Prove Cauchy's Thm: Integral of cos(ax2) from 0 to $\infty$ = $(\pi/8a)^{1/2}$

[metgt4]

Homework Statement

Show that if a is a positive real constant, the function exp(iaz²) --> 0 as |z| --> infinity for 0 < arg(z) < pi/4

By applying Cauchy's theorem to a suitable contour prove that the integral of cos(ax²)dx from 0 to infinity is equal to (pi/8a)^1/2

The Attempt at a Solution

My work thus far is attached as a scanned document since I don't know how to properly insert equations here. I'm not sure how to express cos(nx²) in terms such that I can integrate and find an answer.

Thanks for helping!
Andrew

 

[metgt4]

I'm pretty sure this is wrong but I may as well ask. Would expanding cos(ax^2) as a Taylor series help? If so, where would you go from there? For some reason I just can't seem to get the hang of working with complex numbers.

 

[vela]

Show that if a is a positive real constant, the function exp(iaz²) --> 0 as |z| --> infinity for 0 < arg(z) < pi/4

What you have written down for this part of the problem doesn't make sense.

Consider e^{x+iy}=e^x(\cos y+i\sin y). The imaginary part of the exponent, y, results in the cosine and sine terms. They just oscillate and won't cause the exponential to go to zero. What will cause the exponential to go to zero is if x, the real part of the exponent, goes to -\infty.

In your problem, the exponent is iaz^2. What you want to show is that the real part of this quantity will go to -\infty as |z|\rightarrow\infty when 0&lt;\arg(z)&lt;\pi/4.

By applying Cauchy's theorem to a suitable contour prove that the integral of cos(ax²)dx from 0 to infinity is equal to (pi/8a)^1/2.

Here's a hint. According to Euler's formula, you have

e^{iaz^2}=\cos(az^2)+i\sin(az^2)

The first term happens to look like the function you want to integrate.

 

[metgt4]

That makes perfect sense now. Thanks!