I need help solving this integral

sin(qr)*r/(r^2+d^2)^2dr and its the integral from 0 to infinity

Do you know the theory of residues?

i do a little bit but if i have to get into residues i must be on the wrong track

Are you sure? After a little playing around the integral can be done with a fairly simple application of the residue theorem. The relevant poles are of order two, so the residue will correspond to the first derivative of the analytic function on top.

i am just trying to do this as part of a Born approximation so i figured you wouldn't need residues

The Born approximation is actually often evaluated with the method of residues. However, if this is a physics class, there is no reason why you can't just look up the integral in a table if you can't do it yourself. There isn't any other simple way to do it.

Even if you know very little of residues, you can still get it pretty easy. Try integrating by parts once, then you'll have a very clean integrand with first order poles.

Wow this is an amazing integral. Anyway as it Physics Monkey has said that he has evaluated it using theory of residues, can anyone what it is?

Any chance someone can either solve it or look it up for me since i am rusty with the residues and i just need the value to continue the problem

The integral of [ r sin(qr)/(r^2 + d^2)^2 ] from 0 to infinity is (pi/4)(q/d)exp(-qd). Beware, I did this in about two seconds in my head so I may have an extra factor.

Edit: Woops, I did have an extra factor. The answer should have dimensions of 1/length^2 assuming r is a length.

Mathematica got (pi/4)(q/d)exp(-qd), no squared term on the q factor. But I can't justify it either way.

-Dale

The integral of [ r sin(qr)/(r^2 + d^2)^2 ] from 0 to infinity is (pi/4)(q/d)exp(-qd). Beware, I did this in about two seconds in my head so I may have an extra factor.
Edit: Woops, I did have an extra factor. The answer should have dimensions of 1/length^2 assuming r is a length.Hard to believe that my hero here could be wrong, but looks that way to me. I set d = 0 and then the integral to be done then is:
{sin(qr)}/r^3
which looks very well bounded, a small positive number, probably less than unity if q > 1. Because area under the first half cycle of the sign is not cancelled by the negative contribution of the second half cycle (true for all half cycles pairs, but the r cubed denominator is rapidly killing the higher cycles.)

He has a q/d term that is going to infinity as d --> 0.

I am very nervous "correcting PM" but I guess it is possible I could.

PS - once I knew about pole, residues and contour integrals etc but only enough to get good grades on exams, I never really understood.

If PM really does have clay feet, let me be the first to welcome him to the human race. ;)

On second thought, by edit, my faith in PM is returning. Expanding the sin near r = 0 as something that goes as (qr) + (qr)^3 / (3 or 6? - I forget) + ... and keeping only the first term of the integrand (q/r^2) I bet he is right - That first half cycle could be a whopper! and the exp(-qd) may not be going to unity fast enough with d --> 0 to do much.

Did not discard entire post as Now I am courious. Is the d = 0 case the limit as d --> 0 or a "disconected point"?

Billy T,

The integral should diverge at d=0. To see why look at your expression again, if you set d=0 then for small r you may approximate the integrand sin(qr)/r^3 as q/r^2 thus the small r part of the integral diverges as 1/r.

Billy T,

The integral should diverge at d=0. To see why look at your expression again, if you set d=0 then for small r you may approximate the integrand sin(qr)/r^3 as q/r^2 thus the small r part of the integral diverges as 1/r.Re read my modification made prior to reading your's above.

Hell, I'm just impressed he did it in his head. I would probably get anything more complicated that 2+2 wrong in my head.

-Dale

Mathematica: my yellow sun. Power outages: my kryptonite :)