Convergence of ${\large\int}_{-\infty}^\infty J_0(x)\,J_0(x+a)\,dx$

Question

Consider $$I(a)={\int}_{-\infty}^\infty J_0(x)\,J_0(x+a)\,dx,$$ where $J_0(z)$ is the Bessel Function of the $1^{st}$ kind and $a>0$.

Does this integral converge for any values of $a$? If so, is there a closed form for it? Can we find a value of $a$ such that $I(a)=0$?

Answer 1

The integral converges only when $a$ is an odd multiple of $\pi/2$, and does not seem to vanish even for those $a$.

For large $\left|z\right|$ it is known that $$ J_0(z) = \sqrt{\frac2{\pi\left|z\right|}} \bigl(\cos (\left|z\right|-\frac\pi4) + O(1/\left|z\right|) \bigr). $$ Therefore we have for large $x$ (either $x>0$ and $x<0$): $$ J_0(x+\frac{a}{2}) \, J_0(x-\frac{a}{2}) = \frac2{\pi\left|x\right|} \bigl(\cos (\left|x\right|-\frac\pi4+\frac{a}2) \cos (\left|x\right|-\frac\pi4-\frac{a}2) + O(1/\left|x\right|) \bigr) = \frac1{\pi\left|x\right|} \bigl(\cos (2\left|x\right|-\frac\pi2) + \cos a + O(1/\left|x\right|) \bigr). $$ The integral thus converges iff $\cos a = 0$, as claimed.

Morally speaking, $I(a)$ should be the convolution of $J_0$ with itself (since $J_0$ is an even function), so its Fourier transform should be the square of the Fourier transform of $J_0$. But $J_0(x)$ can be written as a Fourier integral $\frac1\pi \int_{-1}^1 \cos xt \, {\rm d}t/\sqrt{1-t^2}$, so we should have $I(a) = \frac2\pi \int_{-1}^1 \cos at \, {\rm d}t/(1-t^2)$, and now the integrand blows up too quickly at $t = \pm 1$ for the integral to converge $-$ unless $\cos at = 0$ vanishes there, which recovers our criterion $\cos a = 0$. In that case numerical integration corroborates the formula; e.g. for $a = \pm \pi/2$ we should have $I(a) = \frac2\pi \int_{-1}^1 \cos \frac{\pi t}{2} {\rm d}t/(1-t^2)$, and the integral is the Gibbs constant $\int_0^\pi \sin t \, {\rm d}t/t = 1.851937\ldots$ (write $\frac2{1-t^2} = \frac1{1-t} + \frac1{1+t}$, etc.) and indeed $I(a)$ is approximately $1.179$ which is $2/\pi$ times this constant. For larger odd multiples of $\pi/2$ we find that $I(a)$ alternates in sign and approaches $\pm 1$, which again would be consistent with the formula $I(a) = \frac2\pi \int_{-1}^1 \cos at \, {\rm d}t/(1-t^2)$.

To prove this integral formula when $a$ is an odd multiple of $\pi/2$, start by writing $J_0(x-a) + J_0(x+a) = \frac2{\pi}\int_{-1}^1 \cos xt \cos at \, {\rm d}t/\sqrt{1-t^2}$, and apply Parseval's formula to the inner product of $J_0(x)$ with $J_0(x-a) + J_0(x+a)$.

Answer 2

I bet that the answers to the first and third question (about convergence and zeroes) are affirmative, since the function $$ f(z) = \left(1-\frac{x^2}{4}+\frac{x^4}{64}\right)\mathbb{1}_{[0,1]}(x)+\sqrt{\frac{2}{\pi x}}\cos(x-\pi/4)\mathbb{1}_{[1,+\infty)}(x),$$ by following Abramowitz and Stegun, is a very good approximation for $J_0(x)$, but I do not think there is a nice closed-form expression for your integral.