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 !
Best Answer
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.