Special Functions – Bessel functions – Integrals of Bessel functions

Previously, we have seen that \(\displaystyle\frac{d}{dz}\left[z^{\nu}J_{\nu}(z)\right]=z^{\nu}J_{\nu-1}(z)\), which implies that for \(x>0\) and \(\nu>-1\) we have

\[\int_0^x t^{\nu+1}J_{\nu}(t)\,dt=x^{\nu+1}J_{\nu+1}(x).\]

This can also be shown directly:

\begin{align*} \int_0^x t^{\nu+1}J_{\nu}(t)\,dt&=\sum_{k=0}^{\infty}\frac{(-1)^k}{\Gamma(\nu+k+1)\,k!}\frac{1}{2^{\nu+2k}}\int_0^xt^{2\nu+2k+1}\,dt =\sum_{k=0}^{\infty}\frac{(-1)^k}{\Gamma(\nu+k+1)\,k!}\frac{1}{2^{\nu+2k}}\frac{x^{2\nu+2k+2}}{2\nu+2k+2}\\[2.5mm] &=x^{\nu+1}\left(\frac{x}{2}\right)^{\nu+1}\sum_{k=0}^{\infty}\frac{(-1)^k}{\Gamma(\nu+k+2)\,k!}\left(\frac{x}{2}\right)^{2k} =x^{\nu+1}J_{\nu+1}(x). \end{align*}

Now we consider the Laplace transform of \(t^{\mu}J_{\nu}(\alpha t)\):

\[\int_0^{\infty}e^{-st}t^{\mu}J_{\nu}(\alpha t)\,dt=\frac{\Gamma(\mu+\nu+1)\alpha^{\nu}}{\Gamma(\nu+1)2^{\nu}s^{\mu+\nu+1}}\, {}_2F_1\left(\genfrac{}{}{0pt}{}{\frac{\mu+\nu+1}{2},\,\frac{\mu+\nu+2}{2}}{\nu+1}\,;\,-\frac{\alpha^2}{s^2}\right),\quad s>0.\]

This can be proved as follows:

\[\int_0^{\infty}e^{-st}t^{\mu}J_{\nu}(\alpha t)\,dt=\sum_{k=0}^{\infty}\frac{(-1)^k}{\Gamma(\nu+k+1)\,k!}\frac{\alpha^{\nu+2k}}{2^{\nu+2k}} \underbrace{\int_0^{\infty}e^{-st}t^{\mu+\nu+2k}\,dt}_{=\displaystyle\frac{\Gamma(\mu+\nu+2k+1)}{s^{\mu+\nu+2k+1}}} =\frac{\Gamma(\mu+\nu+1)\alpha^{\nu}}{\Gamma(\nu+1)2^{\nu}s^{\mu+\nu+1}}\sum_{k=0}^{\infty}\frac{(\mu+\nu+1)_{2k}}{(\nu+1)_k\,k!} \frac{(-1)^k\alpha^{2k}}{2^{2k}s^{2k}}.\]

Now we use

\[(\mu+\nu+1)_{2k}=2^{2k}\left(\frac{\mu+\nu+1}{2}\right)_k\left(\frac{\mu+\nu+2}{2}\right)_k,\quad k=0,1,2,\ldots\]

to obtain for \(s>0\)

\[\int_0^{\infty}e^{-st}t^{\mu}J_{\nu}(\alpha t)\,dt=\frac{\Gamma(\mu+\nu+1)\alpha^{\nu}}{\Gamma(\nu+1)2^{\nu}s^{\mu+\nu+1}}\sum_{k=0}^{\infty} \frac{\left(\frac{\mu+\nu+1}{2}\right)_k\left(\frac{\mu+\nu+2}{2}\right)_k}{(\nu+1)_k\,k!}\frac{(-1)^k\alpha^{2k}}{s^{2k}} =\frac{\Gamma(\mu+\nu+1)\alpha^{\nu}}{\Gamma(\nu+1)2^{\nu}s^{\mu+\nu+1}}\, {}_2F_1\left(\genfrac{}{}{0pt}{}{\frac{\mu+\nu+1}{2},\,\frac{\mu+\nu+2}{2}}{\nu+1}\,;\,-\frac{\alpha^2}{s^2}\right).\]

The cases \(\mu=\nu\) and \(\mu=\nu+1\) are of special interest. In these cases one of the numerator parameters of the \({}_2F_1\) hypergeometric function equals the denominator parameter. This implies that the \({}_2F_1\) reduces to a \({}_1F_0\) which is a binomial series. For \(\mu=\nu\) we obtain

\[\int_0^{\infty}e^{-st}t^{\nu}J_{\nu}(\alpha t)\,dt=\frac{\Gamma(2\nu+1)\alpha^{\nu}}{\Gamma(\nu+1)2^{\nu}s^{2\nu+1}}\, {}_1F_0\left(\genfrac{}{}{0pt}{}{\nu+\frac{1}{2}}{-}\,;\,-\frac{\alpha^2}{s^2}\right) =\frac{\Gamma(2\nu+1)\alpha^{\nu}}{\Gamma(\nu+1)2^{\nu}s^{2\nu+1}}\,\left(1+\frac{\alpha^2}{s^2}\right)^{-\nu-\frac{1}{2}} =\frac{\Gamma(2\nu+1)\alpha^{\nu}}{\Gamma(\nu+1)2^{\nu}(s^2+\alpha^2)^{\nu+\frac{1}{2}}}\]

and for \(\mu=\nu+1\) we obtain

\[\int_0^{\infty}e^{-st}t^{\nu+1}J_{\nu}(\alpha t)\,dt=\frac{\Gamma(2\nu+2)\alpha^{\nu}}{\Gamma(\nu+1)2^{\nu}s^{2\nu+2}}\, {}_1F_0\left(\genfrac{}{}{0pt}{}{\nu+\frac{3}{2}}{-}\,;\,-\frac{\alpha^2}{s^2}\right) =\frac{\Gamma(2\nu+2)\alpha^{\nu}}{\Gamma(\nu+1)2^{\nu}s^{2\nu+2}}\,\left(1+\frac{\alpha^2}{s^2}\right)^{-\nu-\frac{3}{2}} =\frac{\Gamma(2\nu+2)\alpha^{\nu}\,s}{\Gamma(\nu+1)2^{\nu}(s^2+\alpha^2)^{\nu+\frac{3}{2}}}.\]

Using Pfaff's transformation formula we have

\[{}_2F_1\left(\genfrac{}{}{0pt}{}{\frac{\mu+\nu+1}{2},\,\frac{\mu+\nu+2}{2}}{\nu+1}\,;\,-\frac{\alpha^2}{s^2}\right) =\left(1+\frac{\alpha^2}{s^2}\right)^{-\frac{1}{2}\mu-\frac{1}{2}\nu-\frac{1}{2}} {}_2F_1\left(\genfrac{}{}{0pt}{}{\frac{\mu+\nu+1}{2},\,\frac{\nu-\mu}{2}}{\nu+1}\,;\,\frac{\alpha^2}{s^2+\alpha^2}\right).\]

This implies that we also have

\[\int_0^{\infty}e^{-st}t^{\mu}J_{\nu}(\alpha t)\,dt=\frac{\Gamma(\mu+\nu+1)\alpha^{\nu}}{\Gamma(\nu+1)2^{\nu}(s^2+\alpha^2)^{\frac{1}{2}\mu+\frac{1}{2}\nu+\frac{1}{2}}}\, {}_2F_1\left(\genfrac{}{}{0pt}{}{\frac{\mu+\nu+1}{2},\,\frac{\nu-\mu}{2}}{\nu+1}\,;\,\frac{\alpha^2}{s^2+\alpha^2}\right),\quad s>0.\]

Now, using Gauss's summation formula and Legendre's duplication formula, the limit \(s\downarrow0\) leads to

\[\int_0^{\infty}t^{\mu}J_{\nu}(\alpha t)\,dt=\frac{\Gamma(\mu+\nu+1)\alpha^{\nu}}{\Gamma(\nu+1)2^{\nu}\alpha^{\mu+\nu+1}}\, {}_2F_1\left(\genfrac{}{}{0pt}{}{\frac{\mu+\nu+1}{2},\,\frac{\nu-\mu}{2}}{\nu+1}\,;\,1\right) =\frac{\Gamma(\mu+\nu+1)}{\Gamma(\nu+1)2^{\nu}\alpha^{\mu+1}}\, \frac{\Gamma(\nu+1)\Gamma(\frac{1}{2})}{\Gamma\left(\frac{\nu-\mu+1}{2}\right)\Gamma\left(\frac{\mu+\nu+2}{2}\right)} =\frac{2^{\mu}\Gamma\left(\frac{\mu+\nu+1}{2}\right)}{\alpha^{\mu+1}\Gamma\left(\frac{\nu-\mu+1}{2}\right)}.\]

Another interesting integral is

\[\int_0^{\infty}t^{\mu-1}e^{-\rho^2t^2}J_{\nu}(st)\,dt=\frac{\Gamma\left(\frac{\mu+\nu}{2}\right)}{2^{\nu+1}\Gamma(\nu+1)}\cdot \frac{s^{\nu}}{\rho^{\mu+\nu}}\cdot{}_1F_1\left(\genfrac{}{}{0pt}{}{(\mu+\nu)/2}{\nu+1}\,;\,-\frac{s^2}{4\rho^2}\right),\quad\text{Re}(\mu+\nu)>0.\]

It can be shown that the integral converges for \(\text{Re}(\mu+\nu)>0\). Now we obtain for \(\rho\neq0\)

\[\int_0^{\infty}t^{\mu-1}e^{-\rho^2t^2}J_{\nu}(st)\,dt =\sum_{k=0}^{\infty}\frac{(-1)^k}{\Gamma(\nu+k+1)\,k!}\cdot\frac{s^{\nu+2k}}{2^{\nu+2k}} \cdot\int_0^{\infty}t^{\mu+\nu+2k-1}e^{-\rho^2t^2}\,dt.\]

Using the substitution \(\rho^2t^2=u\) we find that

\begin{align*} \int_0^{\infty}t^{\mu+\nu+2k-1}e^{-\rho^2t^2}\,dt &=\rho^{-\mu-\nu-2k+1}\int_0^{\infty}u^{(\mu+\nu+2k-1)/2}e^{-u}\frac{du}{2\rho\sqrt{u}}\\[2.5mm] &=\frac{1}{2}\,\rho^{-\mu-\nu-2k}\int_0^{\infty}u^{(\mu+\nu+2k-2)/2}e^{-u}\,du\\[2.5mm] &=\frac{1}{2}\,\rho^{-\mu-\nu-2k}\,\Gamma\left(\frac{\mu+\nu+2k}{2}\right). \end{align*}

Hence we have

\begin{align*} \int_0^{\infty}t^{\mu-1}e^{-\rho^2t^2}J_{\nu}(st)\,dt &=\frac{1}{2}\,\rho^{-\mu-\nu}\, \sum_{k=0}^{\infty}\frac{(-1)^k\,\Gamma\left(\frac{\mu+\nu}{2}+k\right)}{\Gamma(\nu+k+1)\,k!} \cdot\frac{s^{\nu+2k}}{2^{\nu+2k}\,\rho^{2k}}\\[2.5mm] &=\frac{\Gamma\left(\frac{\mu+\nu}{2}\right)}{2\,\rho^{\mu+\nu}\,\Gamma(\nu+1)}\cdot \frac{s^{\nu}}{2^{\nu}}\cdot\sum_{k=0}^{\infty}\frac{(-1)^k((\mu+\nu)/2)_k}{(\nu+1)_k\,k!}\cdot\frac{s^{2k}}{2^{2k}\,\rho^{2k}}\\[2.5mm] &=\frac{\Gamma\left(\frac{\mu+\nu}{2}\right)}{2^{\nu+1}\Gamma(\nu+1)}\cdot \frac{s^{\nu}}{\rho^{\mu+\nu}}\cdot{}_1F_1\left(\genfrac{}{}{0pt}{}{(\mu+\nu)/2}{\nu+1}\,;\,-\frac{s^2}{4\rho^2}\right). \end{align*}

The special case \(\mu=\nu+2\) is of special interest: in that case we have \((\mu+\nu)/2=\nu+1\). This implies that the \({}_1F_1\) reduces to a \({}_0F_0\) which is an exponential function. The result is

\[\int_0^{\infty}t^{\nu+1}e^{-\rho^2t^2}J_{\nu}(st)\,dt =\frac{s^{\nu}}{(2\rho^2)^{\nu+1}}\cdot e^{-s^2/4\rho^2},\quad\text{Re}(\nu)>-1.\]

The Hankel transform

The Hankel transform of a function \(f\) is defined by

\[F(s)=\int_0^{\infty}tf(t)J_{\nu}(st)\,dt\]

for functions \(f\) for which the integral converges. The inversion formula is given by

\[f(t)=\int_0^{\infty}sF(s)J_{\nu}(st)\,ds.\]

This pair of integrals is called a Hankel pair of order \(\nu\).

---

Last modified on September 30, 2021