Quantum mechanics Dirac delta representation with integral So I’m doing QM and found bunch of problems for beginners and I’m struggling with this one:
$$\lim_{a\rightarrow 0}\int^{\infty}_{-\infty}e^{\frac{ip x}{\hbar}-a x^2}dx=2\pi\hbar\delta(p).$$
If I swap the limit and the integral I get the correct answer, but can I do that?
 A: Using $\int_{-\infty}^{\infty}e^{-\alpha x^2+\beta x}=\sqrt{\frac{\pi}{\alpha}}\exp{\frac{\beta^2}{4\alpha}}$ 
put $n=1/a$ we get
$\lim_{n \to \infty}\sqrt{\pi n \cdot\frac {\pi 4 \hbar^2}{\pi 4 \hbar^2}}\exp{\frac{-p^2n}{4\hbar^2}}$  Now define $n=>n/4\hbar^2$
Now,$e^{-nx^2}\sqrt{\frac{n}{\pi}}$ defines a generalised function $\delta(x)$ such that
$\int_{-\infty}^{\infty}\delta(x)F(x)=F(o)$ in limit $n\rightarrow\infty$ can be easily proved (ref.M.J.Lighthill Fourier and Generalised Functions for more)
We get your answer $2\pi\hbar\delta(p)$
A: You can just rewrite the integral first(make substitutions $x = -x$, $\frac{p}{\hbar}=k$), say $$I=\int_{-\infty}^{\infty}dx \ e^{-i k.x} e^{-ax^2}$$
Now, this is a like a Fourier transform of a Gaussian function.
Where
$$I = \sqrt{\frac{\pi}{a}}e^{-\frac{k^2}{4a}}$$
Therefore,  you can look at the definitions of dirac delta as limit, and write
$$ 2\pi \ \lim_{a\rightarrow 0} \frac{1}{2\sqrt{{\pi a}}}e^{\frac{-k^2}{4a}} = 2\pi \delta(k) $$
Now, using the property of dirac delta $\delta(ax) = \frac{1}{|a|}\delta(x)$, you can get to your final form.
PS: But, it would be interesting to think of the physical motivation of this manipulation for the context in which OP might be looking at this question.
