Potentially more math than physics here, but I'm looking for the scattering amplitude of the potential:
$$U(r) = \frac{U_o}{(r^2+d^2)^2}.$$
Where $d$ is a constant.
Using the first Born approximation for a spherically symmetric potential leads to:
$$f(q) = \frac{U_0}{q}\int_d^{\infty}\frac{\sin(qr)}{(r^2+d^2)^2}rdr$$
Where $q=|\mathbf{k_i}-\mathbf{k_f}|$ is the momentum transfer. I have yet to find a way to solve this integral analytically (as is suggested is possible) and integral tables have proven only useful in transforming the integral into others that are equally complicated. Any advice or insight is greatly appreciated!
$$\textbf{SOLUTION}$$
In my set up of the Born approximation integral I incorrectly assumed that the polarization potential goes to zero for $r<d$. Given that this is not the case, the integral from above becomes
$$f(q) = \frac{U_0}{q}\int_0^{\infty}\frac{sin(qr)}{(r^2+d^2)^2}rdr$$
This integral can be looked up in an integral table by Dwight, but that's not what I'm here for. To go about solving this, we first set $d^2=z$ notice that
$$\frac{\partial}{\partial z}\Big(\frac{1}{r^2+z}\Big) = -\frac{1}{(r^2+z)^2}$$
so that
$$f(q) = -\frac{U_0}{2iq}\frac{\partial}{\partial z}\Bigg[\int_0^{\infty}\frac{e^{iqr}rdr}{r^2+z} - \int_0^{\infty}\frac{e^{-iqr}rdr}{r^2+z}\Bigg]$$
where I have also expanded the $sin(qr)$ term into it's exponential representation. Letting $r \rightarrow -r$ in the second integral yields
$$f(q) = -\frac{U_0}{2iq}\frac{\partial}{\partial z}\int_{\infty}^{\infty}\frac{e^{iqr}rdr}{r^2+z} = -\frac{U_0}{2iq}\frac{\partial}{\partial z}\int_{\infty}^{\infty}\frac{e^{iqr}rdr}{(r+i\sqrt{z})(r-i\sqrt{z})}$$
revealing two simple poles at $\pm i\sqrt{z}$. Closing the integral in the upper-half plane and using Cauchy's integral formula allows us to evaluate the integral
$$\frac{U_0}{2iq}\int_{\infty}^{\infty}\frac{e^{iqr}rdr}{(r+i\sqrt{z})(r-i\sqrt{z})} = \frac{U_0 \pi}{2q}e^{-q\sqrt{z}}$$
Carrying out the partial derivative and back substituting $z=d^2$ gives us the final result:
$$f(q) = \frac{U_0 \pi}{4d}e^{-qd}$$.
NOTE: This solution was typed up quickly. There may be a factor or a negative, etc. that is missing but the overall technique should hold. Let me know of any errors or improvements you might run across. Thanks!
