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

---

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

Answer 2

Substituting $x \mapsto x/\rho$, the integral becomes

$$ \int_{0}^{\infty} \frac{x J_2(\rho x)}{\eta x^2 + \alpha} \, \mathrm{d}x = \frac{1}{\eta} \int_{0}^{\infty} \frac{x J_2(x)}{x^2 + \lambda^2} \, \mathrm{d}x, $$

where $\lambda = \rho \sqrt{\alpha/\eta}$. To evaluate the integral in the right-hand side, we first recast the integral as the limit of a regularized integral:

\begin{align*} \int_{0}^{\infty} \frac{x J_2(x)}{x^2 + \lambda^2} \, \mathrm{d}x &= \lim_{\varepsilon \to 0^+} \frac{1}{\lambda^2} \int_{0}^{\infty} \left[ x^{1-2\varepsilon} - \frac{x^3}{(x^2 + \lambda^2)^{1+\varepsilon}} \right] J_2(x) \, \mathrm{d}x. \tag{1} \end{align*}

Indeed, noting that the integrand in the right-hand side of $\text{(1)}$ is $\mathcal{O}(x^{-3/2})$ uniformly for $\varepsilon \in (0, 2)$ and $x \geq \lambda$, the limit $\text{(1)}$ is justified by the dominated convergence theorem. Moreover, for $\varepsilon \in (\frac{1}{4}, 2)$, the formulas 10.22.43 and 10.22.46 of DLMF tells that

\begin{align*} \int_{0}^{\infty} x^{1-2\varepsilon} J_2(x) \, \mathrm{d}x &= 2^{1-2\varepsilon} \frac{\Gamma(2-\varepsilon)}{\Gamma(1+\varepsilon)} \tag{2} \end{align*}

and

\begin{align*} \int_{0}^{\infty} \frac{x^3 J_2(x)}{(x^2 + \lambda^2)^{1+\varepsilon}} \, \mathrm{d}x &= \frac{\lambda^{2-\varepsilon}}{2^{\varepsilon}\Gamma(1+\varepsilon)}K_{2-\varepsilon}(\lambda), \tag{3} \end{align*}

where $K_{\nu}$ is the modified Bessel function of the second kind and order $\nu$. This tells that

$$ \frac{1}{\lambda^2} \int_{0}^{\infty} \left[ x^{1-2\varepsilon} - \frac{x^3}{(x^2 + \lambda^2)^{1+\varepsilon}} \right] J_2(x) \, \mathrm{d}x = \frac{1}{\Gamma(1+\varepsilon)} \left[ \frac{1}{\lambda^2} 2^{1-2\varepsilon} \Gamma(2-\varepsilon) - \frac{K_{2-\varepsilon}(\lambda)}{(2\lambda)^{\varepsilon}} \right] $$

holds initially for $\varepsilon \in (\frac{1}{4}, 2)$, and then for all $\varepsilon \in (0, 2)$ by the principle of analytic continuation. Therefore by taking limit as $\varepsilon \to 0^+$, we get

\begin{align*} \int_{0}^{\infty} \frac{x J_2(x)}{x^2 + \lambda^2} \, \mathrm{d}x &= \frac{2}{\lambda^2} - K_2(\lambda), \tag{4} \end{align*}

which is the same as what @Random Variable derived.

Answer 3

Yet another way to the already existing answers is by using the Mellin transform. First rewrite the integral as

$$ \int_0^{\infty} \frac{t}{\beta^2+t^2}J_2(\alpha t)dt = \frac{1}{2} \int_0^{\infty} \frac{1}{1+t}J_2(2\sqrt{xt})dt = \frac{1}{2}\mathcal{H}(x) $$ with $x = (\alpha\beta/2)^2$. With the Mellin transforms

\begin{align} h_1(t) = \frac{1}{1+t} &\to \mathcal{H}_1^*(s) = \Gamma(s)\Gamma(1-s)\\ h_2(t) = J_v(2\sqrt{t}) &\to \mathcal{H}_2^*(s) = \frac{\Gamma(s+v/2)}{\Gamma(v/2+1-s)} \end{align}

the Mellin transform of $\mathcal{H}(x)$ is given by

\begin{align} \mathcal{H}^*(s) &= \mathcal{H}_1^*(1-s) \mathcal{H}_2^*(s) \\ &= x^{-s}\frac{\Gamma(s)\Gamma(s+v/2)\Gamma(1-s)}{\Gamma(v/2+1-s)} \\ &= x^{-s}\frac{\Gamma(s)\Gamma(s+1)\Gamma(1-s)}{\Gamma(2-s)} \end{align}

with $v=2$. There is a pole of order one for $s=0$ and poles of order two for $s=-k-1, k=1,2,3,...$. We have $\text{Res}(\Gamma(s)\Gamma(s+1),s=0)=1$ whereas

$$ \text{Res}(\Gamma(s)\Gamma(s+1),s=-k-1) = -\frac{1}{k!(k+1)!}\left(2\psi(k+1)+\frac{1}{k}\right) $$

using the results from Residue of $\Gamma^{2}$ and $\Gamma^{3}$. Combining this, the inverse Mellin transform is given by

\begin{align} \mathcal{H}(x) &= \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} x^{-s}\frac{\Gamma(s)\Gamma(s+1)\Gamma(1-s)}{\Gamma(2-s)}ds \\ &= 1 - \sum_{k=0}^{\infty} \frac{1}{k!(k+2)!}x^{k+1}( \psi(k+1)+\psi(k+3)-\ln(x)) \\ &= \left(\frac{2}{\alpha\beta}\right)^2 - 2K_2(\alpha\beta) \end{align}

by using the known power series expansion of the modified Bessel function given here. Hence, we have the result

$$ \int_0^{\infty} \frac{t}{\beta^2+t^2}J_2(\alpha t)dt = \frac{2}{(\alpha\beta)^2} - K_2(\alpha\beta) $$

For the slightly more general integral

$$ \mathcal{H}(x) = \int_0^{\infty} \frac{1}{1+t}J_v(2\sqrt{xt})dt $$

assuming that the two poles are not an integer number apart, one has again using the Mellin transform resulting in

$$ \mathcal{H}(x) = \frac{2}{v} {}_1F_2\left(1,1-v/2,v/2+1,x \right) + x^{v/2}\frac{\Gamma(-v/2)\Gamma(v/2+1)}{\Gamma(v+1)}{}_0F_1(v+1,x) $$

which can be further simplified in the case $v\to2$. If only a numerical value is of interest for the asked for integral, one can use this latter expression to high precision.