# Inverse Fourier transform of modified Bessel function

I'm relatively new to Fourier transforms so apologize in advance if this problem seems trivial.

In order to solve a second order PDE I have defined the following sine Fourier transform

$$V(r,\lambda)=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}v(r,z)\sin(\lambda z)\,\textrm{d}z.$$

By doing so I arrive at the following solution for $V$

$$V(r,\lambda)=\sqrt{\frac{2}{\pi}}\frac{1}{\lambda}\frac{I_{1}(\lambda r)}{I_{1}(\lambda)},$$

where $I_{\alpha}$ is the modified Bessel function of the first kind.

My goal is to invert the Fourier transform and obtain the solution for $v$. So far I have that

\begin{align*} v(r,z)&=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}V(r,\lambda)\sin(\lambda z)\,\textrm{d}\lambda \\ &=\frac{2}{\pi}\int_{0}^{\infty}\frac{\sin(\lambda z)}{\lambda}\frac{I_{1}(\lambda r)}{I_{1}(\lambda)}\,\textrm{d}\lambda. \end{align*}

I know that the definition of the modified Bessel function of the first kind gives

$$I_{1}(\lambda r)=\sum_{m=0}^{\infty}\frac{1}{m!\Gamma(m+2)}\left(\frac{\lambda r}{2}\right)^{2m+1}.$$

Therefore

\begin{align*} v(r,z)&=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}V(r,\lambda)\sin(\lambda z)\,\textrm{d}\lambda \\ &=\frac{2}{\pi}\int_{0}^{\infty}\frac{\sin(\lambda z)}{\lambda}\frac{\sum_{m=0}^{\infty}\frac{1}{m!\Gamma(m+2)}\left(\frac{\lambda r}{2}\right)^{2m+1}}{\sum_{m=0}^{\infty}\frac{1}{m!\Gamma(m+2)}\left(\frac{\lambda}{2}\right)^{2m+1}}\,\textrm{d}\lambda. \end{align*}

This looks horrible! Can anyone help me from here on out?

If it's any help I know that the solution should be

$$v(r,z)=2\sum_{\beta_{n}}\frac{1}{\beta_{n}}\frac{J_{1}(r\beta_{n})}{J_{0}(\beta_{n})}e^{-\beta_{n}z},$$

where $J_{\alpha}$ is the Bessel function of the first kind and $\beta_{n}$ are the zeros of $J_{1}(\beta)$.

Thanks!

• Do you suppose that $r\le1$ in order that the integral converges? – Paul Enta Oct 22 '17 at 12:35
• Yes, $r$ is such that $0\le r\le1$. – Juggler Oct 22 '17 at 14:56

A Fourier-Bessel representation for the ratio $$\frac{I_1(\lambda r)}{I_1(\lambda)}=\frac{J_1(i\lambda r)}{J_1(i\lambda)}$$ can be found in Ederlyi, Higher Transcendental functions II, 7.10.4 (56): $$\frac{I_1(\lambda r)}{I_1(\lambda)}=2\sum_{m=1}^\infty\frac{\beta_mJ_1(\beta_m r)}{\left( \beta_m^2+\lambda^2 \right)J_2(\beta_m)}$$ Then, $$v(r,z)=\frac{4}{\pi}\sum_{m=1}^\infty\frac{\beta_mJ_1(\beta_m r)}{J_2(\beta_m)}\int_{0}^{\infty}\frac{\sin(\lambda z)}{\lambda\left( \beta_m^2+\lambda^2 \right)}\,\textrm{d}\lambda$$ This integral can be evaluated easily $$\int_{0}^{\infty}\frac{\sin(\lambda z)}{\lambda\left( \beta_m^2+\lambda^2 \right)}\,\textrm{d}\lambda=\frac{\pi}{2\beta_m^2}\left( 1-e^{-\beta_mz} \right)$$ thus \begin{align} v(r,z)&=-2\sum_{m=1}^\infty\frac{J_1(\beta_m r)\left( 1-e^{-\beta_mz} \right)}{\beta_mJ_0(\beta_m)}\\ &=2\sum_{m=1}^\infty\frac{J_1(\beta_m r)e^{-\beta_mz}}{\beta_mJ_0(\beta_m)}+r \end{align} where we have used $J_0(\beta_m)=-J_2(\beta_m)$ and the Fourier Bessel representation of $r$, which can be found in Ederlyi7.10.4 (54), for example. Obtained result is slightly different from the one you quote, though both forms are very similar. Notice however that the present expression is zero when $z=0$, as expected from the integral expression.

• Thanks for the reply. This is certainly close to the stated result. I can't seem to fault any of the working either! – Juggler Oct 23 '17 at 10:45

Just realized that

$$-2\sum_{m=1}^{\infty}\frac{J_{1}(\beta_{m}r)}{\beta_{m}J_{0}(\beta_{m})}= 2\sum_{m=1}^{\infty}\frac{J_{1}(\beta_{m}r)}{\beta_{m}J_{2}(\beta_{m})}=r.$$

Therefore

$$v(r,z)=r+2\sum_{m=1}^{\infty}\frac{J_{1}(\beta_{m}r)e^{-\beta_{m}z}}{\beta_{m}J_{0}(\beta_{m})}.$$

Again, close to the stated solution but not exactly the same. Starting to think the stated solution is probably incorrect.

• This is what is written at the end of my answer, albeit with a sign error :) – Paul Enta Oct 23 '17 at 13:58
• Indeed, thanks for the help! – Juggler Oct 23 '17 at 14:10
• I edit my answer to correct the error. Again, the stated solution cannot be exact, considering that it does not cancel for $z=0$. – Paul Enta Oct 23 '17 at 14:22
• Another quick question. Erdelyi 7.10.4 (56) gives an expression for $J_{\nu}(xz)/J_{\nu}(z)$. Do you know if there is a corresponding (simple) relation for $J_{\nu-1}(xz)/J_{\nu}(z)$? I can see that the 'opposite' case $J_{\nu+1}(xz)/J_{\nu}(z)$ is stated in the book but can't seem to infer anything from this ... Thanks! – Juggler Nov 6 '17 at 11:28
• You may use the recurrence relation $J_{\nu-1}(xz)=\frac{2\nu}{xz}J_{\nu}(xz)-J_{\nu+1}(xz)$ if the 'opposite' case is known. – Paul Enta Nov 6 '17 at 13:04

Just revisited this problem and have noted a couple of errors that we both made. For completeness:

\begin{align} v(r,z)&=a^{2}\int_{0}^{\infty}\frac{{I}_{1}(\lambda r)}{{I}_{1}(\lambda)}\frac{\sin(\lambda z)}{\lambda}\,\textrm{d}\lambda \\ &=a^{2}\int_{0}^{\infty}\frac{{J}_{1}(\textrm{i}\lambda r)}{{J}_{1}(\textrm{i}\lambda)}\frac{\sin(\lambda z)}{\lambda}\,\textrm{d}\lambda \\ &=2a^{2}\int_{0}^{\infty}\sum_{n=1}^{\infty}\frac{\beta_{n}{J}_{1}(\beta_{n}r)}{(\beta_{n}^{2}+\lambda^{2}){J}_{2}(\beta_{n})}\frac{\sin(\lambda z)}{\lambda}\,\textrm{d}\lambda \\ &=2a^{2}\int_{0}^{\infty}\sum_{n=1}^{\infty}\left[\frac{{J}_{1}(\beta_{n}r)}{\beta_{n}{J}_{2}(\beta_{n})}-\frac{\lambda^{2}{J}_{1}(\beta_{n}r)}{\beta_{n}(\beta_{n}^{2}+\lambda^{2}){J}_{2}(\beta_{n})}\right]\frac{\sin(\lambda z)}{\lambda}\,\textrm{d}\lambda \\ &=r-2a^{2}\int_{0}^{\infty}\sum_{n=1}^{\infty}\frac{\lambda^{2}{J}_{1}(\beta_{n}r)}{\beta_{n}(\beta_{n}^{2}+\lambda^{2}){J}_{2}(\beta_{n})}\frac{\sin(\lambda z)}{\lambda}\,\textrm{d}\lambda \\ &=r-2a^{2}\sum_{n=1}^{\infty}\frac{{J}_{1}(\beta_{n}r)}{\beta_{n}{J}_{2}(\beta_{n})}\int_{0}^{\infty}\frac{\lambda\sin(\lambda z)}{\beta_{n}^{2}+\lambda^{2}}\,\textrm{d}\lambda \\ &=r+2\sum_{n=1}^{\infty}\frac{{J}_{1}(\beta_{n}r)e^{-\beta_{n}z}}{\beta_{n}{J}_{0}(\beta_{n})}. \end{align}