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 $\frac{e^{-u|r|}}{|r|}$.

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

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}]                   

 Out[2]= ------
            r u
         2 E
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  • $\begingroup$ 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

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