Integral involving Bessel functions of the first kind

Question

I am stuck with the following integral. Does it converge?

$$ \int_{0}^{\infty}\left(J_1(x)^2+J_1(x)J_1(x)^{''}\right)\text{d}x $$

According to tables I find that the first term is divergent, so I assume it is overall divergent but it could very well be that the second term tames it.

Edit: It seems that only for the second term to be $a J_1(x)J_1(x)^{''}$ with $a=1$ the integral can be shown to converge, in other cases the divergent part $J_1(x)^2$, for example when $a=1/2$, cannot be substracted by the proposed identities, right?

Answer 1

Since $J_1$ is a solution of the Bessel differential equation: $$ x^2 f'' + x f' + x^2 f = f \tag{1}$$ by exploiting integration by parts we have that: $$ \int_{0}^{+\infty}\left(J_1(x)^2+J_1(x)\,J_1''(x)\right)\,dx = \frac{1}{2}\int_{0}^{+\infty}\left(\frac{J_1(x)}{x}\right)^2\,dx \tag{2} $$ so we just need to recall that the Fourier transform of $\frac{J_1(x)}{x}$ is given by: $$ \mathcal{F}\left(\frac{J_1(x)}{x}\right)(t) = \sqrt{\frac{2}{\pi}}\sqrt{1-t^2}\cdot\mathbb{1}_{(-1,1)}(t)\tag{3}$$ to be able to state: $$ \int_{0}^{+\infty}\left(J_1(x)^2+J_1(x)\,J_1''(x)\right)\,dx = \frac{1}{\pi}\int_{0}^{1}\left(1-t^2\right)\,dt = \frac{2}{3\pi}\tag{4}$$ as a consequence of Parseval's theorem.

Answer 2

Mathematica says it converges to $2/(3\pi)$. You can probably get this from this identity: $$ \int_{0}^{\infty}\frac{J_{\mu}(at)J_{\nu}(at)}{t^{\lambda}}dt= \frac{(\frac{1}{2}a)^{\lambda-1} \Gamma\left(\frac{1}{2}\mu+\frac{1}{2}\nu-\frac{1}{2}\lambda+\frac{1}{2}\right) \Gamma(\lambda)}{ 2\Gamma\left(\frac{1}{2}\lambda+\frac{1}{2}\nu- \frac{1}{2}\mu+\frac{1}{2}\right) \Gamma\left(\frac{1}{2} \lambda+\frac{1}{2}\mu-\frac{1}{2}\nu+\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\lambda+\frac{1}{2}\mu+\frac{1}{2}\nu+\frac{1}{2}\right)} $$ with careful limits as $\lambda \to 0$, and the expression $$ J_1''(x) = \frac{1}{4} (J_3(x)-3 J_1(x)) $$ reduces to the integral to one of the form $$ \frac{1}{4}\int_0^{\infty} J_1(x) (J_1(x)+J_3(x)) \, dx, $$ which you can then do with the linked identity.

Answer 3

Sometimes, simpler is better ...

So, let's just start with the identity that Chappers provided, namely

$$\int_{0}^{\infty} (J_1(x)^2+J_1(x)J_1(x)'')\,dx= \frac{1}{4}\int_0^{\infty} J_1(x) (J_1(x)+J_3(x)) \, dx $$

Then, let's substitute the recurrence relation

$$J_2(x)=\frac{x}{4} (J_1(x)+J_3(x))$$

to find that

$$\int_{0}^{\infty} (J_1(x)^2+J_1(x)J_1(x)'') \, dx = \int_0^{\infty} \frac{J_1(x)J_2(x)}{x} \, dx $$

Finally, let's exploit the orthogonality relationship

$$\int_0^{\infty} \frac{J_{\alpha}(x)J_{\beta}(x)}{x} \, dx =\frac{2}{\pi}\frac{\sin (\frac{\pi}{2}(\alpha-\beta))}{\alpha^2-\beta^2}$$

with $\alpha =2$ and $\beta =1$ to obtain

$$\int_0^{\infty} \frac{J_{2}(x)J_{1}(x)}{x} \, dx =\frac{2}{\pi}\frac{\sin (\frac{\pi}{2}(2-1))}{2^2-1^2}=\frac{2}{3\pi}$$

which recovers the expected result!