I'm trying to prove this integral representation:

$$J_\nu = \frac{({x/2})^\nu}{\Gamma(\nu+1/2)\sqrt{\pi}}\int_{-1}^{1}(1-t^2)^{\nu-1/2}e^{itx}dt$$

I think countour integration would be useful, but in this case I'm trying to avoid direct integration.

I've tried changing variables of the integral and used the generating function:

$$\int_{-1}^{1}(1-t^2)^{\nu-1/2}e^{itx}dt=\int_{-\pi/2}^{\pi/2}(\cos \theta)^{2\nu}e^{ix\sin\theta}d\theta=\int_{-\pi/2}^{\pi/2}(\cos \theta)^{2\nu}\left(\sum _{n=-\infty}^\infty J_ne^{in\theta}\right)d\theta$$

I think I need to use somewhere the Beta function.

I would prefer a hint, not the whole proof.

  • 1
    $\begingroup$ The way I've done it in the past is to solve Bessel's differential equation using the Laplace transform. That integral arises when calculating the inverse Laplace transform using an appropriate Bromwich contour. $\endgroup$ – Antonio Vargas Mar 22 '14 at 13:55
  • $\begingroup$ @AntonioVargas I'll give that method a try. But I know there's a way without using it. $\endgroup$ – jinawee Mar 22 '14 at 14:03

The typical way to derive it is to expand $e^{itx}$ in its Maclaurin series, switch the order of summation and integration, integrate, and then apply the duplication formula for the gamma function.

$$ \begin{align} \int_{-1}^{1}(1-t^2)^{\nu-1/2}e^{itx} \, dt &= \int_{-1}^{1} (1-t^{2})^{\nu-1/2} \sum_{n=0}^{\infty}\frac{(itx)^{n}}{n!} \, dt \\ &= \sum_{n=0}^{\infty} \frac{(ix)^{n}}{n!} \int_{-1}^{1} t^{n} (1-t^{2})^{\nu-1/2} \, dt \\ &= 2 \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!} \int_{0}^{1} t^{2n} (1-t^{2})^{\nu-1/2} \, dt \tag{1} \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!} \int_{0}^{1} u^{n-1/2} (1-u)^{\nu-1/2} \, du \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!} B \left(n + \frac{1}{2}, \nu + \frac{1}{2} \right) \\ &=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!} \frac{\Gamma(n+ \frac{1}{2}) \Gamma(\nu+\frac{1}{2})}{\Gamma(n+\nu+1)} \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!} \frac{2^{1-2n} \sqrt{\pi} \ \Gamma(2n) \Gamma(\nu+\frac{1}{2})}{\Gamma(n) \Gamma(n+\nu+1)} \frac{n}{n} \tag{2} \\ &= \sqrt{\pi} \, \Gamma \left(\nu + \frac{1}{2} \right)\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n! \Gamma(n+\nu+1)} \left(\frac{x}{2} \right)^{2n} \\ &= \sqrt{\pi} \, \Gamma \left( \nu+\frac{1}{2} \right) \left( \frac{2}{x}\right)^{v} J_{\nu}(x)\end{align}$$

$(1)$ The integrand is even if $n$ is even and odd if $n$ is odd.

$(2)$ Duplication formula for the gamma function


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