Integrating $\int_0^\infty\frac{u^2}{\sqrt{u^2+a^2}}J_1(ru)e^{-z\sqrt{u^2+a^2}}du$

Question

I came across the following non-trivial improper integral while I was elaborating on a fluid mechanical problem involving porous media: $$ f(r,z,a) = \int_0^\infty \frac{u^2}{\sqrt{u^2+a^2}} J_1(ru) e^{-z\sqrt{u^2+a^2}} \mathrm{d}u , $$ wherein $r$, $z$, and $a$ are positive real numbers. Physically, $a$ is a measure of the resistance of the porous medium against flow (impermeability coefficient.) Using a different solution route of the original problem (which does not involve integral transforms), I was able to find that $f(r,z,a)$ is actually given by $$ f(r,z,a) = \frac{r(1+aR)}{R^3} e^{-aR} , $$ with $R = \sqrt{r^2+z^2}$. I can check that this expression is correct by numerical substitution but I have no clue how this could be rigorously proven.

What I tried is to use Poisson's representation of the Bessel function but this does not lead to the desired result. It would be highly appreciated if someone here could be of help to clarify how this could be the case. Thank you!

In particular, it can readily be checked that in the limit case $a=0$ both expressions lead to the same value, namely, $$ f(r,z,0) = \frac{r}{\left( r^2+z^2\right)^{3/2}} . $$

Answer 1

Differentiating the integral formula $$\int_{0}^{\infty} \exp \left(-\alpha x^{2}- \frac{\beta}{4x^{2}}\right) \, \mathrm dx = \frac{\sqrt{\pi}}{2\sqrt{\alpha}} \, e^{-\sqrt{\alpha \beta}} \, , \quad (\alpha, \beta > 0)$$ with respect to the parameter $\beta$ twice, we get $$ \int_{0}^{\infty} \frac{1}{x^{4}} \, \exp \left(-\alpha x^{2}- \frac{\beta}{4x^{2}} \right) \, \mathrm dx = 2 \sqrt{\pi} \, \frac{1+\sqrt{\alpha \beta}}{ \beta^{3/2}} \, e^{-\sqrt{\alpha \beta}}.$$

(Differentiation under the integral sign is justified since for any positive $\delta < \beta$, $ \frac{1}{x^{2}}\exp \left(-\alpha x^{2}- \frac{\beta}{4x^{2}}\right) $ is dominated by the integrable function $\frac{1}{x^{2}}\exp \left(-\alpha x^{2}- \frac{\delta}{4x^{2}}\right) $, and $\frac{1}{x^{4}} \, \exp \left(-\alpha x^{2}- \frac{\beta}{4x^{2}}\right) $ is dominated by the integrable function $\frac{1}{x^{4}} \, \exp \left(-\alpha x^{2}- \frac{\delta}{4x^{2}}\right) $.)

We also have $$\begin{align} \int_{0}^{\infty} u^{2} J_{1}(ru) e^{-t^{2}u^{2}} \, \mathrm du &= \int_{0}^{\infty} u^{2} e^{-t^{2}u^{2}} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!(m+1)!} \left(\frac{ru}{2} \right)^{2m+1} \, \mathrm du \\ &= \sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!(m+1)!} \left(\frac{r}{2} \right)^{2m+1} \int_{0}^{\infty} u^{2m+3} e^{-t^{2}u^{2}} \, \mathrm du \\ &= \sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!(m+1)!} \left(\frac{r}{2} \right)^{2m+1} \frac{(m+1)!}{2t^{2m+4}} \\ &= \frac{r}{4t^{4}}\sum_{m=0}^{\infty} \frac{(-1)^{m}}{m!} \left(\frac{r^{2}}{4t^{2}} \right)^{m} \\ &= \frac{r}{4t^{4}} \, \exp \left(-\frac{r^{2}}{4t^{2}} \right). \end{align}$$

Therefore,

$ \begin{align} \int_{0}^{\infty} \, \frac{u^2}{\sqrt{u^2+a^2}} \, J_1(ru) \, e^{-z \sqrt{u^{2}+a^{2}}} \, \mathrm du &= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} u^{2} J_{1}(ru) \int_{0}^{\infty}\exp \left(-(u^{2}+a^{2})t^{2}-\frac{z^{2}}{4t^{2}} \right) \, \mathrm dt \, \mathrm du \\ &= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \exp \left(-a^{2}t^{2}- \frac{z^{2}}{4t^{2}} \right) \int_{0}^{\infty} u^{2} J_{1}(ru) e^{-t^{2}u^{2}} \, \mathrm du \, \mathrm dt \\ &= \frac{r}{2\sqrt{\pi}} \int_{0}^{\infty} \frac{1}{t^{4}} \, \exp \left(-a^{2}t^{2} - \frac{r^{2}+z^{2}}{4t^{2}} \right) \, \mathrm dt \\ &= \frac{r}{2\sqrt{\pi}} 2 \sqrt{\pi} \, \frac{1+a\sqrt{r^{2}+z^{2}}}{(r^{2}+z^{2})^{3/2}} e^{-a\sqrt{r^{2}+z^{2}}} \\ &= r \, \frac{1+a \sqrt{r^{2}+z^{2}}}{(r^{2}+z^{2})^{3/2}} \, e^{-a \sqrt{r^{2}+z^{2}}} . \end{align}$

Answer 2

$$\int_0^\infty x\operatorname{J}_0(rx)\frac{e^{-z\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}dx=\frac{e^{-a\sqrt{r^2+z^2}}}{\sqrt{r^2+z^2}}$$ $$\overset{\large \frac{d}{dr}}\Rightarrow \int_0^\infty x^2\operatorname{J}_1(rx)\frac{e^{-z\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}dx=\frac{r\left(1+a\sqrt{r^2+z^2}\right)}{(r^2+z^2)^{3/2}}e^{-a\sqrt{r^2+z^2}}$$

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In order to obtain the first result we can start by using the following identity: $$\int_0^\infty \frac{\cos t}{t^2+a^2}dt=\frac{\pi}{2}\frac{e^{-a}}{a}$$ $$\Rightarrow \mathcal J=\int_0^\infty x\operatorname{J}_0(rx)\color{blue}{\frac{e^{-z\sqrt{x^2+a^2}}}{\sqrt{x^2+a^2}}}dx$$ $$=\color{blue}{\frac{2z}{\pi}}\int_0^\infty x\operatorname{J}_0(rx) \color{blue}{\int_0^\infty \frac{\cos t}{t^2+z^2(x^2+a^2)}dt}dx$$ $$=\frac{2}{z\pi} \int_{0}^\infty \cos t\int_0^\infty \frac{\color{red}{x}\color{green}{\operatorname{J}_0(rx)}}{\color{red}{t^2/z^2+a^2+x^2}}dxdt$$

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Now, for the inner integral we will employ the following property of the Laplace transform: $$\int_0^\infty f(x)g(x) dx = \int_0^\infty \mathcal L(f(x))\mathcal L^{-1}(g(x))dx$$ $$\Rightarrow \mathcal J=\frac{2}{z\pi} \int_{0}^\infty \cos t\int_0^\infty\frac{\color{red}{\cos\left(x\sqrt{t^2/z^2+a^2}\right)}}{\color{green}{\sqrt{r^2+x^2}}}dxdt$$ $$\overset{x\sqrt{t^2/z^2+a^2}\to x}=\frac{2}{z\pi} \int_{0}^\infty \cos t\int_0^\infty \frac{\cos x}{\color{blue}{\sqrt{r^2(t^2/z^2+a^2)+x^2}}}dxdt$$

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Next, we will get rid of the square root from the denominator by using: $$\int_0^\infty e^{-ay^2}dy = \frac{\sqrt \pi}{2\sqrt a}$$

$$\Rightarrow \mathcal J=\frac{\color{blue}{2}\cdot 2}{z\pi\color{blue}{\sqrt \pi}} \int_{0}^\infty \cos t\int_0^\infty \cos x \color{blue}{\int_0^\infty \exp\left({-\left(r^2t^2/z^2+r^2a^2+x^2\right)y^2}\right)dy}dxdt$$

$$=\frac{4}{z\pi\sqrt \pi} \int_{0}^\infty \cos t\int_0^\infty \exp\left({-\left(r^2t^2/z^2+r^2a^2\right)y^2}\right)\color{red}{\int_0^\infty \cos x e^{-x^2y^2}dx}dydt$$

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Finally, to get rid of the cosine terms, we will use the following identity (twice in a row): $$\int_0^\infty \cos x e^{-y^2 x^2}=\frac{\sqrt \pi}{2y}e^{-1/(4y^2)}$$ $$\Rightarrow \mathcal J=\frac{2}{z\pi}\int_{0}^\infty \cos t\int_0^\infty \exp\left({-\left(r^2t^2/z^2+r^2a^2\right)y^2}\color{red}{-\frac{1}{4y^2}}\right)\color{red}{\frac{1}{y}}dydt$$ $$=\frac{2}{z\pi}\int_{0}^\infty\exp\left({-r^2a^2y^2}-\frac{1}{4y^2}\right)\frac{1}{y}\color{blue}{\int_0^\infty \cos t e^{-r^2y^2t^2/z^2}dt}dy$$ $$=\frac{1}{\sqrt \pi}\int_{0}^\infty\exp\left({-r^2a^2y^2}-\frac{1}{4y^2}\color{blue}{-\frac{z^2}{4r^2y^2}}\right)\frac{1}{y}\color{blue}{\frac{1}{ry}}dy$$ $$\overset{\large y\to\frac{1}{y}}=\frac{1}{r\sqrt \pi}\int_0^\infty \exp\left(-\frac{y^2}{4}\left(1+\frac{z^2}{r^2}\right)-\frac{r^2a^2}{y^2}\right) dy=\frac{e^{-a\sqrt{r^2+z^2}}}{\sqrt{r^2+z^2}}$$ Where the last result was obtained using Glasser's Master Theorem.