Show that $J_0(x)=\frac{2}{\pi}\int_0^\infty\frac{\sin\left(x+y\right)}{x+y}J_0(y)dy$

I've been trying to show, without success, that: \begin{align*} J_0(x)=\frac{2}{\pi}\int_0^\infty\frac{\sin\left(x+y\right)}{x+y}J_0(y)dy\quad\text{for }0\leq x<\infty, \end{align*} where $$J_0$$ is the Bessel function of the first kind of order $$0$$. The result has been verified numerically.

One of the things I tried is to use the integral representation $$\frac{1}{x+y}=\int_0^\infty e^{-t(x+y)}dt$$ an apply Fubini's. After simplying the first integral I arrive at: \begin{align*} \frac{1}{i\pi}\int_0^\infty e^{-tx}\left(\frac{\cos x+i\sin x}{\sqrt{(t-i)^2+1}}+\frac{-\cos x+i\sin x}{\sqrt{(t+i)^2+1}}\right)dt, \end{align*} which I doubt has a simple closed-form.

• +1. Looking forward to seeing how this one develops. Another appearance of the $\operatorname{sinc}$ kernel. Jul 2, 2021 at 4:17
• +1 for this nice question Mar 12, 2023 at 17:40

Let $$I(x)$$ be represented by the integral

\begin{align} I(x)&=\frac2\pi\int_0^\infty \frac{\sin(x+y)}{x+y}J_0(y)\,dy\tag1 \end{align}

Letting $$t=-x$$ and $$I(-t)=J(t)$$ in $$(1)$$ we see that

\begin{align} J(t)&=\frac2\pi\int_{-\infty}^\infty \frac{\sin(t-\tau)}{t-\tau}J_0(\tau)H(\tau)\,d\tau\\\\ &=\frac2\pi (\text{sinc}*J_0H)(t)\tag2 \end{align}

Using the convolution theorem, the Fourier transform of $$J(t)$$ in $$(2)$$, as given by $$\mathscr{F}\{J\}(\omega)=\int_{-\infty}^\infty J(t)e^{i\omega t}\,dt$$, can be expressed as

\begin{align} \mathscr{F}\{J\}(\omega)&=\frac2\pi\underbrace{\mathscr{F}\{\text{sinc}\}(\omega)}_{=\pi \text{Rect}(\omega/2)}\mathscr{F}\{J_0H\}(\omega)\\\\ &=2\text{rect}(\omega/2)\int_0^\infty J_0(t) e^{i\omega t}\,dt\\\\ &=2\text{rect}(\omega/2) \left(\int_0^\infty J_0(t) \cos(\omega t)\,dt+i\int_0^\infty J_0(t) \sin(\omega t)\,dt\right)\\\\ &=2\text{rect}(\omega/2) \left(\frac12\int_{-\infty}^\infty J_0(t) e^{i\omega t}\,dt+i\int_0^\infty J_0(t) \sin(\omega t)\,dt\right)\tag3 \end{align}

In THIS ANSWER, I showed that the Fourier Transform of $$J_0(t)$$ is equal to $$\mathscr{F}\{J_0\}(\omega)=\frac{2\text{rect}(\omega/2)}{\sqrt{1-\omega^2}}$$ while using the result found HERE, we see that $$\int_0^\infty J_0(t)\sin(\omega t)\,dt=0$$ when $$|\omega|<1$$.

Using these results in $$(3)$$ reveals

\begin{align} \mathscr{F}\{J\}(\omega)&= \frac{2\text{rect}(\omega/2)}{\sqrt{1-\omega^2}}\tag4 \end{align}

Since the inverse Fourier Transform of the right-hand side of $$(4)$$ is $$J_0(t)$$ and $$J_0(t)=J_0(-t)=J_0(x)$$, we are done!.

• How did we conclude that: \begin{align*} \mathscr{F}\{J_0H\}(\omega)=\frac1{2\pi}\left(\mathscr{F}\{J_0\}*\mathscr{F}\{H\}\right)(\omega)? \end{align*} Jul 2, 2021 at 21:29
• We appealed to the convolution theorem again. Jul 3, 2021 at 0:16
• @AndrésSantos I've edited the answer to streamline the analysis, which now avoids the second application of the convolution theorem. Jul 6, 2021 at 3:33
• Excellent, the argument seems to be more clear. We may also evaluate \begin{align*} \int_0^\infty J_0\left(t\right)e^{i\omega t}dt \end{align*} directly by using the integral representation \begin{align*} J_0\left(x\right)=\frac{1}{\pi}\int_0^\pi\cos\left(x\sin\theta\right)d\theta,\qquad \text{Re }x>0. \end{align*} Which is shown in this question. Jul 6, 2021 at 20:46
• Yes, I believe I covered that. Jul 6, 2021 at 22:12

If we integrate over the entire real line, we get the peculiar result $$\int_{-\infty}^{\infty}\frac{\sin(x+y)}{x+y} \, J_{0}(y) \, \mathrm dy = \pi J_{0}(x) =2 \int_{0}^{\infty} \frac{\sin(x+y)}{x+y} \, J_{0}(y) \, \mathrm dy.$$

It's peculiar because the integrand is not even.

More generally, we can show that $$\int_{-\infty}^{\infty} \frac{\sin(x+y)}{x+y} \, J_{n}(y) \, \mathrm dy= \pi (-1)^{n} J_{n}(x)$$ for $$x \ge 0$$ and $$n \in \mathbb{Z}_{\ge 0}$$.

Consider the function $$f(z) = \frac{e^{i(x+z)}}{x+z} \, J_{n}(z).$$

For $$0 < \arg(z) < \pi$$, $$|f(z)| \sim \frac{1}{\sqrt{2 \pi}} \, \frac{1}{|z|^{3/2}}$$ as $$|z| \to \infty$$. (See here.)

Therefore, if we integrate counterclockwise around a contour consisting of the real axis (indented at $$z=-x$$) and the semicircle above it, it follows from the estimation lemma that $$\operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{i(x+y)}}{x+y} \, J_{n}(y) \, \mathrm dy - i \pi \operatorname{Res}[f(z), -x] =0,$$

where $$\operatorname{Res}[f(z), -x] = e^{i(x-x)} J_{n}(-x) = (-1)^{n} J_{n}(x).$$

Equating the imaginary parts on both sides of the equation, we have $$\int_{-\infty}^{\infty} \frac{\sin(x+y)}{x+y} \, J_{n}(y) \, \mathrm dy = \pi (-1)^{n}J_{n}(x).$$