Is there a way to calculate this difficult integral analytically?

[bruno67]

I am trying to calculate an integral of the form


$$I=\int_{0}^{\infty}\exp(iax)\frac{(x^2+b^2)^{-c}}{x+id}dx,$$


where $$a,b,d\in\mathbb{R}$$ and $$c>0$$. I don't think it's possible to do it analytically (at least I couldn't do it). Is there a way to calculate the "exact" value by using a series expansion, perhaps?

 

[nonequilibrium]

I think i can do it analytically in closed form. Are you still interested (as you posted it in april)? (otherwise I won't be bothered to type it out)

 

[bruno67]

I think i can do it analytically in closed form. Are you still interested (as you posted it in april)? (otherwise I won't be bothered to type it out)

Yes, I am still interested, and I haven't been able to find the solution. How do you think it can be done?

 

[nonequilibrium]

My apologies, I was too hasty, my reasoning only works for b = 0... :(

 

[bruno67]

My apologies, I was too hasty, my reasoning only works for b = 0... :(

Still, I would be curious to know how you did it, even for b=0.

 

[nonequilibrium]

shoot, seems I'm wrong again: my idea was to play with contour integrals.
For example, if b = 0, you can lay the branchcut of z^c on the positive real axis and then integrate of a keyhole contour (google image "keyhole contour" if you don't know what shape i mean), then on the bottom line you'd get $$-e^{-2 \pi i c} I$$ (with I the integral you're interested in), and on the top line obviously I itself. The contribution of the origin could be controlled, I think (for the simple case of 0 < c < 1 it will go to zero, for c > 1 i think you can also get the answer but haven't thought it through). The sum of the contour integral would then equal $$2 \pi i \frac{e^{ad-ic3\pi/2}}{d^c}$$ (where I used the residue of the integrand in z = -id for the case of d > 0; analogous expression for d < 0). Anyway the problem is that the contribution from the outer circle doesn't go to zero. That's the problem. Stupid of me, posted too hastily.