# Definite integration of f(x)/sqrt(-(x-q1)(x-q2))

1. Jan 18, 2007

### ianbell

Recently I fruitlessly asked if anybody could help with the definite integral of
sin(a x) / ( x sqrt(-(x-q1)(x-q2)) )
from x=q1 to x=q2 where a,x, q1 and q2 are all real.

If the sqrt wasn't there one could use contour integration and consider residues at q1,q2 and 0 but with the sqrt I am somewhat stumped.

Yet this seems an obvious area to me: integrations between the roots of a quadratic Q(x) of functions of the form f(x) / Q(x)^alpha .

Is this really virgin territory?

2. Jan 18, 2007

### mjsd

I have to say I probably have never done such integrals before (i may have but if it is must be some time ago), have you tried expanding the bit without the sin as a Laurent series and see if you can do term by term integration? Note: this is just a suggestion, not a hint.

3. Jan 18, 2007

### ianbell

Thanks for that. I've not tried it. I'm not that keen on tackling this analytically since my calculus is very rusty and I was hoping it would be a standard problem, already tackled.

The definite integral between the quadratic roots of f(x) / sqrt(Q(x)) for f(x) ammoung the basic fundamental functions (x^n, sin(a x), exp(x) etc) feels
like a field that somebody would have addressed.

If, for example, you integrate V(sqrt(x^2 + r^2 + R^2 - 2 x R cos(theta)))
from theta=0 to 2 pi corresponding to the potential at (x,0,z)
of an origin centred ring of radius R of matter generating radial potential V(r)
and make the obvious substitution you face the integral from
s=s1= z^2+(r-R)^2 to s=s2= z^2+(r+R)^2 of
V(sqrt(s)) / sqrt(Q(s))
where Q(s) = -(s-s1)(s-s2)
so with f(x) = 1/sqrt(x) we have the Coulomb potential of a charged ring.

t arises as the potential of a circular ring of integration arounnd a ring

4. Jan 18, 2007

### mjsd

Putting the limits of the integral aside, if you can analytically do the indefinite integral then no worries about the limits at all. My concern is that if you wanna do it as a contour integral, don't you need at least one limit to go to $$\pm\infty$$? So that you can actually close the contour, Or is there a neat trick somewhere to avoid that? I must admit I haven't thought about this carefully so far.... but if you are not keen on doing it analytically, you may as well do it numerically. or try using Mathematica to see whether you actual have analytic solution first.

5. Jan 19, 2007

### ianbell

Obviously if you can do the indefinite integral the definite is easy but I suspect the definite integral (corresponding to the potential of a full ring) is likely to be easier. To do it with contour you'd obviously have to close the contour from q1 to q2 somehow.

I don't have Mathematica. The Wolfram online integrator fails for indefinite integrals of sin(a x)/ sqrt(-(x-b)(x-c)) and sin(a x)/x / sqrt(-(x-b)(x-c))