Show the diffusion equation is a normalised distribution. The diffusion equation is defined to be $$P(x,t) = \dfrac{1}{\sqrt{4D\pi t}} \exp \left(-\dfrac{x^2}{4Dt}\right),$$ where $D$ is a physical constant.
Show that the reaction diffusion equation is a normalised distribution.
I take that this means that I need to show that $$\int_{-\infty}^\infty P(x,t) \,dx = 1$$ (the sum of probabilities is $1$), which I have shown, and that $$\int_{-\infty}^{\infty} x P(x,t) \,dx = 0$$ or that the expectation is zero.
However, previous parts of the question make us show that $$\int_{-\infty}^\infty x^2 e^{-\alpha x^2} \,dx = 1.$$
Have I got the definition of the expectation wrong? Or am I trying to show the wrong thing? If so, why?
I don't think I'll need help with calculations, but if I do, I'll ask again.

I am not quite sure what the problem is. But the expextation of your distribution is indeed zero, since
\begin{equation}
\frac{x}{\sqrt{4D\pi t}}e^{-\frac{x^2}{4Dt}}
\end{equation}
is an odd function. Moreover
\begin{equation}
\int_{-\infty}^{\infty}x^2e^{-\alpha x^2}dx = \sqrt{4\alpha\pi}\int_{-\infty}^{\infty}\frac{1}{\sqrt{4\alpha\pi}}x^2e^{-\alpha x^2} dx= \sqrt{4\alpha\pi}\left(\frac{1}{2\alpha}\right),
\end{equation}
where we used that the second moment of the normal$(0,\frac{1}{\sqrt{2\alpha}})$ distribution equals $(\frac{1}{\sqrt{2\alpha}})^2 + 0^2$.
