PDE Fourier transform(heat equation) I solved the heat equation using the fourier transform method and got my solution to be 
$$u(x,t)= \int_{-\infty}^{\infty}\frac{1}{\sqrt{4\pi kt}}\exp(\frac{-(x-y)^2}{4kt})\tag{1}$$ 
however I have intial condition $u(x,0) =\phi(x)$ . I can't seem to show how the intial condition is satisfied
 A: 
Here, we present a way forward that does not rely on knowledge of the theory or generalized functions (i.e., distributions).  Rather, we apply standard tools from real analysis.

Starting from the integral expression for $u(x,t)$, we have
$$u(x,t)=\frac{1}{\sqrt{4\pi kt}}\int_{-\infty}^\infty \phi(y) e^{-\frac{(x-y)^2}{4kt}}\,dy$$
Enforcing the substitution $y\mapsto x+\sqrt{4kt}y$ yields
$$\begin{align}
u(x,t)&=\frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty \phi(x+\sqrt{4kt}y) e^{-y^2}\,dy\\\\
&=\phi(x)+\frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty \left(\phi(x+\sqrt{4kt}y) -\phi(x)\right)\,e^{-y^2}\,dy\\\\
\end{align}$$
If $\phi(x)$ is continuous and bounded, then the Dominated Convergence Theorem guarantees that
$$\begin{align}
\lim_{t\to 0^+}\int_{-\infty}^\infty \left(\phi(x+\sqrt{4kt}y) -\phi(x)\right)\,e^{-y^2}\,dy&=\int_{-\infty}^\infty \lim_{t\to 0^+}\left(\phi(x+\sqrt{4kt}y) -\phi(x)\right)\,e^{-y^2}\,dy\\\\
&=0
\end{align}$$
Therefore, we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{t\to 0^+}u(x,t)=\phi(x)}$$
as was to be shown!
A: You should have gotten $$u(x,t) = \int^\infty_{-\infty} \frac{1}{\sqrt{4\pi k t}} e^{\frac{-(x-y)^2}{4kt}} \phi(y) dy;$$ i.e., the initial condition should be included in there. Then $u(x,0) = \phi(x)$ since $\frac{1}{\sqrt{4\pi k t}} e^{\frac{-(x-y)^2}{4kt}} \to \delta(x)$ as $t \to 0^+$ in the distributional sense, where $\delta(x)$ is the Dirac mass centered at $x$ (any decent early graduate level PDE book should have a proof that the heat kernel tends to the Dirac mass).
