Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm given $\Psi(x,t)$ as a proposal for a wave function. $\Psi(x,t)=\int_{1}^{1+\Delta k} e^{i(kx-wt)} k^2 dk$

Now I try to compute $\Psi^*(x,t)\Psi(x,t)$ wich is the product

$(\int_{1}^{1+\Delta k} e^{-i(kx-wt)} k^2 dk) (\int_{1}^{1+\Delta k} e^{i(kx-wt)}k^2 dk)$

In wich way should I transform this to a double integral? Taking into account that $w=w(|k|)$

Thanks for your time.

share|cite|improve this question
There is no comment about $\Delta k$ so I assumed is a real number because $k$ is. – Jorge Sep 25 '11 at 12:48
This is a product of Fourier integrals, so you can obtain its Fourier transform as the convolution of the two Fourier transforms. – joriki Sep 25 '11 at 13:01
In Fourier Transforms, as far as I know, the range of integration is $\mathbb{R}$ not just $(1,1+\Delta k)$ – Jorge Sep 25 '11 at 13:04

It is easier to solve the one dimensional integral, and then to perform the multiplication:

$\int_1^{1+\Delta} e^{ikx-i\omega t} k^2 dk = - e^{-i\omega t} \frac{\partial}{\partial x^2}( \int_1^{1+\Delta} e^{ikx} dk) = - \Delta e^{-i\omega t} \frac{\partial}{\partial x^2}\left ( e^{[i(1+\Delta/2)x]} \frac{sin(\frac{\Delta x}{2})}{\frac{\Delta x}{2}}\right) $

All is left is to perform the differentiationwith respect to x.

share|cite|improve this answer
Thank you very much, nice idea! – Jorge Sep 25 '11 at 15:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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