0
$\begingroup$

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 $k$.

$\endgroup$
  • $\begingroup$ $h(x) = sin(\alpha x) \star e^{-|x|}$ where $\star$ is convolution due the conversion between product and convolution. $\endgroup$ – Srini Mar 5 '17 at 23:56
0
$\begingroup$

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 $\beta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.