Integration of $r J_0(r)$ Question
Let $J_0(r)$ denote the Bessel function of the first kind of order zero, I'd like to compute the following integration:
$$ I = \int_{0}^{\infty} rJ_0(r) dr $$
I tried using the Laplace transformation to compute this but I'm not sure if I'm doing right. Is my approach correct? Do you have any other approches to prove this?
My attempt
Using the differential property of Laplace transform, we have
$$\mathcal{L}\{tf(t)\} = \int_0^\infty tf(t) e^{-st} dt = -\frac{d \mathcal{L}\{f(x)\}}{ds} $$
Based on the Proof Wiki, the Laplace transform of the Bessel function is:
$$\mathcal{L}\{J_0(t)\}  = \frac{1}{\sqrt{s^2+1}}$$
Therefore, it follows that
$$\mathcal{L}\{rJ_0(r)\} =\frac{d}{ds}\left( \frac{1}{\sqrt{s^2+1}}\right) = \frac{s}{(s^2+1)^{3/2}} $$
In other words, we have
$$\int_0^\infty rJ_0(r) e^{-sr} dr = \frac{s}{(s^2+1)^{3/2}} $$
Thus, by considering the limit: $s \rightarrow 0$, we obtain the following:
$$ I = \int_{0}^{\infty} rJ_0(r) dr = 0 $$
 A: As other users noted, your integral is indeterminate but at the limit of some transformation (Laplace, Mellin) is zero:
This is more clear with the Mellin transform. Consider the series expansion of $J_{0}$:
$\displaystyle J_{0}(x) = \sum_{j=0}^{\infty} \frac{(-x^2/4)^{j}}{j!^2}$
then
$\displaystyle I= \int_{0}^{\infty} x J_{0}(x)dx = \int_{0}^{\infty} x\sum_{j=0}^{\infty} \frac{(-x^2/4)^{j}}{j!^2} dx$
Let $\displaystyle w = x^2 \Longrightarrow dw=2xdx$
then
$\displaystyle I =  \frac{1}{2}\int_{0}^{\infty} \sum_{j=0}^{\infty} \frac{(-w/4)^{j}}{j!^2} dw$
Now, denote $\displaystyle f(w) = \frac{1}{2}\sum_{j=0}^{\infty} \frac{(-w/4)^{j}}{j!^2} dx $
and consider the Mellin transform of the $f(w)$ function:
$\displaystyle \left\{ \mathscr{M} f \right\}(s) = \int_{0}^{\infty}w^{s-1} f(w) dw =\frac{1}{2}\int_{0}^{\infty}w^{s-1} \sum_{j=0}^{\infty} \frac{(-w/4)^{j}}{j!^2} dw $
your integral is the Mellin transform at $s=1$
By Ramanujan's master theorem:
If we denote $\displaystyle \phi(j)= \frac{1}{4^j\Gamma(j+1)}$
therefore
$\displaystyle \left\{\mathscr{M} f \right\}(s)  =\frac{1}{2}\int_{0}^{\infty}w^{s-1} \sum_{j=0}^{\infty} \frac{(-w/4)^{j}}{j!^2} dw = \frac{1}{2}\Gamma(s)\phi(-s) = \frac{1}{2}\frac{4^s\Gamma(s)}{\Gamma(1-s)} $
If we put $\displaystyle s=1$ clearly the right hand side is  indeterminate
But if we take the limit as $\displaystyle s \to 1$
$\displaystyle \lim _{s\to 1} \frac{1}{2}\int_{0}^{\infty}w^{s-1} \sum_{j=0}^{\infty} \frac{(-w/4)^{j}}{j!^2} dw = \lim_{s\to 1} \frac{1}{2}\frac{4^s\Gamma(s)}{\Gamma(1-s)} = 0 $
The conclusion is that the integral is zero only if we express it as a limit of the transformation (you cannot interchange the limit and integral operation)
A: $$
\int{xJ_0\left( x \right) dx}
\\
=xJ_1\left( x \right) 
\\
\int_0^{\infty}{xJ_0\left( x \right) dx}
\\
=xJ_1\left( x \right) |_{0}^{\infty}
$$
But we know
$$
J_{\alpha}\left( x \right) \approx \sqrt{\frac{2}{\pi x}}\cos \left( x-\frac{\alpha \pi}{2}-\frac{\pi}{4} \right) 
$$
and
$$
J_1(0)=0
$$
We can conclude from these two facts that the integral is divergent
