2
$\begingroup$

The inverse Fourier transform is defined as:

$$\mathcal{F}^{-1}[g](x) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} g(k) e^{i k x} d k$$

I can't get an inverse Fourier Transform to

Q1: $$g(k)=\cos (\sqrt{ k^2 d -a^2 })$$ Q2:
$$g(k)=\frac{1}{ \sqrt{ k^2 d -a^2 }}\sin (\sqrt{ k^2 d -a^2 })$$

I would really appreciate some any help .

$\endgroup$
3
  • $\begingroup$ why would you need a closed form expression for the Fourier transform of those complicated function ? $\endgroup$ – reuns Apr 17 '16 at 19:38
  • $\begingroup$ have you tried mathematica or stuff like that? $\endgroup$ – user190080 Apr 17 '16 at 19:42
  • $\begingroup$ No. Is Mathematica obtain the result ? $\endgroup$ – Hamada Al Apr 20 '16 at 10:11
1
$\begingroup$

Q1: $$ \mathcal{F}^{-1}[g](x) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \cos (\sqrt{ k^2 d -a^2 }) e^{i k x} d k $$ I see no reason to think it may be written any simpler than that.
Also note that the integrand involves cosine of an imaginary argument, at least on part of the range.

$\endgroup$
3
  • $\begingroup$ Do you mean that i should write $2 \cos(\theta)= Re \quad( e^{i \theta})$ or what ? $\endgroup$ – Hamada Al Apr 17 '16 at 23:23
  • $\begingroup$ For example when $k=0$ you are talking about $\cos(\sqrt{-a^2})$. $\endgroup$ – GEdgar Apr 18 '16 at 0:55
  • $\begingroup$ But K is the variable of integration . $\endgroup$ – Hamada Al Apr 20 '16 at 10:12

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.