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Who likes integrals? e^(trig function)

1. Homework Statement
I'm working on a physics problem and I'm winding up with the following integral. Anyone have any clues? Contour?

[tex]\int_{0}^{2\pi} e^{i k cos\left(\theta + k_\theta\right)} d\theta [/tex]

2. Homework Equations



3. The Attempt at a Solution

I tried subbing in a bunch of u's and whatever else. I'm just plain stuck. I've been trying to get it in terms of something I can just look up in a table, but I can't. Can anyone give me a direction? I don't know whether this is a contour integral or what, but I'm fairly certain it is. If it is, it's been awhile since I've done any complex stuff and I don't have a book with me to help. So, if that's the case I'll just bail on the problem. Thanks!
 

Answers and Replies

Well I doubt this is going to be much help but according to my maths program it's insoluble in elementary functions in a general integral and it pops out -infinities as the answer to the definite integral in the range of 0pi and 2pi, and given theta=0pi,1pi...2pi. given anything more of a range or negative thetas or both, just increases the number of series of -infinities. If it is soluble my maths program isn't powerful enough to do it, unless your happy with the answer -infinity of course. :smile:

This is in need of renormalisation as

[tex]\int_0^{2\pi}e^{ik\cos({\theta}+k_\theta})\;d\theta=-\infty [/tex]


does not converge I think or converges to -infinity :). Or if it does converge my maths program is broken.

Hehe do you add a constant to infinity, doesn't seem much point. :biggrin: since it's still infinity.


Anyone else got any thoughts? Is my maths program screwed again. :smile:

Have you tried converting it to Cartesian co-ordinates, instead of polars, or is that a stupid suggestion, probably pointless? Is there any solution in Cartesians, if not I doubt there's any solution in reality either.

Sorry I can't delete the above post so I'll have to double post.

I'm also assuming k is a constant.
 
Last edited:
Yep, k is constant. Well, it does take the form of a delta function, so I'm thinking it may just be 1 when Theta plus ktheta is pi/2,3pi/2, but I'm not sure. Actually, this problem was in cartesian coordinates, but it suggested to convert to spherical before integrating, haha. So, that much is a no go.
 
Dick
Science Advisor
Homework Helper
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If the limits are really 0 to 2pi, then it's not a delta function and it's not -infinity. It's a complex number depending on k and k_theta. E.g. if k=0 then it's value is 2pi. I don't think you are going to find an elementary expression for it's value. Are you sure you changed coordinates correctly?
 
Yep, i'm certain about the change of coordinates. If it's a complex number, that's probably fine, especially if it depends on k and k_theta, that makes perfect sense, because later I have to integrate this with respect to k both k and k_theta. Would I be doing a contour integral or something to get the result that you got?

Thanks.
 
P.S. I realize that to evaluate the k=0 case you are just integrating 1.
 
Dick
Science Advisor
Homework Helper
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I really didn't get much, I just set k=0 so the integrand is 1. That was just a 'for instance'. I don't know how to do any other value exactly. I do know it's not divergent or a delta function, just because it's the integral of a continuous function over a closed interval. Contours won't help. If you really needed to you could integrate it numerically, but I suspect something else has gone wrong if this is just an exercise. What is the original problem?
 
I've been working on my stats, going through some problems wherever I can find them. I ran into this in Reif 1.29.

Problem statement:
(a) Using an appropriate Dirac-delta function, find the probability density w(s) for displacements of uniform length l, but in any random direction of three-dimensional space. (Hint: Remember that the function w(s) must be such that [tex]\int\int\int w(\vec{s})d\vec{s}=1[/tex] when integrated over all space.)

For this part I got [tex]w(\vec{s})=\frac{\delta\left(\rho - l)}{4\pi l^2}[/tex]. This made sense to me, since (1) it integrated to 1 over all space and, (2) it seemed to me that the probability should be uniform over a sphere of radius l. rho is the radius in spherical coordinates.

(b) Use the result of part (a) to calculate [tex]Q(\vec{k})[/tex]. (Perform the integral in spherical coordinates.)

This is where I'm winding up stuck.

[tex]Q(\vec{k})=\int w(\vec{s}) e^{i\vec{k}\bullet \vec{s}} d^3\vec{s} [/tex]
You can see that due to the dot product and spherical coordinates, I'm getting trig functions in the exponential. Obviously rho is easy to handle, but once I get to theta and phi I get quite stuck. The constants in the original problem above aren't exactly what I'm looking at here, but it seemed like if there was a way to compute that, I could handle the rest from there. Thanks!

P.S. I'm willing to accept that my density is just wrong.
 

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