In order to get the cosine transform of a Marcum Q function of order 1 (see this), I ended up with this series:
$$e^{-\alpha}\,\sum_{n=0}^{\infty}\frac{\alpha^n}{n!}\,\sum_{k=0}^n\frac{\beta^k}{k!}(-1)^k\frac{\partial^k}{\partial \beta^k}\left(\frac{\beta}{\omega^2+\beta^2}\right) $$
Does anyone have any suggestion? Thanks!
$\textbf{EDIT}$
According to the suggestion:
$$\frac{\partial^k}{\partial \beta^k}\left(\frac{\beta}{\omega^2+\beta^2}\right)=\frac{1}{2}\frac{\partial^k}{\partial \beta^k}\left(\frac{1}{\beta+i\omega}+\frac{1}{\beta-i\omega}\right)=\frac{1}{2}(-1)^k\,\Gamma(k+1)\left(\frac{1}{(\beta+i\omega)^{k+1}}+\frac{1}{(\beta-i\omega)^{k+1}}\right)$$
Therefore the initial series is:
$$\frac{1}{2}e^{-\alpha}\,\left[\frac{1}{\beta+i\omega}\sum_{n=0}^{\infty}\frac{\alpha^n}{n!}\,\sum_{k=0}^n\left(\frac{\beta}{\beta+i\omega}\right)^k+\frac{1}{\beta-i\omega}\sum_{n=0}^{\infty}\frac{\alpha^n}{n!}\,\sum_{k=0}^n\left(\frac{\beta}{\beta-i\omega}\right)^k\right]=\\= \frac{1}{2}e^{-\alpha}\,\left[\frac{1}{i\omega}\sum_{n=0}^{\infty}\frac{\alpha^n}{n!}\left(1-\left(\frac{\beta}{\beta+i\omega}\right)^{n+1}\right)-\frac{1}{i\omega}\sum_{n=0}^{\infty}\frac{\alpha^n}{n!}\left(1-\left(\frac{\beta}{\beta-i\omega}\right)^{n+1}\right)\right]=\\= \frac{1}{2i\omega}e^{-\alpha}\,\left[\frac{\beta}{\beta-i\omega}\sum_{n=0}^{\infty}\frac{\alpha^n}{n!}\left(\frac{\beta}{\beta-i\omega}\right)^{n}-\frac{\beta}{\beta+i\omega}\sum_{n=0}^{\infty}\frac{\alpha^n}{n!}\left(\frac{\beta}{\beta+i\omega}\right)^{n}\right]=\\= \frac{1}{2i\omega}e^{-\alpha}\,\left[\frac{\beta}{\beta-i\omega}\exp\left(\frac{\alpha\beta}{\beta-i\omega}\right)-\frac{\beta}{\beta+i\omega}\exp\left(\frac{\alpha\beta}{\beta+i\omega}\right)\right]=\\= \frac{e^{-\frac{\alpha\omega^2}{\omega^2+\beta^2}}}{\omega^2+\beta^2}\,\left[\frac{\beta^2}{\omega}\sin\left(\frac{\alpha\beta\omega}{\omega^2+\beta^2}\right)+\beta\cos\left(\frac{\alpha\beta\omega}{\omega^2+\beta^2}\right)\right]$$
Did I do everything right? Is there anything that I can simplify further??
