How Can You Solve This Three Dimensional Integration in Spherical Polars?

[Brewer]

In the course of working out a question, I came across a three dimensional integration in spherical polars. No while this shouldn't be a problem for me, I've been working through it until I got to this bit (which agrees with class mates - we all seem to be stuck at this point!):

\frac{\hbar}{R^2q}\int\sin(\frac{qr}{\hbar})e^{-r/R}dr with limits 0 and infinity.

I think that the obvious thing to do would be to integrate by parts, but it appears that this will go on indefinately, just with sin changing with cos and vice versa.

Is there a better way to do this, or is there no solution for this problem (I assume that there is, as its part of a show that question)?

Thanks for any help given.

 

[MathematicalPhysicist]

well the question is actually what is the integral of sin(x)e^(-x)dx
well there was one way shown in the forums that you could use the identity:
e^ix=cos(x)+isin(x) i'll let you think of it or check it in the forums here.
another way is to integrate by parts, which will get you:
-cos(x)e^(-x)-int(cos(x)e^(-x))=-cos(x)e^(-x)-sin(x)e^(-x)-int(sin(x)e^-x)
which after rearranging gives you the answer.

 

[D H]

LQG showed you two ways, one using Euler's expression and the other integrating by parts twice. The latter is a standard trick when integrating the product of a trigonometric function and an exponential. Note: LQG left out the factors q/\hbar and 1/R. You will have to re-solve with those factors.

A third way is to recognize the solution must be of the form

\left(\alpha\sin(qr/\hbar)+\beta\cos(qr/\hbar)\right)\text{e}^{-r/R}

Differentiate and solve for \alpha and \beta.

 

[MathematicalPhysicist]

well, i thought the big problem was the integral itself not the coefficients.

 

[D H]

Your'e correct, LQG. Solving the basic integral with unitary constants is the big problem. Everything after is just busy work.

 

[Brewer]

I've integrated by parts twice and it appears that I've now gotten to this:

\frac{\hbar}{qR}\int e^{\frac{-r}{R}}\sin(\frac{qr}{\hbar}) dr = R-\frac{Rq}{\hbar}\int e^{\frac{-r}{R}} \sin(\frac{qr}{\hbar}) dr

which is just the original integration multiplied by the inverse of the original constant subtracted from R. Is this the right way of doing it? Can I then get rid of the integrals somehow as they appear on both sides of the equations? I can't just subtract them or anything I don't think, but from the form of the answer I think that they should disappear somehow.

Am I missing something pretty obvious and fundamental here?

 

[HallsofIvy]

I've integrated by parts twice and it appears that I've now gotten to this:

\frac{\hbar}{qR}\int e^{\frac{-r}{R}}\sin(\frac{qr}{\hbar}) dr = R-\frac{Rq}{\hbar}\int e^{\frac{-r}{R}} \sin(\frac{qr}{\hbar}) dr

which is just the original integration multiplied by the inverse of the original constant subtracted from R. Is this the right way of doing it? Can I then get rid of the integrals somehow as they appear on both sides of the equations? I can't just subtract them or anything I don't think, but from the form of the answer I think that they should disappear somehow.

Am I missing something pretty obvious and fundamental here?

Basically, then you have an equation that says:
\frac{\hbar}{qR}X = R-\frac{Rq}{\hbar}X
(X is the integral you want)
How do you solve for X?

 

[Brewer]

Yes its since dawned on me, that I can just rearrange to solve for the integral. I actually found that I'd made a mistake with some of the constants (although I'm not sure for this part I was working on here it makes any difference), but I have now solved this problem and gotten to the required answer (which makes me think that I have it correct!)

Thanks for all your help guys.

Brewer