How to solve $\int_0^{+\infty}\,ax\,J_0(ax)\,dx$

Question

From some equalities I ended up with understanding that:

$$\int_0^{+\infty}\,ax\,J_0(ax)\,dx = 1$$

with $J_0(ax)$ the bessel function of the first kind and $a>0$. But I don't know how to demonstrate it. I tried using the series representation of $J_0(ax)$, without any success!

Thanks!

$\mathbf{EDIT}$

I had to calculate the following double integral:

$$2b\int_0^{+\infty}dR\int_0^{+\infty}dk\,J_0(k\sqrt{R})\,k\,\exp(-bk^2)$$

with $b>0$. So if I first integrate in $k$, I obtain:

$$\int_0^{+\infty}dR\,\exp\left(-\frac{R}{4b}\right)=4b$$

since

$$2b\int_0^{+\infty}dk\,J_0(k\sqrt{R})\,k\,\exp(-bk^2)=\exp\left(-\frac{R}{4b}\right)$$.

Now, if I integrate first in $R$, I have:

$$2b\int_0^{+\infty}dk\,\left[\int_0^{+\infty}dRJ_0(k\sqrt{R})\right]\,k\,\exp(-bk^2)=4b$$

meaning that

$$\int_0^{+\infty}dk\,\left[\int_0^{+\infty}dRJ_0(k\sqrt{R})\right]\,k\,\exp(-bk^2)=2$$

From this it follows that $\int_0^{+\infty}dRJ_0(k\sqrt{R})\neq0$???

Answer 1

The statement is false - the integral as stated does not converge. To see this, use the differential equation defining $y = J_0(x)$:

$$x y'' + y' + x y = 0$$

so that

$$(x y')' = -x y \implies x y' = -\int dx \, x y$$

or

$$\int dx \, x J_0(x) = x J_1(x) + C$$

or

$$a \int dx \, a x J_0(a x) = a x J_1(a x) + C$$

The RHS goes to $\infty$ as $x \to \infty$.

EDIT

As was pointed out, the above statement is not quite right. The integral does not converge, because the RHS is oscillatory with a divergent amplitude.

Answer 2

$$J_0(x) \approx \sqrt{\frac{2}{\pi x}}\cos(x-\pi/4) \quad \text{For large } x $$

The integrand is going to blow up at $\infty$ so I suspect that your integral diverges.