While preparing for an exam I came across an integral of the form(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int_0^\infty dx\;e^{-\alpha x}\sin{q x}[/tex]

with [itex]q,\alpha>0[/itex].

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

[tex]\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} \\

&= \sum_n (-1)^{n+1}\dfrac{q^{2n+1}}{(2n+1)!}\dfrac{d^{2n+1}}{d\alpha^{2n+1}}e^{-\alpha x}\end{align}[/tex]

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

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

Here comes the troubling part. For [itex]q/\alpha<1[/itex] this geometric series converges nicely to

[tex]\dfrac{q}{q^2+\alpha^2}.[/tex]

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

I guess my question is a) Where have I gone wrong in my solution such that it is only valid for the case [itex]q/\alpha<1[/itex]? 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.

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# Yukawa Potential Integral

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