here is the question:

For each positive integer $n$, the Bessel function $J_n(x)$ may be defined by

$$J_n(x) = \frac{x^n}{1\cdot 3\cdot 5\cdots(2n-1)\pi}\int^1_{-1}(1-t^2)^{n-1/2}\cos(xt) \, dt$$

Prove that $J_n(x)$ satifies Bessel's differential equation:

$$J^{''}_n+ \frac{1}{x}J^{'}_n + \left(1-\frac{n^2}{x^2}\right)J_n = 0$$

I understand that I have to plug $J_n$ into the differential equation, but my problem lies in taking the first and second derivatives of a function based on n... ie, if I were to say n=0 this problem would be fairly easy for me.

Perhaps someone could give me some hints?

Thanks in advance !


1 Answer 1


Just remember that $n$ is a parameter, not a variable, so when you take derivatives you'll treat it as such. Also notice that you'll be applying the product rule:

$$ J_n(x)=u(x)v(x) $$ where

$$ u(x)=\frac{x^n}{1\cdot 3\cdots (2n-1)\pi}\quad v(x)=\int_{-1}^1(1-t^2)^{n-1/2}\cos(xt)dt $$ So, $J_n^\prime(x)=u^\prime v+v^\prime u$, where "prime" means $\frac{\partial}{\partial x}$. I'll get you started with the derivatives:

$$ u^\prime=\frac{nx^{n-1}}{1\cdot 3\cdots(2n-1)\pi}\\ v^\prime=\frac{\partial}{\partial x}\int_{-1}^1(1-t^2)^{n-1/2}\cos(xt)dt=-\int_{-1}^1t(1-t^2)^{n-1/2}\sin(xt)dt $$ The rest is "just" algebra. You can always check your derivatives with Wolfram Alpha, as well.

  • $\begingroup$ OH, I thought "prime" meant d/dt. Thats why i was having trouble checking and doing it... ok thanks! $\endgroup$
    – Neo
    Commented Apr 12, 2013 at 5:47
  • $\begingroup$ Ah, that would definitely cause problems. $t$ is just the "dummy" integration variable. Glad to help. $\endgroup$
    – icurays1
    Commented Apr 12, 2013 at 5:49

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