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

Question

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.

Answer 1

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!.

Answer 2

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.

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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 $$\int_{-\infty}^{\infty} \frac{\sin(x+y)}{x+y} \, J_{n}(y) \, \mathrm dy = \pi (-1)^{n}J_{n}(x). $$