I'm trying to evaluate a Laplace transform that is equivalent to the following integral. According to Wolfram Alpha, it evaluates to
$$I = \int_0^\infty {\operatorname{erfc}\left(\frac{a}{\sqrt{t}}\right)e^{-st}}\mathrm dt = \frac{e^{-2a\sqrt{s}}}{s}$$
for positive 'a' and 's'. These assumptions are valid for my application. My approach so far has been to integrate by parts and then introduce the variable substitutions $u=\sqrt{t}$, $v=\frac{\sqrt{s}u^2 + a}{u}$, $w=v^2$, and $x=w-2a\sqrt{s}$ to achieve the following form:
$$I = \frac{2a}{s\sqrt{\pi s}} \int_0^\infty \frac{e^{-x}}{\sqrt{x^2 – 4a^2s}}\mathrm dx$$
At this point, Wolfram Alpha says that the integrand has no elementary antiderivative and exceeds standard computation time trying to evaluate the definite integral. I'm not very experienced using methods involving complex variables but since the integrand is not real-valued for $\lvert x\rvert<2a\sqrt{s}$ I suspect that is what's needed. How should I proceed?
Best Answer
Ok, I managed to evaluate the integral.
Let's call our integral $I(s,a)$. We start by integrating by parts with $u=e^{-st}$. $\DeclareMathOperator\erf{erf}$.
$$\begin{align}I(s,a) &= \left.-\frac1se^{-st}\erf\Big(\frac a{\sqrt t}\Big)\right|_0^\infty -\int_0^\infty -\frac1se^{-st}\cdot \frac{-a}{\sqrt\pi t^{3/2}}\exp\Big(-\frac{a^2}t\Big)\,\mathrm dt \\ &= \frac1s-\frac a{s\sqrt\pi} \int_0^\infty \exp\Big(-st-\frac{a^2}t\Big)\frac{\mathrm dt}{t^{3/2}} \end{align}$$
Now, we substitute $t=\frac1{u^2}$. As the integrand is even,
$$\begin{align}I(s,a) &=\frac1s-\frac a{s\sqrt\pi}\int_{-\infty}^\infty \exp\Big(-a^2u^2-\frac s{u^2}\Big) \,\mathrm du \\ &= \frac1s-\frac {ae^{-2a\sqrt s}}{s\sqrt\pi}\int_{-\infty}^\infty \exp\Big(-a^2\Big(u-\frac{\sqrt s}{au}\Big)^2\Big)\,\mathrm du\end{align}$$
Now using the Master Theorem, $$\begin{align} I(s,a) &= \frac1s-\frac {ae^{-a\sqrt s}}{s\sqrt\pi}\underbrace{\int_{-\infty}^\infty e^{-a^2u^2}\,\mathrm du }_{=\frac{\sqrt\pi}a}\\I(s,a) &= \frac{1-e^{-2a\sqrt s}}s \end{align}$$
For sanity check, $I(10,5) = \frac{1-e^{-10\sqrt{10}}}{10}$, which is confirmed by Wolfram|Alpha.
Now noting that $\text{erfc}(x)=1-\erf(x)$ and $\int_0^\infty e^{-st}\, dt=\frac1s$, we get
$$\int_0^\infty\text{erfc}\Big(\frac a{\sqrt t}\Big)e^{-st}\,\mathrm dt =\frac{e^{-2a\sqrt s}}s $$