considering just the even part (the integral is taken over the symmetric interval $[-1,1]$), we find $$I_n=\int_{-1}^{1}(x^{2}-1)^{n}e^{i\sigma x}dx=\int_{-1}^{1}(x^{2}-1)^{n}\cos(\sigma x)dx.$$ According to $I_n= \int_{-1}^1 (1 − x^2 )^n \cos(ax) \mathrm dx$, for $\sigma\not=0$, $I_n$ satisfies the following recurrence for $n\geq 2$, $$I_n = -\frac{2n}{\sigma^2}\left((2n-1)I_{n-1}+(2n-2)I_{n-2}\right).$$ I don't think there is a simple closed formula for $I_n$ when $\sigma\not=0$.

On the other hand, when $\sigma=0$, it is easy to see that $$I_{n}=\int_{-1}^{1}(x^{2}-1)^{n}dx=-\frac{2n}{2n+1}I_{n-1}=2\frac{(-2)^{n} n!}{(2n+1)!!}.$$