Inverse Laplace transform of $\frac{\exp(\frac{\lambda s}{1 - 2s})}{(1 - 2s)^{k/2}}$ (MGF of noncentral chi-squared distribution) I am trying to use the countour integral to calculate the inverse Laplace transform of the function
$$F(s) = \frac{\exp(\frac{\lambda s}{1 - 2s})}{(1 - 2s)^{k/2}} \hspace{1cm}\mathrm{for} \hspace{1cm} s<\frac{1}{2}$$
The above function is the MGF of the noncentral chi-squared function Wikipedia: noncentral chi-squared distribution.
From Bromwich's integral, we have
\begin{align}
f(x) = & \frac{1}{2\pi i}\int_{c - \infty i}^{c + \infty i} F(s)e^{sx}\,ds \\
     = & \frac{1}{2\pi i}\int_{c - \infty i}^{c + \infty i} \frac{\exp(\frac{(\lambda + x -2xs) s}{1 - 2s})}{(1 - 2s)^{k/2}} \, ds \\
     = & \frac{1}{2\pi i}\oint_{C} \frac{\exp(\frac{(\lambda + x -2xs) s}{1 - 2s})}{(1 - 2s)^{k/2}} \, ds - \frac{1}{2\pi i}\int_{C_{R}} \frac{\exp(\frac{(\lambda + x -2xs) s}{1 - 2s})}{(1 - 2s)^{k/2}} \, ds
\end{align}
The difficult part for me is how to choose the contour since $s$ exists in both the nominator and denominator of the exponential function. I have no idea about how to choose a contour such that the integral along $C_{R}$ vanishes.
The answer to the question is 
\begin{align}
f(x) = & 1 - Q_{k/2}(\sqrt{\lambda}, \sqrt{x}) \\
     = & e^{-\frac{\lambda}{2}} {}_{0}F_{1}(;\frac{k}{2};\frac{\lambda x}{4}) \frac{e^{-\frac{x}{2}} x^{\frac{k}{2}-1}}{2^{\frac{k}{2}}\Gamma(\frac{k}{2})}
\end{align}
with $Q_{m}(a, b)$ denoting the Marcum Q-function. That's the PDF of the noncentral Chi-square distribution. But how to get this answer from its MGF $F(s)$???
 A: It might be easier to solve by series approach instead:
$f(x)=\mathcal{L}_{-s\to x}^{-1}\left\{\dfrac{e^\frac{\lambda s}{1-2s}}{(1-2s)^\frac{k}{2}}\right\}$
$=\mathcal{L}_{-s\to x}^{-1}\left\{\sum\limits_{m=0}^\infty\dfrac{\lambda^ms^m}{m!(1-2s)^{m+\frac{k}{2}}}\right\}$
$=\mathcal{L}_{-s\to x}^{-1}\left\{\sum\limits_{m=0}^\infty\dfrac{\lambda^m(1-(1-2s))^m}{2^mm!(1-2s)^{m+\frac{k}{2}}}\right\}$
$=\mathcal{L}_{-s\to x}^{-1}\left\{\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{\lambda^mC_n^m(-1)^n(1-2s)^n}{2^mm!(1-2s)^{m+\frac{k}{2}}}\right\}$
$=\mathcal{L}_{-s\to x}^{-1}\left\{\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^n\lambda^m}{2^mn!(m-n)!(1-2s)^{m-n+\frac{k}{2}}}\right\}$
$=\mathcal{L}_{s\to x}^{-1}\left\{\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^n\lambda^m}{2^mn!(m-n)!(2s+1)^{m-n+\frac{k}{2}}}\right\}$
$=\sum\limits_{m=0}^\infty\sum\limits_{n=0}^m\dfrac{(-1)^n\lambda^mx^{m-n+\frac{k}{2}-1}e^{-\frac{x}{2}}}{2^{2m-n+\frac{k}{2}}n!(m-n)!\Gamma\left(m-n+\dfrac{k}{2}\right)}$
$=\sum\limits_{n=0}^\infty\sum\limits_{m=n}^\infty\dfrac{(-1)^n\lambda^mx^{m-n+\frac{k}{2}-1}e^{-\frac{x}{2}}}{2^{2m-n+\frac{k}{2}}n!(m-n)!\Gamma\left(m-n+\dfrac{k}{2}\right)}$
$=\sum\limits_{n=0}^\infty\sum\limits_{m=0}^\infty\dfrac{(-1)^n\lambda^{m+n}x^{m+\frac{k}{2}-1}e^{-\frac{x}{2}}}{2^{2m+n+\frac{k}{2}}n!m!\Gamma\left(m+\dfrac{k}{2}\right)}$
$=\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^n\lambda^{m+n}x^{m+\frac{k}{2}-1}e^{-\frac{x}{2}}}{2^{2m+n+\frac{k}{2}}n!m!\left(\dfrac{k}{2}\right)_m\Gamma\left(\dfrac{k}{2}\right)}$
$=\sum\limits_{m=0}^\infty\dfrac{\lambda^mx^{m+\frac{k}{2}-1}e^{-\frac{x+\lambda}{2}}}{2^{2m+\frac{k}{2}}m!\left(\dfrac{k}{2}\right)_m\Gamma\left(\dfrac{k}{2}\right)}$
$=\dfrac{x^{\frac{k}{2}-1}e^{-\frac{x+\lambda}{2}}}{2^\frac{k}{2}\Gamma\left(\dfrac{k}{2}\right)}{}_0F_1\left(;\dfrac{k}{2};\dfrac{\lambda x}{4}\right)$
