How an I evaluate the following integral: $$\frac{u_0}{\pi}\int_{-\infty}^{\infty}\frac{\sin( \alpha)\cos(\alpha x)}{\alpha}e^{-k \alpha^2 t}d \alpha \ \ ?$$
It arises as the solution $u(x,t)$ of the following PDE IVP:
$$ku_{xx}=u_t \\ u(x,0) = \begin{cases} u_0, \ \ |x|< 1 \\ 0, \ \ \ \ |x|>1 \end{cases} \\ -\infty<x<\infty \\ t>0$$
Thank you.
Using a CAS, I (it) found that $$\int_{-\infty}^{\infty}\frac{\sin( \alpha)\cos(\alpha x)}{\alpha}e^{-k \alpha^2 t}d \alpha =\frac{\pi}{2} \left(\text{erf}\left(\frac{x+1}{2 \sqrt{k t}}\right)-\text{erf}\left(\frac{x-1}{2 \sqrt{k t}}\right)\right)$$ provided that $$\Re(k)\neq 0\lor k\notin \mathbb{R})\land \Re(k t)>0$$
I hope and wish this can help you.