I need to show that
If we complexify the integrand and make a sort of pacman-contour
with
with
with
with
As we let
This contour contains the 2 second order singularities
And we can evaluate with the Cauchy integral formula. I was somewhat surprised when I found . It's easy to check that
and
As a result we get
If all steps were ok, somehow I want to conclude that tends to
But I'm not sure how. Also I found working with log(z) somewhat trouble-some. I believe I need to use some nice property of log(z)
Can anyone offer some suggestions?
I was afraid so. I believe it's called the key-hole contour.I don't understand what your contour is; there may be several mistakes in your parameterization... Furthermore, what are ? Could you describe your contour in words, or with a sketch?
http://upload.wikimedia.org/wikipedi...ontour.svg.png
When you see this picture, you understand why I called it a 'pacman' contour ;p
With I meant small numbers, actually dependant on R, but it's somewhat hard to define this contour it terms of R
I made a pretty sad mistake though. I forgot to multiply with .
Thanks for the precision. Note that is not defined for all ; you must exclude a half-line, usually . Then the contour includes values where the function is not defined.
However, it seems to works if you symmetrize the picture, so that Pacman "looks to the left". This way the contour avoids the negative real axis .
We have . When converges toward from the upper-half plane, then converges toward and when it converges from the lower-half plane, converges toward . Thus, the integral along is almost equal to (real parts cancel). My computation shows that the residue at is , and at . The conclusion is , which is cool, but not what we need...
Another contour will be necessary. For instance only the upper-half of the Pacman. Like a parliamentary hemicycle. Integral on the horizontal part on the right of 0 converges to your integral, and the integral on the horizontal part on the left converges to . By the way you'll recover the value of the above integral by looking at the imaginary part.
Hi. I don't think Cauchy's Integral Theorem can be used on this problem since the contour contains two poles. Also, just skip that "almost on the real axis thing" with the and just go for "exactly on the real axis" in the limit. Then instead of using , use and the same contour so that when you sum the branch-cut contours, the parts cancel and you're left with:
but keep in mind of the branch you're using: over the contour above the real axis you have and over the contour below the real axis you have and
I did it quick. May need a little fixin' up but I think it's close. That's my way of sayin', you work through it and in the process, learn how to do it .