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

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

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 $q/\alpha<1$ 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 $q/\alpha<1$? 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 $|z|<1$. For these values you have
$$\sum_{k=0}^{\infty} z^k=\frac{1}{1-z}.$$
This shows, why the series diverges for $|z|>1$: The analytic function defined by the series only for $|z|<1$ 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 $\text{Re} \; \alpha>0$.

 

[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.