Why does the integral of the Yukawa potential converge for q/alpha >= 1?

[conana]

While preparing for an exam I came across an integral of the form

\int_0^\infty dx\;e^{-\alpha x}\sin{q x}

with q,\alpha>0.

My question will be regarding my solution to the integral which I present as follows:

I expand the sine function as a Taylor series and differentiate with respect to alpha to yield

\begin{align}e^{-\alpha x}\sin{q x} &amp;= \sum_{n=0}^\infty (-1)^n\dfrac{q^{2n+1}}{(2n+1)!}x^{2n+1}e^{-\alpha x} \\<br /> &amp;= \sum_n (-1)^{n+1}\dfrac{q^{2n+1}}{(2n+1)!}\dfrac{d^{2n+1}}{d\alpha^{2n+1}}e^{-\alpha x}\end{align}

After integrating with respect to x and differentiating with respect to alpha I arrive at

\int_0^\infty dx\;e^{-\alpha x}\sin{q x}=\dfrac{q}{\alpha^2}\sum_n \left(i\dfrac{q}{\alpha}\right)^{2n}.

Here comes the troubling part. For q/\alpha&lt;1 this geometric series converges nicely to

\dfrac{q}{q^2+\alpha^2}.

However, Mathematica tells me that the integral, unlike my geometric series above, will still converge for q/\alpha\geq 1.

I guess my question is a) Where have I gone wrong in my solution such that it is only valid for the case q/\alpha&lt;1? b) Is there a more straightforward way of performing this integral?

Thanks in advance for any insight you all may offer.

I realize this is more of a math question, but it came up while performing the Fourier transform of the Yukawa potential and I thought that the physics community here would be well acquainted with this integral.

[Edit]: I want to make clear that this is not a homework problem. I was simply curious if I could perform the integral by hand.

 

[The_Duck]

As a simpler route, I think you can just write $\sin(qx) = [e^{iqx} - e^{-iqx}]/(2i)$, which turns the integral into a sum of two easy integrals of exponentials.

 

[conana]

As a simpler route, I think you can just write $\sin(qx) = [e^{iqx} - e^{-iqx}]/(2i)$, which turns the integral into a sum of two easy integrals of exponentials.

Oh wow... I was originally trying to do just that except that I was thinking I would use the residue theorem... which makes no sense for this problem. Wow that is easy. Thanks, The_Duck.

However, I am still curious as to where my overly-complicated solution goes wrong. Any thoughts?

 

[vanhees71]

That's a trick any physicist has learn at a point. You have a series, converging for some values, and then after summing you do an analytic continuation. Most prominent examples is renormalization theory in quantum field theory: You evaluate an integral in d dimensions and analytically continuate to the physical value d=4 after subtracting divergent terms of the type 1/(d-4) (dimensional regularization) and so on.

You example is a nice example. Of course the geometric series converges only for |z|&lt;1. For these values you have
\sum_{k=0}^{\infty} z^k=\frac{1}{1-z}.
This shows, why the series diverges for |z|&gt;1: The analytic function defined by the series only for |z|&lt;1 has a pole at z=1, and a well-known theorem from complex function theory tells you that the Taylor series around a point z_0 has the radius of divergence determined by the largest disk not containing any singularities of the corresponding function. Here z_0=0 and the closest (and in this case only) singularity is at z=1.

The function as an analytic (meromorphic) function is uniquely defined on the entire complex plane except at the pole z=1. Thus the function is valid on this much larger domain than the convergence region of the original power series tells you.

The integral is obviously convergent for any real q for \text{Re} \; \alpha&gt;0.

 

[conana]

Thank you vanhees71 for your response. I believe I understand the main idea of what you are saying and will do some playing around with this idea.