Inverse Laplace Hi how to verify the following 
I tried substitution and integration by parts but can bot figure it out..
$$\int_0^{\infty} \exp(- \lambda t )   \frac{x}{\sqrt{2\pi t^3}}\exp(-\frac{x^2}{2t}) dt = \exp(-\sqrt{2\lambda}  x)$$
Thank you...
 A: Let $a = \sqrt{2\lambda} x$. Making a change of variables $2\lambda t =a s$:
$$
  \int_0^{\infty} \exp(- \lambda t ) \frac{x}{\sqrt{2\pi t^3}}\exp\left(-\frac{x^2}{2t}\right) \mathrm{d}t = \int_0^\infty \mathrm{e}^{-\frac{a}{2}\left(s + s^{-1}\right)} \frac{\sqrt{a}}{\sqrt{2 \pi s^3}} \mathrm{d}s
$$
Now split the integral as follows, using $s^{-3/2} = \frac{1}{2} \left(s^{-3/2}-s^{-1/2} \right) + \frac{1}{2} \left(s^{-3/2}+s^{-1/2} \right)$:
$$ \begin{eqnarray}
  \int_0^\infty \mathrm{e}^{-\frac{a}{2}\left(s + s^{-1}\right)} \frac{\sqrt{a}}{\sqrt{2 \pi s^3}} \mathrm{d}s &=& \int_0^\infty \mathrm{e}^{-\frac{a}{2}\left(s + s^{-1}\right)} \frac{\sqrt{a}}{\sqrt{2 \pi}} \frac{1}{2} \left(s^{-3/2} -s^{-1/2}\right) \mathrm{d}s \\ && +  \int_0^\infty \mathrm{e}^{-\frac{a}{2}\left(s + s^{-1}\right)} \frac{\sqrt{a}}{\sqrt{2 \pi}} \frac{1}{2} \left(s^{-3/2} + s^{-1/2}\right) \mathrm{d}s = \\
 &=& -\int_0^\infty \mathrm{e}^{a-\frac{a}{2}\left(s^{1/2} + s^{-1/2}\right)^2} \frac{\sqrt{a}}{\sqrt{2 \pi}} \frac{\mathrm{d}}{{\mathrm{d}s}} \left(s^{1/2} +s^{-1/2}\right) \mathrm{d}s + \\
 && \int_0^\infty \mathrm{e}^{-a-\frac{a}{2}\left(s^{1/2} - s^{-1/2}\right)^2} \frac{\sqrt{a}}{\sqrt{2 \pi}} \frac{\mathrm{d}}{{\mathrm{d}s}} \left(s^{1/2} -s^{-1/2}\right) \mathrm{d}s 
\end{eqnarray}
$$
Now making changes of variables $u_1 = \sqrt{a} \left(\sqrt{s} +\frac{1}{\sqrt{s}}\right)$ in the first integral, and  $u_2 = \sqrt{a} \left(\sqrt{s} -\frac{1}{\sqrt{s}}\right)$ in the second gives us
$$
 \int_0^\infty \mathrm{e}^{-\frac{a}{2}\left(s + s^{-1}\right)} \frac{\sqrt{a}}{\sqrt{2 \pi s^3}} \mathrm{d}s = \left[ -\mathrm{e}^{-a} \Phi\left(\sqrt{a} \left(s^{-1/2} - \sqrt{s}\right)\right) + \mathrm{e}^{a} \Phi\left(-\sqrt{a} \left(s^{-1/2} + \sqrt{s}\right) \right) \right]_{s \downarrow 0}^{s \uparrow \infty}
$$
where $\Phi(x)$ is the cumulative density function of the standard normal distribution, in particular, $\lim_{x \to -\infty} \Phi(x) = 0$ and  $\lim_{x \to +\infty} \Phi(x) = 1$. Hence
$$
 \int_0^\infty \mathrm{e}^{-\frac{a}{2}\left(s + s^{-1}\right)} \frac{\sqrt{a}}{\sqrt{2 \pi s^3}} \mathrm{d}s = 0 - \left(- \mathrm{e}^{-a}\right) = \mathrm{e}^{-a}
$$
