Fourier Transform of product $g(x)e^{-x}$

Question

I know that the Fourier transform for $f(x)$ gives $\hat{f}(k)=\int_R f(x)e^{-ikx}dx$ but I am unsure how to compute the Fourier transform of a function such as: $g(x)e^{-x}$

Thank you in advance.

Answer 1

Let's try to just do it: $$\begin{align} \hat f(k) &= \int_{-\infty}^\infty f(x) e^{-ikx}\,dx \\ &= \int_{-\infty}^\infty g(x) e^{-x} e^{-ikx}\,dx \\ &= \int_{-\infty}^\infty g(x) e^{-ikx-x}\,dx \\ &= \int_{-\infty}^\infty g(x) e^{-i(k-i)x}\,dx \\ &\stackrel{?}{=} \hat g(k-i) \end{align}$$ That last step is dubious. The reason is that the ordinary Fourier transform is defined only for $k$ and $x$ both being real. However, if the Fourier transform of $g(x)$, that is $\hat g(k)$, can be analytically continued to the lower half of the complex $k$-plane, then the result makes sense.

An example where this works out simply is $g(x)=e^{-x^2}$, whose Fourier transform is $\hat g(k) = \sqrt\pi e^{-k^2/4}$. It is easy to confirm that indeed the Fourier transform of $f(x)=e^{-x^2-x}$ is $\hat f(k)=\sqrt{\pi}e^{-(k-i)^2/4}=\hat g(k-i)$.

Where you can go wrong is if $g(x)$ does not decay to zero rapidly enough as $x\to-\infty$. This is because as $x\to-\infty$, $e^{-x}$ diverges rapidly. So if $\lim_{x\to-\infty}g(x)e^{-x}\ne 0$, then Fourier transform integral does not converge. For instance, if you try $g(x)=\cos(x)$, you can compute its Fourier transform easily enough: $\hat g(k)=\pi[\delta(k-1)+\delta(k+1)]$. However, it is not valid to then say that $f(x)=\cos(x)e^{-x}$ has a Fourier transform $\hat f(k)=g(k-i)=\pi[\delta(k-i-1)+\delta(k-i+1)]$. The reason is that if you actually try do the integral, you will find $$\hat f(k)=\int_{-\infty}^\infty \cos(x)e^{-x}e^{-ikx}\,dx $$ does not converge due to the limit as $x\to-\infty$.

In contrast, the example with $g(x)=e^{-x^2}$ worked out well because as $x\to-\infty$, $e^{-x^2}$ decays to zero more rapidly than $e^{-x}$ diverges to $\infty$.