Fourier transform with trigonometric and exponential functions

Question

My question states:

A function $h(x)$ has its Fourier transform given by
$$\mathcal{F}[sin(\alpha x)]\mathcal{F}[e^{-|x|}]$$ Show that $h(x)$ is of the form $A\sin({\alpha x})$ and find the constant $A$.

Hence, without directly computing the integral, show that
$$\int_{-\infty}^{\infty} \sin(\beta p) e^{-|x-p|}dp=\frac{2sin(\beta x)}{1+\beta^{2}}$$

I can calculate the integrals and derive the equation
$$\mathcal{F}[h(x)]=i \sqrt{\frac{\pi}{2}} (\delta(\alpha -k)-\delta (\alpha + k)) \sqrt{\frac{2}{\pi}} \frac{1}{1+k^{2}} = \frac{i}{1+k^{2}} (\delta(\alpha -k)-\delta (\alpha + k))$$ but I can't move from this. I am unsure how to calculate the inverse Fourier transform from this, and also I don't know why in the equation given in the question, there is constant $\beta$ in the denominator, while I have similar equation with variable $k$.

Answer 1

NOTE: I would advise you not to you write $\mathcal F[h(x)]$ for the Fourier transform of a function $h(x)$, as when you calculate the Fourier transform you integrate over $x$. It's better to write, for example $\mathcal F[h](k)$, instead.

I think there is a sign missing when calculating the Fourier transform of $\sin \alpha x$, and the Fourier transform of $h$ should be

$$\mathcal F[h](k) = -\frac{i}{1+k^2}(\delta(a-k)-\delta(a+k))$$

the Fourier transform of $h$. The inverse Fourier transform (I might have put the wrong constant in front as I am not sure what normalisation you are using: you can easily fix that) gives you back $h$ as

$$ h(x) =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \hat h(k) e^{+ i kx}dk =-\frac{i}{1+a^2}\left( e^{i\alpha x}-e^{-i\alpha x}\right) = \frac{2}{1+\alpha^2} \sin (\alpha x). $$

So your constant $A$ would be equal to $2/(1+\alpha^2)$.

For the last part, call $f(x) = \sin \alpha x$ and $g(x) = e^{-|x|}$. Then, Since

$$ \mathcal F[h] = \mathcal F[f] \mathcal F[g], $$

you have $$ h = \mathcal F^{-1} \left( \mathcal F[f] \mathcal F[g]\right) = \left(\mathcal F^{-1}\mathcal F\right)[f] * \left(\mathcal F^{-1}\mathcal F\right)[g]= f * g, $$

where $*$ denote the convolution product. In other words, we get $$\frac{2 \sin (\alpha x)}{1+\alpha^2} = h(x)= (f*g)(x) =\int_{-\infty}^{+\infty} \sin(\alpha p) e^{-|x-p|} dp,$$

which holds for all $\alpha$, which you may as well call $\beta$.