So math stack exchange isn't really helping much with this. So initially, I'm proving the inverse laplace transform using contour integration. This is a good starting point for my research when I eventually need to find the inverse laplace transform when the functions could not be found in tables. I need to prove that: $$\DeclareMathOperator{\erfc}{erfc} \mathscr{L}^{-1} \bigg[ \frac{1}{s}\big(\exp (- \sqrt{s/k}x)\big) \bigg] = \erfc\left(\frac{x}{2\sqrt{kt}}\right) $$ This inverse laplace transform can be found using a table of Laplace Transforms. Using the following contour:
Source: https://tex.stackexchange.com/questions/269684/hankel-bromwich-contour-problem
Then, after considering all contributions of this contour to get: $$ \mathscr{L}^{-1} \bigg[ \frac{1}{s}\big(\exp (- \sqrt{s/k}x)\big) \bigg] = 1 - \frac{1}{\pi} \int_{0}^{\infty} \exp(-ut) \sin(\sqrt{u/k} x) \frac{du}{u} $$ Here, we can simplify the integral by letting: $v^{2} = ut$ and $y = x/\sqrt{kt}$ to get: $$ \mathscr{L}^{-1} \bigg[ \frac{1}{s}\big(\exp (- \sqrt{s/k}x)\big) \bigg] = 1 - \frac{2}{\pi} \int_{0}^{\infty} \exp(-v) \sin(yv) \frac{dv}{v}. $$ How do I continue from here to eventually get to : $$ 1 - \erf\left(\frac{y}{2}\right) = 1 - \erf\left(\frac{x}{2\sqrt{kt}}\right) $$$$ 1 - erf\left(\frac{y}{2}\right) = 1 - erf\left(\frac{x}{2\sqrt{kt}}\right) $$
