We are given that: $$u(x,t)=\frac{2}{\pi} \int_0^\infty e^{-k^2t}G_s(k)\sin {kx}\space\text{d}k,$$ where $G_s(k)$ is the Fourier sine transform of $g(x)$.
Find the solution $u(x,t)$ when $g(x)=\delta(x-2)$ and show that when $x/t\ll1$ that
$$u(x,t)=\frac{x}{(\pi t^3)^{1/2}}e^{-(x^2+4)/[4t]}+...$$
I know for the Dirac delta function, when $g(x)=\delta(x-2)$, then $$G_s(k)=\int_0^\infty\delta(x-2)\sin {kx}\space\text{d}x,$$ $$G_s(k)=\sin {2k}.$$ And then I have no idea how to do the rest. Could anyone help me out please?
Hint: This is apparently just Gaussian integrations in the complex plane. For $t>0$, we have:
$$ u(x,t)~=~\frac{2}{\pi} \int_0^\infty \mathrm{d}k~ \exp\left[-k^2t\right]G_s(k)\sin kx $$ $$~=~\frac{1}{\pi} \int_{\mathbb{R}} \mathrm{d}k~ \exp\left[-k^2t\right]\sin 2k \sin kx $$ $$ ~=~ \frac{1}{2\pi} \int_{\mathbb{R}} \mathrm{d}k~ \exp\left[-k^2t\right]\sum_{\sigma=\pm 1}\sigma \cos (x-2\sigma)k $$ $$~=~ \frac{1}{2\pi} \sum_{\sigma=\pm 1}\sigma\int_{\mathbb{R}} \mathrm{d}k~ \exp\left[-k^2t +i(x-2\sigma)k \right] $$ $$~=~ \frac{1}{2\sqrt{\pi t}} \sum_{\sigma=\pm 1}\sigma \exp\left[-\frac{(x-2\sigma)^2}{4t} \right] . $$