Using Cauchy's Theorem to Solve the Complex Integral of cos(ax^2)

metgt4
Messages
34
Reaction score
0

Homework Statement



By applying Cauchy's theorem to a suitable contour, prove that the integral of cos(ax2) = (pi/8a)1/2


Homework Equations



Cauchy's integral formula:

http://en.wikipedia.org/wiki/Cauchy's_integral_formula

The Attempt at a Solution



I'm not sure where to go after what I've got scanned. I've tried a few other things, but I can't seem to get anywhere that looks like something I could solve.
 

Attachments

  • scan0002.jpg
    scan0002.jpg
    8.6 KB · Views: 436
Physics news on Phys.org
Hint: Consider the function exp(-a z^2).
 
What I do know about eiaz^2 is that within 0 < argz < pi/4 it is analytic and
eiaz^2-->0 as |z|-->infinity

Could I expand that in terms of a Taylor series and use the method of residues to solve the integral? Would this be the correct expansion for cos(ax2):

ex = 1 + x + x2/2! + x3/3!

Taylor series for x = iaz2 added to the taylor series for x = -iaz2
= 1/2(eiaz^2 + e-iaz^2) = 1/2[(1 + iaz2 - a2z4/2 - a3z6/6 +...) + (1 - iaz + a2z4/2 + a3z6/6 +...) = 1/2(2) = 1?

and then would I express 1 as a complex exponential?

Is that right?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top