<s>[INST] Heat equation with initial value I have a 1-dimensional homogeneous heat equation:
$$ u''(x, t) = \dot u(x, t)$$
The initial value is $u(x, 0) = \exp\left(-x^2\right)$.
I plugged this into the solution formula:
$$ u(x, t) = \frac{1}{\sqrt{4 \pi t}} \int_{-\infty}^\infty \mathrm dy \, \exp\left(-\frac{(x-y)^2}{4t} - y^2\right)$$
Since I am not sure how to calculate this integral, I typed it into Mathematica and got:
1/Sqrt[4 Pi t] Integrate[
  Exp[((y - x)^2)/(4 t) - y^2], {y, -Infinity, Infinity}]

$$\text{ConditionalExpression}\left[\frac{\exp\left(\frac{x^2}{4 t-1}\right)}{\sqrt{4-\frac{1}{t}} \sqrt{t}},\left(4
   \Re(t)\neq 1\lor \Re\left(\frac{x}{t}\right)>0\right)\land \Re\left(\frac{1}{t}\right)<4\right]$$
Plotting this looks strange at best:
http://wstaw.org/m/2012/11/10/1_1.png
For $t=0$, it is not even close to the initial value.


*

*How do I solve this integral?

*Is what Mathematica gave me correct in some sort?
 [/INST] Setting
$$
\xi=\sqrt{\frac{1+4t}{4t}}\left(y-\frac{x}{1+4t}\right),
$$
we have:
\begin{eqnarray}
\frac{(x-y)^2}{4t}+y^2&=&\frac{(1+4t)y^2-2xy+x^2}{4t}\\
&=&\frac{1+4t}{4t}\left[y^2-\frac{2x}{1+4t}y+\frac{x^2}{1+4t}\right]\\
&=&\frac{1+4t}{4t}\left[\left(y-\frac{x}{1+4t}\right)^2+\frac{x^2}{1+4t}-\frac{x^2}{(1+4t)^2}\right]\\
&=&\frac{1+4t}{4t}\left[\left(y-\frac{x}{1+4t}\right)^2+\frac{4x^2t}{(1+4t)^2}\right]\\
&=&\xi^2+\frac{x^2}{1+4t}.
\end{eqnarray}
It follows that
\begin{eqnarray}
u(x,t)&=&\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^\infty\exp\left(-\frac{(x-y)^2}{4t}-y^2\right)\, dy\\
&=&\frac{1}{\sqrt{4\pi t}}\sqrt{\frac{4t}{1+4t}}\exp\left(-\frac{x^2}{1+4t}\right)\int_{-\infty}^\infty\exp(-\xi^2)\, d\xi\\
&=&\frac{1}{\sqrt{\pi(1+4t)}}\exp\left(-\frac{x^2}{1+4t}\right)\int_{-\infty}^\infty\exp(-\xi^2)\, d\xi.
\end{eqnarray}
Using the fact that
$$
\int_{-\infty}^\infty\exp(-\xi^2)\, d\xi=\sqrt{\pi},
$$
we get
$$
u(x,t)=\frac{1}{\sqrt{1+4t}}\exp\left(-\frac{x^2}{1+4t}\right).
$$
</s>