Integration of abs(k)e^(ikx)dk

[rmiller70015]

Homework Statement

This is for a quantum free particle problem, doing the Fourrier transform and I just want to make sure I've got the integral correct.

Relevant Equations

$$\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty} \phi (k) e^{ikx}dk$$


Where $\phi$ is:
$$

f\left(k\right) = \left\{

\begin{array}{lr}

A(a-|k|) & : |k| \le a\\

0 & : |k| > a

\end{array}

\right.\\

$$

Split the integral
$$\frac{Aa}{\sqrt{2\pi}}\int^{\infty}_{-\infty}e^{ikx}dk - \frac{A}{\sqrt{2\pi}}\int^{\infty}_{-\infty}|k|e^{ikx}dk$$

Apply the boundary conditions, this is where my biggest source of uncertainty comes from I doubled the integral and integrated from 0 to a instead of from -a to +a to get rid of that absolute value. I plotted the |k|e^k function and it appears to have even parity.

$$\frac{Aa}{\sqrt{2\pi}}\int^{a}_{-a}e^{ikx}dk - \frac{2A}{\sqrt{2\pi}}\int^{a}_{0}|k|e^{ikx}dk$$

$$\frac{Aa}{\sqrt{2\pi}ix}(e^{iax} - e^{-iax}) + \frac{2Ai}{\sqrt{2\pi}x}[ke^{ikx}-\frac{1}{ix}e^{ikx}]^a_0$$

$$\frac{Aa}{\sqrt{2\pi}ix}(e^{iax} - e^{-iax}) - \frac{2A}{\sqrt{2\pi}ix}(ae^{iax}-0 - \frac{1}{ix}(e^{iax}-1))$$

$$\frac{Aa}{\sqrt{2\pi}ix}(e^{iax}-e^{-iax}-2e^{iax}) - \frac{2A}{\sqrt{2\pi}(ix)^2}(e^{iax}-1)$$

Clean up and use Euler
$$\frac{2iAa}{\sqrt{2\pi}x}cos(ax) + \frac{2A}{\sqrt{2\pi}x^2}(e^{iax}-1)$$

 

[BvU]

Hi,

Apply the boundary conditions, this is where my biggest source of uncertainty comes from I doubled the integral and integrated from 0 to a instead of from -a to +a to get rid of that absolute value

If you have to split anyway, why not write $\phi = a+k$ from -a to 0 and $\phi = a-k$ from 0 to a ?

Are you somewhat familiar with the FT ? Since your $\phi$ is a convolution of a boxcar in time (6) with itself, I would expect (14) something like a $\displaystyle {\ {sin^2 x\over x^2}}\$ for $\psi$.

$\$

 

[vela]

Mathematica yields a simpler result (this one's wrong):
$$\frac{2A}{\sqrt{2\pi}} \frac{1-e^{iax}+iax}{x^2}.$$ On the other hand, if I evaluate the integral using the InverseFourierTransform function, I get a result like @BvU's expectation.

You can't double the integral from 0 to $a$ because while $f(k)$ is even, the integrand as a whole isn't.