I am trying to answer the following question:
Consider an infinitely long metal rod. The one dimensional heat equation is $\frac{\partial u}{\partial t}=a^2\frac{\partial^2 u}{\partial x^2}$. Let the temperature at $t=0$ be the delta function. Show that $u(x,t)\propto \frac{1}{a\sqrt{2t}}\exp({\frac{-x^2}{4a^2t}})$.
In lectures we were shown that $u(x,t)$ has the following solutions (we are using the convention that the forward Fourier transform uses $e^{+ixy}$ and the inverse uses $e^{-ixy}$).
I applied these solutions to and got:
Which is the correct formula, but I am getting and equality rather than a proportionaliy. I think the error must be somewhere in the inverse fourier transform, but I just can't figrue out where.
I think you just have a careless mistake in your last line. You have $\sqrt{2\pi}^2$ in the denominator of the first equality, but in the second you just have $\sqrt{2}\sqrt{ \pi}$, with the square having been dropped. If you fix this you find your answer ought to have a constant factor of $\frac{1}{\sqrt{2 \pi}}$ in front, which would agree with the formula for the standard heat kernel when $a= \frac{1}{\sqrt{2}}$.
(The minus sign in front of the second equality also seems to be a mistake.)