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!

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.
