Evaluating analytically the improper integral $ \int_0^\infty \frac{x}{x^2 + \beta^2}\,J_2(\alpha x) \,\mathrm{d}x$ for $\alpha,\beta\in\mathbb{R}_+$

Question

While I was elaborating on a physical problem involving fluids and interfaces, I came across the following integral, which seems at a first glance, like very easy to solve, but it turned out that Maple fails, unfortunately, to provide correct expressions. I am thinking of using the residue theorem but no notable progress has been made so far. Any help or suggestions are very much appreciated:
$$
\int_0^\infty \frac{x}{\eta x^2 + \alpha} \, J_2(\rho x) \, \mathrm{d}x \, ,
$$
wherein $J_2$ stand for Bessel function of the first kind of order 2.
Here, $\alpha$, $\rho$, and $\eta$ are positive real numbers.

Answer 1

Using the Laplace transform of the Bessel function of the first kind and Basset's integral representation of the modified Bessel function of the second kind, we have

$$ \begin{align} \int_{0}^{\infty} \frac{x}{\beta^{2}+x^{2}} \, J_{2}(\alpha x) \, \mathrm dx &= \int_{0}^{\infty} J_{2}(\alpha x) \int_{0}^{\infty} \cos(\beta t)e^{- x t} \, \mathrm dt \, \mathrm dx \\ &= \int_{0}^{\infty} \cos(\beta t) \int_{0}^{\infty} J_{2}(\alpha x) e^{-tx} \, \mathrm dx \, \mathrm dt \\ &= \frac{1}{\alpha^{2}} \int_{0}^{\infty} \cos(\beta t) \left(\frac{\alpha^{2}}{\sqrt{\alpha^{2}+t^{2}}} + \frac{2t^{2}}{\sqrt{\alpha^{2}+t^{2}}}-2t \right) \, \mathrm dt \\ &=K_{0}(\alpha \beta) + \frac{2}{\alpha^{2}} \int_{0}^{\infty} \cos (\beta t) \left(\frac{t^{2}}{\sqrt{a^{2}+t^{2}}}-t \right) \, \mathrm dt. \end{align} $$

But notice that $$ \begin{align} K_{0}(\alpha \beta) &= \int_{0}^{\infty} \frac{\cos(\beta t)}{\sqrt{\alpha^{2}+t^{2}}} \, \mathrm dt \\ &= \int_{0}^{\infty} \cos(\beta t) \left(\frac{1}{\sqrt{\alpha^{2}+t^{2}}} -\frac{ \boldsymbol{1}_{t > 1}}{t} \right) \, \mathrm dt + \int_{0}^{\infty} \boldsymbol{1}_{t > 1} \, \frac{\cos(\beta t)}{t} \, \mathrm dt \\ &= \int_{0}^{\infty} \cos(\beta t) \left(\frac{1}{\sqrt{\alpha^{2}+t^{2}}} -\frac{\boldsymbol{1}_{t > 1}}{t} \right) \, \mathrm dt + \int_{1}^{\infty} \frac{\cos(\beta t)}{t} \, \mathrm dt \\ &= \int_{0}^{\infty} \cos(\beta t) \left(\frac{1}{\sqrt{\alpha^{2}+t^{2}}} -\frac{\boldsymbol{1}_{t > 1}}{t} \right) \, \mathrm dt -\operatorname{Ci}(\beta) . \end{align}$$

Differentiating both sides of the above equation with respect to $\beta$ twice, we get

$ \begin{align} \frac{\alpha^{2}}{2} \left(K_{0}(\alpha \beta) + K_{2} (\alpha \beta) \right) &= -\int_{0}^{\infty} \cos(\beta t) \left(\frac{t^{2}}{\sqrt{\alpha^{2}+t^{2}}} -\boldsymbol{1}_{t > 1} \, t \right) \, \mathrm dt + \frac{\cos(\beta) + \beta \sin(\beta)}{\beta^{2}} \\ &= \small -\int_{0}^{\infty} \cos(\beta t) \left(\frac{t^{2}}{\sqrt{\alpha^{2}+t^{2}}} -t \right) \, \mathrm dt - \int_{0}^{1} t \cos(\beta t) \, \mathrm dt + \frac{\cos(\beta) + \beta \sin(\beta)}{\beta^{2}} \\ &= -\int_{0}^{\infty} \cos(\beta t) \left(\frac{t^{2}}{\sqrt{\alpha^{2}+t^{2}}} -t \right) \, \mathrm dt + \frac{1}{\beta^{2}}. \end{align}$

Therefore, $$ \begin{align} \int_{0}^{\infty} \frac{x}{\beta^{2}+x^{2}} \, J_{2}(\alpha x) \, \mathrm dx &= K_{0}(ab) + \frac{2}{\alpha^{2}} \left(\frac{1}{\beta^{2}}- \frac{\alpha^{2}}{2} \left(K_{0}(\alpha \beta) + K_{2} (\alpha \beta) \right) \right) \\ &= \frac{2}{\alpha^{2} \beta^{2}} - K_{2}(\alpha \beta). \end{align}$$

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Original answer:

The following answer is similar to the answer I posted here.

Let $H_{2}^{(1)}(z)$ be the Hankel function of the first kind of order $2$, defined as $$H_{2}^{(1)}(z)=J_{2}(z)+iY_{2}(z). $$

The function $H_{2}^{(1)}(z)$ has a branch cut along the negative real axis.

On the the upper side of the branch cut, $$H_{2}^{(1)}(-x) = -J_{2}(x) + i Y_{0}(x) , \quad x >0.$$

Since $\frac{x}{x^{2}+\beta^{2}}$ is an odd function, we can write $$\int_{0}^{\infty} \frac{x}{x^{2}+\beta^{2}} \, J_{2}(\alpha x) \, \mathrm dx = \frac{1}{2}\, \Re \operatorname{PV} \int_{-\infty}^{\infty} \frac{x}{x^{2}+\beta^{2}} \, H_{2}^{(1)}(\alpha x) \, \mathrm dx. $$

Using the fact that $H_{2}^{(1)}(z) \sim \sqrt{\frac{2}{\pi z}}e^{i\left(z- \frac{3 \pi}{4}\right)}$ as $|z| \to \infty$ in the upper half-plane, we can integrate the function $$f(z) = \frac{z}{z^{2}+\beta^{2}} \, H_{2}^{(1)}(\alpha z)$$ around a closed semicircular contour in the upper half-plane and conclude that $$\operatorname{PV} \int_{-\infty}^{\infty} f(x) \, \mathrm dx + \lim_{\epsilon \to 0} \int_{C_\epsilon} f(z) \, \mathrm dz = 2 \pi i \operatorname{Res}[f(z), i \beta],$$ where $C_{\epsilon}$ is a small clockwise-oriented semicircle of radius $\epsilon$ around the branch point at the origin.

Since $$\begin{align} \lim_{z \to 0} zf(z) &= \lim_{z \to 0} \frac{z^{2}}{z^{2} + \beta^{2}} J_{2 }(\alpha z) + i \lim_{z \to 0} \frac{z^{2}}{z^{2} + \beta^{2}} Y_{1}(\alpha z) \\ &= 0 + i \lim_{z \to 0} \frac{z^{2}}{z^{2}+\beta^{2}} \left(- \frac{4}{\pi \alpha^{2}z^{2}} + \mathcal{O}(1) \right) \\ &= -\frac{4i}{\pi \alpha^{2} \beta^{2}} ,\end{align} $$

we have $$\lim_{\epsilon \to 0} \int_{C_\epsilon} f(z) \, \mathrm dz = - i \pi \left(-\frac{4i}{\pi \alpha^{2} \beta^{2}} \right) = - \frac{4}{\alpha^{2} \beta^{2}}.$$

And similar to what I showed in my other answer, $$\operatorname{Res}[f(z), i \beta] = \frac{1}{2}H_{2}^{(1)} (i\alpha \beta) = \frac{iK_{2}(\alpha \beta)}{\pi}. $$

Therefore, $$\begin{align} \int_{0}^{\infty} \frac{x}{x^{2}+\beta^{2}} \, J_{2}(\alpha x) \, \mathrm dx &= \frac{1}{2} \, \Re \left(2 \pi i \left(\frac{iK_{2}(\alpha \beta)}{\pi} \right) + \frac{4}{\alpha^{2} \beta^{2}} \right) \\ &= \frac{2}{\alpha^{2} \beta^{2}} - K_{2}(\alpha \beta). \end{align}$$