Is there a simple way to show the following: $\int_0^\pi \cos(x\sin(\theta)) \ d\theta$ satisfies $x^2y''+xy'+x^2y=0$.

I know this is the integral representation of the Bessel function, I am wondering if there is a simple method to prove this, without making use that the integral is the integral representation of the Bessel function.


With $y=\int_0^{\pi} \cos{(x\sin{\theta})} \, d\theta $, $$ y' = -\int_0^{\pi} \sin{\theta}\sin{(x\sin{\theta})} \, d\theta \\ y''= -\int_0^{\pi} \sin^2{\theta}\cos{(x\sin{\theta})} \, d\theta = - y + \int_0^{\pi} \cos^2{\theta} \cos{(x\sin{\theta})} \, d\theta . $$ Multiplying by $x^2$ and integrating by parts, we therefore have $$ \begin{align} x^2(y''+y) &= \int_0^{\pi} (x\cos{\theta})^2 \cos{(x\sin{\theta})} \, d\theta \\ &= [ x\cos{\theta} \sin{(x\sin{\theta})} ]_0^{\pi} + \int_0^{\pi} x \sin{\theta} \sin{(x\sin{\theta})} \, d\theta \\ &= 0 - 0 - xy', \end{align} $$ as required.

  • $\begingroup$ When doing the integration by parts, isn't there a factor of $2$ in the second term because of the derivative of $\cos^2(\theta)$?? $\endgroup$ – babynewton Sep 17 '19 at 7:51
  • 1
    $\begingroup$ No, I’m integrating $x\cos{\theta} \cos{(x\sin{\theta})}$, which leaves only $x\cos{\theta}$ to differentiate. $\endgroup$ – Chappers Sep 17 '19 at 10:42

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