Hello,
So I want to evaluate \int_0^{\infty} e^{i z^2}. One of the central steps is to evaluate the integral around the contour line z = R e^{i \theta}. So I substitute in z = R e^{i \theta} so that I get e^{i R^2 e^{i 2 \theta}} = e^{i R^2 cos(2 \theta) - R^2 sin(2 \theta)}.. I have to show convergence to 0. This equals e^{i R^2 cos(2 \theta)} e^{- R^2 sin(2 \theta)} = (cos(R^2 cos(2 \theta)) + i sin(R^2 cos(2 \theta))) e^{- R^2 sin(2 \theta)}. So my sources say that the absolute value of this becomes e^{ - R^2 sin(2 \theta)}, implying that |(cos(R^2 cos(2 \theta)) + i sin(R^2 cos(2 \theta)))| = 1 throughout the entire interval. Of course, the absolute value of e^{i \theta} is 1 as long as the number is of this form. But then, why can't the absolute value of e^{i R^2 e^{i 2 \theta}} also be 1 throughout the range? What properties make a the absolute value of a number of the form e^{i \theta} NOT converge to 1 for ANY value of \theta?
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Another question: Since I have to evaluate \int_0^{\infty} e^{i z^2} dz, where dz = i e^{i \theta} d \theta - I have to multiply this equation by that factor. I probably have to do this for the final solution. But since this is bounded by i, and my only objective is to show that this integral converges to 0, this factor of dz should be effectively negligible, right?
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MAIN QUESTION:
Why does |(cos(R^2 cos(2 \theta)) + i sin(R^2 cos(2 \theta)))| = 1 UNIFORMLY WHEN
the absolute value of e^{i R^2 e^{i 2 \theta}} is NOT equal to 1?
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