I am trying to solve the following problem from my textbook:
If $f(x) \in L^{2}(-\infty, \infty)$ has Fourier transform $F(\alpha)$, then what is the function of $x$ whose transform is the function $H(\alpha)=e^{-t\alpha ^{2}}F(\alpha)$? ($t$ is a positive real constant).
The answer given in the back of the book is $\displaystyle h(x) = \left[4 \pi t\right]^{-1/2}\int_{-\infty}^{\infty}\exp \left[\frac{-(x-y)^{2}}{4t} \right]f(y)dy$, but I have no idea how they got that.
I attempted the problem by writing that $\displaystyle h(x) = e^{-t\alpha^{2}}\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ix\alpha}dx = \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x) e^{-t\alpha^{2}-ix\alpha} dx$, but from that point on, I don't know what to do.
Is this possibly a situation where a convolution is involved? Or is it some kind of a translation? I am really lost, and am trying to teach these things to myself. Could somebody please help me understand what's going on here, and how to get the solution the book has?
Thank you.
Note that we can write the Fourier Transform of the convolution, $\mathscr{F}\{f*g\}=\mathscr{F}\left(\int_{-\infty}^\infty f(x')g(x-x')\,dx'\right)$, between two functions $f$ and $g$ as
$$\begin{align} \mathscr{F}\{f*g\}(\alpha)&=\int_{-\infty}^\infty \left(\int_{-\infty}^\infty f(x')g(x-x')\,dx'\right)\,e^{i\alpha x}\,dx\\\\ &=\int_{-\infty}^{\infty}f(x')\left(\int_{-\infty}^\infty g(x-x')e^{i\alpha x}\,dx\right)\,dx'\\\\&=\int_{-\infty}^{\infty}f(x')\left(\int_{-\infty}^\infty g(x'')e^{i\alpha (x'+x'')}\,dx''\right)\,dx'\\\\ &=\left(\int_{-\infty}^{\infty}f(x')e^{i\alpha x'}\right)\left(\int_{-\infty}^{\infty}g(x'')e^{i\alpha x''}\,dx''\right)\\\\ &=\left(\mathscr{F}\{f\}(\alpha)\right)\left(\mathscr{F}\{g\}(\alpha)\right)\tag 1 \end{align}$$
Let $F(\alpha)=\mathscr{F}\{f\}(\alpha)$ and $e^{-t\alpha^2}=G(\alpha)=\mathscr{F}\{g\}(\alpha)$ in $(1)$. Therefore, the inverse Fourier Transform $\mathscr{F}^{-1}\{F(\alpha)G(\alpha)\}$ of $F(\alpha)G(\alpha)$ is given by $f*g$. Since $g$ is the inverse Fourier Transform of $e^{-t\alpha^2}$, then $g(x)=\frac{1}{\sqrt{4\pi t}}e^{-x^2/(4t)}$ and we have
$$(f*g)(x)=\frac{1}{\sqrt{4\pi t}}\int_{-\infty}^{\infty}f(x')e^{-(x-x')^2/(4t)}\,dx'$$
as was to be shown!