1
$\begingroup$

It can be proved that the potential $\frac{e^{-u|r|}}{|r|}$ has Fourier transform $\frac{4\pi}{u^2+q^2}$. Now, I'm trying to go backwards and do the inverse Fourier transform but I'm running into trouble. We have the inverse Fourier transform below

$$\int dq d\theta d\phi \frac{1}{2\pi}q^2 \sin\theta e^{iqr\cos\theta} \frac{4\pi}{u^2+q^2}$$

$$=\pi\int \frac{2\sin(qr)}{qr}\frac{q^2}{u^2+q^2}dq $$ after solving the angular part.

The limits of the $q$ integral are $0$ and $\infty$ but weirdly, Mathematica refuses to solve this claiming it is divergent. But the limit of the integrand at infinity is zero. How can I get the correct result which I know is $\frac{e^{-u|r|}}{|r|}$.

$\endgroup$
  • $\begingroup$ You have a Mathematica stack exchange site. $\endgroup$ – Trimok Jun 28 '14 at 11:24
2
$\begingroup$

I) One mathematical problem is that the function

$$\tag{1} f(q)~:=~ \frac{q\sin(rq)}{q^2+u^2}, \qquad q,r,u~>~0, $$

is not integrable $f\notin {\cal L}^{1}(\mathbb{R}_{+})$, because the integral over the absolute value of the integrand is infinite:

$$\tag{2} \int_{\mathbb{R}_{+}} \! dq~|f(q)| ~=~\infty.$$

However it is still possible to define the integral $\int_{\mathbb{R}_{+}} \! dq~f(q)$ as an improper integral, i.e. as a limit:

$$\tag{3} \lim_{Q\to\infty} \int_0^Q \! dq~f(q)~=~\frac{\pi}{2}e^{-ur}.$$

II) Nevertheless, I got Mathematica to evaluate the improper integral $\int_{\mathbb{R}_{+}} \! dq~f(q)$ via this command:

 In[1]:= f[q_] := q Sin[r q]/(q^2+u^2)                                           

 In[2]:= Simplify[ Integrate[f[q], {q,0,Infinity}], {r>0,u>0}]                   

           Pi
 Out[2]= ------
            r u
         2 E
$\endgroup$
  • $\begingroup$ Thanks, will redirect the question of how to get Mathematica to solve these types of integrals on the relevant site. $\endgroup$ – user1936752 Jun 29 '14 at 6:47

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.