3
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ This appears to be a restricted version of the Fourier transform ${\cal F}_{[-1,1]}(f)(\sigma/2\pi)$ where $f(x)=(x^2-1)^n$. I don't think there's a particularly nice expression for it. $\endgroup$
    – TheSimpliFire
    Apr 30, 2021 at 8:07

2 Answers 2

5
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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 $$I_{n}=\int_{-1}^{1}(x^{2}-1)^{n}dx=-\frac{2n}{2n+1}I_{n-1}=2\frac{(-2)^{n} n!}{(2n+1)!!}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.