Find the Fourier transformation of $e^{-ax^2}$, for any constant $a>0$. In general, $$\int \frac{1}{\sqrt{(2\pi \sigma^2)}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} e^{-iλx }dx = e^{-i\muλ-\frac{1}{2}\sigma^2t^2 }$$ holds.
Apparently, the result can be used to solve the heat equation. I have tried to do it as below $$ \frac {1}{2\pi} \int_{-\infty}^{\infty} e^{-ax^2}e^{iλx} dx $$ After some calculations, I got $\frac {1}{\sqrt{4\pi a}}e^{-\frac{λ^2}{4a}} $. But I looked up on the internet and there are some notes show that it should be $\frac {1}{\sqrt{2 a}}e^{-\frac{λ^2}{4a}} $.
Edit: I found that the notes I saw on the internet use $ \frac {1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ax^2}e^{iλx} dx $ instead of $ \frac {1}{2\pi} $. Maybe that's the reason. But why the notes provided by my lecturer used $ \frac {1}{2\pi} $? Are there no standard form?
By linear transformations it's sufficient to check the case $\mu=0,\,\sigma=1$, so we want to prove $$\phi(t):=\int_{\Bbb R}\frac{1}{\sqrt{2\pi}}\exp\bigg(-\frac{x^2}{2}+itx\bigg)dx=\exp-\frac{t^2}{2}.$$By inspection $\phi(0)=1$, so it suffices to prove the following is $0$: $$\frac{d\phi}{dt}+t\phi(t)=\int_{\Bbb R}\frac{1}{\sqrt{2\pi}}(ix+t)\exp\bigg(-\frac{x^2}{2}+itx\bigg)dx=\Bigg[-\frac{i}{\sqrt{2\pi}}\exp\bigg(-\frac{x^2}{2}+itx\bigg)\Bigg]_{-\infty}^\infty.$$