Another thing you could have done, to avoid so many calculations was to note that $\displaystyle v(x,t)= \frac{1}{2\sqrt{\pi kt}} e^{\frac{-x^2}{4kt}}$ solves the heat equation $\displaystyle v_t=kv_{xx}$ on $\displaystyle \mathbb{R} \times (0,\infty )$ and noting that $\displaystyle u(x,t)=u^t \ast f$ where the $\displaystyle u^t$ means simply that the convolution is with respect to the first variable. Now properties of this operation give you the result based on that for $\displaystyle v$. The fact that it satisfies the initial (or boundary, depending on how you view it) condition on $\displaystyle t=0$ is more delicate since you can't just plug that value.