Inverse Laplace

Question

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...

Answer 1

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 $$ \int_0^\infty \mathrm{e}^{-\frac{a}{2}\left(s + s^{-1}\right)} \frac{\sqrt{a}}{\sqrt{2 \pi s^3}} \mathrm{d}s = 0 - \left(- \mathrm{e}^{-a}\right) = \mathrm{e}^{-a} $$