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Cauchy-Goursat Theorem

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Given: Integral from zero to infinity of cos(x^2) dx = integral from zero to infinity of sin(x^2) dx = 1/2 * (square root of pi/2).

These can be evaluated by considering cos(x^2) = Re(e^ix^2) and sin(x^2) = Im(e^ix^2).

1.) Integrate the function f(z) = e^i(z)^2 around the positively oriented boundary of the sector 0 <= r <= R, 0 <= theta <= pi/4, and using the Cauchy-Goursat Theorem show that:

the integral from 0 to R: cos(x^2)dx =

1 / (2)^1/2 * [integral from 0 to R: e^-r^2 dr] - Re * [integral on CR: e^i(z)^2 dz]

2.) Show that the value of the integral along the arc CR tends to zero as R tends to infinity by obtaining the inequality:

| integral on CR of e^i(z^2) dz | <= (R/2) * [integral from 0 to (pi/2) of e^(-R^2) * sin(theta) d(theta).

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Solution Summary

Cauchy-Goursat Theorem is utilized.

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