$\int_0^\infty \frac{u}{u^2+\sigma^2} \, J_{4m} (u) \, \mathrm{d} u = K_{4m}(\sigma) + ?$

Question

As I delved into the intricacies of a fluid mechanics problem that revolved around Green's functions in porous media, I stumbled upon the following intriguing infinite integral: $$ \phi_m (\sigma) = \int_0^\infty \frac{u}{u^2+\sigma^2} \, J_{4m} (u) \, \mathrm{d} u \, , $$ wherein $\sigma \ge 0$ and $m \in \mathbb{N}$. The first four terms can be obtained as \begin{align} \phi_0 &= K_0(\sigma) \, , \\ \phi_1 &= K_4(\sigma) + \frac{4}{\sigma^4} \left( \sigma^2-12\right) \, , \\ \phi_2 &= K_8(\sigma) + \frac{8}{\sigma^8} \left( \sigma^6-60\sigma^4+2\,880\sigma^2-80\,640 \right) \, , \\ \phi_3 &= K_{12} (\sigma) + \frac{12}{\sigma^{12}} \left( \sigma^{10}-140\sigma^8 + 17\,920\sigma^6 - 1\,935\,360\sigma^4 + 154\,828\,800\sigma^2-6\,812\,467\,200 \right) \, . \end{align} It appears that the overall expression will begin with the first term, denoted as $K_{4m}(\sigma)$, accompanied by a generalized hypergeometric function for which I have yet to discover a general expression. Any assistance in this matter would be greatly valued. Thank you.

Answer 1

Assume that $\sigma >0$ and that $m \in \mathbb{N}_{\ge 0}.$

Using the same basic approach I used here, I will show that $$\int_{0}^{\infty} \frac{x}{x^{2} + \sigma^{2}} \, J_{4m}(x) \, \mathrm dx = K_{4m} (\sigma) - \mathcal{P} \, K_{4m}(\sigma),$$

where $\mathcal{P} \, K_{4m}(\sigma) $ is the principal part of the series expansion of $K_{4m} (\sigma)$ about $\sigma =0$.

By "principal part" I mean the negative-power part of the series expansion of $K_{4m}(\sigma)$ about $\sigma =0$.

For example, the principal part of $K_{4}(\sigma)$ is $\frac{48}{\sigma^{4}} - \frac{4}{\sigma^{2}}.$

The series is a Laurent series plus the term $(-1)^{n+1} \log\left(\frac{\sigma}{2}\right) I_{4m}(\sigma)$.

Wolfram Alpha expands $I_{4m}(\sigma)$ at $\sigma =0$.

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Let $H_{4m}^{(1)}(z) $ be the principal branch of the Hankel function of the first kind of order $4m$ defined as $$H_{4m}^{(1)}(z) = J_{4m}(z) + i Y_{4m}(z). $$

The principal branch of the Hankel function of the first kind has a branch cut on the negative real axis.

On the upper side of the branch cut, $$H_{4m}^{(1)}(-x) = \color{red}{-}J_{4m}(x) + i Y_{4m}(x), \quad x >0.$$

(For $x>0$, $J_{\nu}(xe^{i \pi}) = e^{i \pi \nu} J_{\nu}(x)$ and $Y_{\nu}(xe^{i \pi})= e^{- i \pi \nu} Y_{\nu}(x) + 2i \cos(\pi \nu) J_{\nu}(x)$.)

Therefore, since $\frac{x}{x^{2}+ \sigma^{2}} $ is an odd function, we have $$\int_{0}^{\infty} \frac{x}{x^{2}+\sigma^{2}} \, J_{4m}(x) \, \mathrm dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{x}{x^{2}+ \sigma^{2}} \, \Re \left( H_{4m}^{(1)} (x) \right) \, \mathrm dx, $$ where the integration form $-\infty$ to $0$ is on the upper side of the branch cut.

From the series expansion of $Y_{\nu}(z)$ about $z=0$, it becomes apparent that the series expansion of $$\frac{z}{z^{2}+\sigma^{2}} \, H_{4m}^{(1)}(z) $$ about $z=0$ is of the form $$\sum_{k=m}^{-1}a_{2k+1} z^{2k+1} + h(z), $$ where $h(z)$ is a function such that $\lim_{z \to 0} h(z) = 0$.

Since there are no even negative power terms in the expansion, the Cauchy principal value of $$\int_{-\infty}^{\infty} \frac{x}{x^{2}+\sigma^{2}} \, H_{4m}^{(1)}(x) \, \mathrm dx $$ exists, and we can write $$\int_{0}^{\infty} \frac{x}{x^{2}+\sigma^{2}} \, J_{4m}(x) \, \mathrm dx = \frac{1}{2} \, \Re \, \operatorname{PV} \int_{-\infty}^{\infty}\frac{x}{x^{2}+\sigma^{2}} \, H_{4m}^{(1)}(x) \, \mathrm dx.$$

Now let's integrate the function $$f(z) = \frac{z}{z^{2}+\sigma^{2}} \, H_{4m}^{(1)}(z) $$ around a contour that consists of the upper side of the branch cut from $-R$ to $-\epsilon$, a small clockwise-oriented semicircle of radius $\epsilon$ about the origin ($C_{\epsilon}$), the positive real axis from $\epsilon$ to $R$, and the upper half of the circle $|z|=R$.

The asymptotic form of $H_{4m}^{(1)}(z)$ for large argument tells us that the integral vanishes on the upper half of the circle $|z|=R$ as $R \to \infty$ since $$|f(z)| \sim \sqrt{\frac{2}{\pi }} \frac{e^{- \Im(z)}}{|z|^{3/2}}$$ as $|z| \to \infty$ in the upper half plane.

Then letting $\epsilon \to 0$, we have $$\operatorname{PV} \int_{-\infty}^{\infty} \frac{x}{x^{2}+ \sigma^{2}} H_{4m}^{(1)} (x) \, \mathrm dx + \lim_{\epsilon \to 0} \int_{C_{\epsilon}} f(z) \, \mathrm dz = 2 \pi i \operatorname{Res}[f(z),i \sigma] \overset{\diamondsuit}{=} 2 K_{4m}(\sigma). $$

( $\diamondsuit$ See entry 10.27.8 and my previous answer.)

As stated previously, the series expansion of $f(z)$ at $z=0$ is of the form $$\sum_{k=m}^{-1}a_{2k+1} z^{2k+1} + h(z), $$ where $h(z)$ is a function such that $\lim_{z \to 0} h(z) = 0 $.

This is similar to the form of the expansion in this question, the difference being that $h(z)$ is not necessarily analytic at $z=0$.

However, the result from that question still holds. That is, as $\epsilon \to 0$, the integral $\int_{C_{\epsilon}} f(z) \, \mathrm dz $ goes to $-i \pi$ times the coefficient of the $\frac{1}{z}$ term of $f(z)$.

For simplicity, I'm going to refer to the coefficient as the residue even though it's technically not a residue since the series is not a Laurent series.

So we have $$ \operatorname{PV} \int_{-\infty}^{\infty} \frac{x}{x^{2}+ \sigma^{2}} H_{4m}^{(1)} (x) \, \mathrm dx - i \pi \operatorname{Res} \left[f(z), 0 \right] = 2 K_{4m}(\sigma), $$

where $$ \begin{align} \operatorname*{Res}_{z=0} f(z) &= \operatorname*{Res}_{z=0} \frac{z}{z^{2}+\sigma^{2}} \, H_{4m}^{(1)} (z) \\ &= \frac{1}{2} \, \operatorname*{Res}_{z=0} \left( \frac{1}{z+ i \sigma} H_{4m}^{(1)}(z) + \frac{1}{z-i\sigma } H_{4m}^{(1)}(z) \right) \\ &= \frac{1}{2} \left( -\operatorname*{Res}_{z=0} \frac{1}{ - i\sigma-z } H_{4m}^{(1)}(z) - \operatorname*{Res}_{z=0} \frac{1}{i \sigma-z} H_{4m}^{(1)}(z)\right). \end{align} $$

From the answers to this question, we know that if a function $p(z)$ is analytic at $z=w$ with a pole at $z=a$, then the residue of $\frac{p(z)}{w-z}$ at $z=a$ is the principle part of the Laurent series expansion of $p(z)$ about $z=a$ evaluated at $z=w$.

It would seem that we can generalize this theorem: If a function $p(z)$ is analytic at $z=w$ and has a series expansion at $z=a$ of the form $p(z) = \sum_{k=-n}^{-1} c_{k} (z-a)^{k} + r(z)$, where $\lim_{z \to a} r(z) =0$, then the "residue" of $\frac{p(z)}{w-z}$ at $z=a$ is the principal part of the series expansion of $p(z)$ about $z=a$ evaluated at $z=w$.

Using $\mathcal{P}$ to denote the principal part of the series expansion at the origin, we then have

$$\begin{align} \operatorname*{Res}_{z=0} f(z) &= \frac{1}{2} \left(-\mathcal{P} H_{4m}^{(1)}(-i \sigma) -\mathcal{P} \, H_{4m}^{(1)}(i \sigma)\right) \\ &\overset{\spadesuit}{=} \frac{1}{2} \left(-\mathcal{P} \left(2I_{4m}(\sigma) - \frac{2i}{\pi} \, K_{4m}(\sigma) \right) - \mathcal{P} \, \left( -\frac{2i}{\pi} \, K_{4m}(\sigma) \right)\right) \\ &\overset{\clubsuit}{=} \frac{2i}{\pi} \, \mathcal{P} \, K_{4m}(\sigma). \end{align}$$

($\spadesuit$ $H_{\nu}^{(1)}(e^{- i \pi/2} \sigma ) = 2e^{-i \pi \nu/2} I_{\nu}(\sigma)- \frac{2i}{\pi} e^{i \pi \nu/2} K_{\nu}(\sigma)$, which can be derived from the identity $H_{\nu}^{(1)}(z) = \frac{J_{-\nu}(z) - e^{- i \pi \nu}J_{\nu}(z)}{i \sin(\pi \nu)} $.)

($\clubsuit$ $I_{4m}(\sigma)$ doesn't have a principal part.)

Therefore, $$\operatorname{PV} \int_{-\infty}^{\infty} \frac{x}{x^{2}+ \sigma^{2}} H_{4m}^{(1)} (x) \, \mathrm dx - i \pi \left(\frac{2i}{\pi} \, \mathcal{P} \, K_{4m}(\sigma) \right)= 2 K_{4m}(\sigma), $$ and

$$\begin{align} \int_{0}^{\infty} \frac{x}{x^{2}+\sigma^{2}} \, J_{4m}(x) \, \mathrm dx &= \frac{1}{2} \, \Re \, \operatorname{PV} \int_{-\infty}^{\infty}\frac{x}{x^{2}+\sigma^{2}} \, H_{4m}^{(1)}(x) \, \mathrm dx \\ &= \frac{1}{2} \left(2 K_{4m}(\sigma) - 2 \, \mathcal{P} \, K_{4m} (\sigma)\right) \\ &= K_{4m}(\sigma) - \mathcal{P} \, K_{4m}(\sigma) \\&= K_{4m}(\sigma)- \frac{1}{2} \left(\frac{\sigma}{2} \right)^{-4m} \sum_{k=0}^{2m-1} \frac{(4m-k-1)!}{k!} \left(- \frac{\sigma^{2}}{4} \right)^{k} . \end{align}$$

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We also have

$$\int_{0}^{\infty} \frac{J_{4m+1}(x)}{x^{2} + \sigma^{2}} \, \mathrm dx = \frac{1}{\sigma} \left(-K_{4m+1}(\sigma)+ \mathcal{P} \, K_{4m+1}(\sigma) \right), $$

$$\int_{0}^{\infty} \frac{x}{x^{2}+\sigma^{2}} \, J_{4m+2}(x) \, \mathrm dx = - K_{4m+2}(\sigma) + \mathcal{P} \, K_{4m+2} (\sigma),$$

and

$$\int_{0}^{\infty} \frac{J_{4m+3}(x)}{x^{2} + \sigma^{2}} \, \mathrm dx = \frac{1}{\sigma} \left(K_{4m+3}(\sigma)- \mathcal{P} \, K_{4m+3}(\sigma) \right). $$

Answer 2

It can be checked that $$ \phi_m = K_{4m}(\sigma) + \frac{1}{\sigma^{4m}} \left( S_m - D_m \right), $$ where $$ S_m = \sum_{k=0}^{m-1} \frac{4^{2k+1} (2m+2k)!}{2 (2m-2k-1)!} \sigma^{4m-4k-2}, $$ and $$ D_m = \sum_{k=0}^{m-1} \frac{4^{2k+2} (2m+2k+1)!}{2 (2m-2k-2)!} \sigma^{4m-4k-4}. $$