How can I solve this integral involving the Bessel function?

Question

How do I compute the following integral which includes the Bessel function? $$\int_0^\infty \frac{y}{y^2-1}J_0(ty)dy, \quad t\in \mathbb R$$

There are singularities inside the integral, which I don't know how to deal with. I tried the residue theorem but failed.

I know that the following holds:

$$\int_0^\infty \frac{y}{y^2\color{red}{+}1}J_0(ty)dy = K_0(t), \quad t\in \mathbb R$$

But it is unclear to me if this helps at all in solving this problem, as I tried manipulating the integral using some substitutions to get it into a similar form to this, but ultimately failed.

Answer 1

Hint

Let $y=\frac{x}{t}$ to face $$\int_0^\infty \frac{x }{(x+t)(x-t)}\, J_0(x)\,dx$$ Use partial fraction decomposition to face two similar integrals $$I_\pm=\int_0^\infty \frac{J_0(x)}{x\pm t}\,dt$$ I suppose that each of them would express in terms of other Bessel and Struve functions. If lucky, the sum of the two integrals should simplify.

Answer 2

EDIT: I simplified my answer by changing from a semicircular contour to a quarter-circle contour.

---

Assume that $t>0$.

The integral doesn't converge, but we can show that $$ \operatorname{PV}\int_{0}^{\infty} \frac{y}{y^{2}-1} \, J_{0}(ty) \, \mathrm dy = - \frac{\pi}{2} \, Y_{0}(t).$$

Let $H^{(1)}_{0}(x)$ be the Hankel function of the first kind of order zero defined as $$H^{(1)}_{0}(z) = J_{0}(z) + i Y_{0}(z). $$

The principal branch of $H_{0}^{(1)}(z)$ is analytic in the right half-plane.

Let's integrate the function $$f(z) = \frac{z}{z^{2}-1} \, H_{0}^{(1)} (tz) $$ around the quarter-circle contour $[r, R] \cup Re^{i[0, \pi/2]} \cup [iR, i r] \cup r e^{i[\pi/2, 0]}$.

As $r \to 0$, the integral along the small quarter-circle about that origin vanishes since $$\begin{align} \lim_{z \to 0} zf(z) &= \lim_{z \to 0} \frac{z^{2}}{z^{2} - 1} J_{0 }(t z) + i \lim_{z \to 0} \frac{z^{2}}{z^{2} -1} Y_{0}(tz) \\ &\overset{(1)}{=} 0 + i \lim_{z \to 0} \frac{z^{2}}{z^{2}+\beta^{2}} \left(\frac{2}{\pi} \log(tz) + \mathcal{O}(1) \right) \\ &= 0.\end{align} $$

And since $$ |f(z)| \sim \sqrt{\frac{2}{\pi}} \frac{1}{|z|^{3/2}} e^{- \Im(z)} \tag{2} $$ as $|z| \to \infty$ in the upper half-plane, the integral along the large quarter-circle vanishes as $R \to \infty$.

So we have $$\operatorname{PV} \int_{0}^{\infty} \frac{y}{y^{2}-1} H_{0}^{(1)}(ty) \, \mathrm dy -i \pi \operatorname{Res} \left[f(z), 1 \right] - \int_{0}^{\infty} \frac{x}{x^{2}+1} \, H_{0}^{(1)} \left(e^{i \pi/2} tx \right) \, \mathrm dx =0, $$

where $$ \operatorname{Res} \left[f(z), 1 \right] = \frac{1}{2}H_{0}^{(1)} (t) = \frac{1}{2} \left(J_{0}( t) + i Y_{0}(t) \right).$$

Since $H_{0}^{(1)} \left(e^{i \pi/2} tx \right) = \frac{2}{i \pi} K_{0}(tx)$ (see here), the integral $ \int_{0}^{\infty} \frac{x}{x^{2}+1} \, H_{0}^{(1)} \left(e^{i \pi/2} tx \right) \, \mathrm dx $ is purely imaginary.

Therefore, $$ \begin{align} \operatorname{PV} \int_{0}^{\infty} \frac{y}{y^{2}-1} J_{0}(ty) \, \mathrm dy &= \Re \, \operatorname{PV} \int_{0}^{\infty} \frac{y}{y^{2}-1} H_{0}^{(1)} (ty) \, \mathrm dy \\ &= \Re \left( \frac{i \pi }{2} \left(J_{0}(t) + i Y_{0}(t) \right)\right) \\ &= - \frac{\pi}{2} Y_{0}(t), \end{align}$$ which is the result Wolfram Alpha returns for specific values of $t$.

---

$(1)$ https://dlmf.nist.gov/10.8.E2

$(2)$ https://en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms