dourtuntellorvl

2022-06-30

We know the Hankel transform of order 0 is defined as

${F}_{0}(k)={\int}_{0}^{\mathrm{\infty}}f(r){J}_{0}(kr)\phantom{\rule{thinmathspace}{0ex}}r\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}r}.$

In this regard, I am now trying to calculate the Hankel transform of the function

$f(r)=\frac{{e}^{-a\sqrt{{r}^{2}+{z}^{2}}}}{\sqrt{{r}^{2}+{z}^{2}}},$

with $a\in \mathbb{R}$ . Unfortunately, I was only able to obtain the solution for $a=0$ . Any thoughts on how to solve this?

${F}_{0}(k)={\int}_{0}^{\mathrm{\infty}}f(r){J}_{0}(kr)\phantom{\rule{thinmathspace}{0ex}}r\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}r}.$

In this regard, I am now trying to calculate the Hankel transform of the function

$f(r)=\frac{{e}^{-a\sqrt{{r}^{2}+{z}^{2}}}}{\sqrt{{r}^{2}+{z}^{2}}},$

with $a\in \mathbb{R}$ . Unfortunately, I was only able to obtain the solution for $a=0$ . Any thoughts on how to solve this?

Paxton James

Beginner2022-07-01Added 25 answers

$\begin{array}{}\text{(1)}& F(k,a,z):={\int}_{0}^{\mathrm{\infty}}\frac{\mathrm{exp}{\textstyle (}\phantom{\rule{negativethinmathspace}{0ex}}-a\sqrt{{r}^{2}+{z}^{2}}{\textstyle )}}{\sqrt{{r}^{2}+{z}^{2}}}{J}_{0}(k\text{}r)\text{}rdr=\frac{\mathrm{exp}{\textstyle (}\phantom{\rule{negativethinmathspace}{0ex}}-z\sqrt{{a}^{2}+{k}^{2}}{\textstyle )}}{\sqrt{{a}^{2}+{k}^{2}}}\end{array}$

$\frac{\mathrm{exp}{\textstyle (}\phantom{\rule{negativethinmathspace}{0ex}}-a\sqrt{{r}^{2}+{z}^{2}}{\textstyle )}}{a\sqrt{{r}^{2}+{z}^{2}}}={\int}_{1}^{\mathrm{\infty}}{e}^{-a\text{}r\text{}t}{J}_{0}(a\text{}z\sqrt{{t}^{2}-1})dt$

Insert into left-hand side of (1) and interchange $\int .$ The innermost integral has a closed form,

${\int}_{0}^{\mathrm{\infty}}{e}^{-a\text{}r\text{}t}{J}_{0}(k\text{}r)rdr=\frac{a\text{}t}{{\textstyle (}(at{)}^{2}+{k}^{2}{{\textstyle )}}^{3/2}}$

which Mathematica knows. So we now have

$\begin{array}{}\text{(2a)}& F(k,a,z)={a}^{2}{\int}_{1}^{\mathrm{\infty}}{J}_{0}(a\text{}z\sqrt{{t}^{2}-1})\frac{t\text{}dt}{{\textstyle (}(at{)}^{2}+{k}^{2}{{\textstyle )}}^{3/2}}\end{array}$

$\begin{array}{}\text{(2b)}& =\frac{1}{2a}{\int}_{1}^{\mathrm{\infty}}\frac{{J}_{0}(a\text{}z\sqrt{t-1})}{{\textstyle (}t+(k/a{)}^{2}{{\textstyle )}}^{3/2}}dt=\frac{1}{2a}{\int}_{0}^{\mathrm{\infty}}\frac{{J}_{0}(a\text{}z\sqrt{t})}{{\textstyle (}t+p{{\textstyle )}}^{3/2}}dt\text{with}p=1+(k/a{)}^{2}\end{array}$

where from line 2a to 2b we substituted ${t}^{2}\to t$ and in the last step the limits of the integrand have been shifted with a subsequent change of the parameter. Now use the integral relationship

${\int}_{0}^{\mathrm{\infty}}{J}_{0}(c\sqrt{u})(u+p{)}^{-3/2}du=2\frac{\mathrm{exp}{\textstyle (}\phantom{\rule{negativethinmathspace}{0ex}}-c\sqrt{p}{\textstyle )}}{\sqrt{p}}.$

Mathematica knows this integral with $c=1,$ and it is easy to work in the scaling factor. Algebra completes the proof.

$\frac{\mathrm{exp}{\textstyle (}\phantom{\rule{negativethinmathspace}{0ex}}-a\sqrt{{r}^{2}+{z}^{2}}{\textstyle )}}{a\sqrt{{r}^{2}+{z}^{2}}}={\int}_{1}^{\mathrm{\infty}}{e}^{-a\text{}r\text{}t}{J}_{0}(a\text{}z\sqrt{{t}^{2}-1})dt$

Insert into left-hand side of (1) and interchange $\int .$ The innermost integral has a closed form,

${\int}_{0}^{\mathrm{\infty}}{e}^{-a\text{}r\text{}t}{J}_{0}(k\text{}r)rdr=\frac{a\text{}t}{{\textstyle (}(at{)}^{2}+{k}^{2}{{\textstyle )}}^{3/2}}$

which Mathematica knows. So we now have

$\begin{array}{}\text{(2a)}& F(k,a,z)={a}^{2}{\int}_{1}^{\mathrm{\infty}}{J}_{0}(a\text{}z\sqrt{{t}^{2}-1})\frac{t\text{}dt}{{\textstyle (}(at{)}^{2}+{k}^{2}{{\textstyle )}}^{3/2}}\end{array}$

$\begin{array}{}\text{(2b)}& =\frac{1}{2a}{\int}_{1}^{\mathrm{\infty}}\frac{{J}_{0}(a\text{}z\sqrt{t-1})}{{\textstyle (}t+(k/a{)}^{2}{{\textstyle )}}^{3/2}}dt=\frac{1}{2a}{\int}_{0}^{\mathrm{\infty}}\frac{{J}_{0}(a\text{}z\sqrt{t})}{{\textstyle (}t+p{{\textstyle )}}^{3/2}}dt\text{with}p=1+(k/a{)}^{2}\end{array}$

where from line 2a to 2b we substituted ${t}^{2}\to t$ and in the last step the limits of the integrand have been shifted with a subsequent change of the parameter. Now use the integral relationship

${\int}_{0}^{\mathrm{\infty}}{J}_{0}(c\sqrt{u})(u+p{)}^{-3/2}du=2\frac{\mathrm{exp}{\textstyle (}\phantom{\rule{negativethinmathspace}{0ex}}-c\sqrt{p}{\textstyle )}}{\sqrt{p}}.$

Mathematica knows this integral with $c=1,$ and it is easy to work in the scaling factor. Algebra completes the proof.

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