On integrals related to $\int^{+\infty}_{-\infty} e^{-x^2} dx = \sqrt{\pi}$ You are given the result that $$\int^{+\infty}_{-\infty} e^{-x^2} dx = \sqrt{\pi}$$
a. Use this result to find $$\int^{+\infty}_{-\infty} e^{-ax^2} dx$$
b. Use the above results to find $$\int^{+\infty}_{-\infty} x^2e^{-ax^2} dx$$
[Hint: Consider $\frac{d}{d\alpha} \int^{+\infty}_{-\infty}e^{-ax^2}dx$]
c. Use the above results to to show that $$P(x,t)=\frac{1}{\sqrt{4 \pi D t}} \exp \left( -\frac{x^2}{4Dt}\right) $$ is a normalized distribution.
(Note: I'm still learning - this problem is somewhat advanced for my level, so if someone could write out an explicit and complete solution, that would be the most helpful answer for me.)
As of now, @matt has successfully helped me to understand the solutions for (a) and (b) in his response below; but I still need to have the solution for (c) explained to me.  I do not understand what needs to be done. for (c).
 A: a. We make the substitution $t=\sqrt{a}x$, $a>0$, thus:
 $$
  \int_{-\infty}^{+\infty}e^{-ax^2}dx
  =\lim_{\substack{n\rightarrow\infty\\m\rightarrow\infty}}\int_{-n}^{+m}e^{-ax^2}dx
  =\lim_{\substack{n\rightarrow\infty\\m\rightarrow\infty}}\int_{-n\sqrt{a}}^{+m\sqrt{a}}e^{-t^2}\frac{dt}{\sqrt{a}}
  =\frac1{\sqrt{a}}\int^{+\infty}_{-\infty} e^{-x^2} dx
  = \sqrt{\frac{\pi}{a}}
 $$
b. We differentiating our result from part (a) with respect to $a$.
 $$
  \frac{d}{da} \int^{+\infty}_{-\infty}e^{-ax^2}dx
  = \frac{d}{da}\left(\sqrt{\frac{\pi}{a}}\right)
  \quad\Rightarrow\quad
  \int^{+\infty}_{-\infty}\frac{d}{da}e^{-ax^2}dx
  = \frac{d}{da}\left(\sqrt{\frac{\pi}{a}}\right)
 $$
Note that moving $\frac{d}{da}$ inside the integral is justified since both $e^{-ax^2}$ and $\frac{d}{da}(e^{-ax^2})$ are continuous.
Simplifying gives:
 $$
 -\int_{-\infty}^{+\infty}x^2e^{-ax^2}dx = -\frac{\sqrt{\pi}}{2a^{3/2}}
  \quad\Rightarrow\quad
 \int_{-\infty}^{+\infty}x^2e^{-ax^2}dx = \frac{\sqrt{\pi}}{2a^{3/2}}
 $$
c. We must show: (1) $P(x,t)$ is non-negative; and (2) $\int P(x,t)=1$. We establish (1) by noting that $\exp(-x^2/4DT)\geq 0$ for all $x\in\mathbb{R}$.  To establish (2), observe the following result which follows from part (a) (we set $a=1/(4Dt)$):
 $$
  \int_{-\infty}^{+\infty}P(x,t)\;dx
  =\frac{1}{\sqrt{4 \pi D t}} \int_{-\infty}^{+\infty}  \exp \left( -\frac{x^2}{4Dt}\right)\;dx
  =\frac{1}{\sqrt{4 \pi D t}} \sqrt{\frac{\pi}{1/(4Dt)}} = 1
 $$
EDIT: Added some clarification for part (c).
