# Why does this curious equality between two integrals seem to hold?

I was messing around on Wolfram Alpha, checking the numerical values of the following ratio for the first several values of $$n$$:

$$\frac { | \int_0^\infty{e^{i(x^n)}} | } { \int_0^\infty{e^{-x^n}}}$$

(the numerator is the complex magnitude after integration). I tried it for all integer values of $$n$$ from 2 to 10 inclusive, and in all cases the ratio came out to 1 (within computational error of order $$10^{-5}$$ or less each time). I also tried it for a few random non-integer powers greater than 1, with the same result.

Unfortunately, it doesn't seem to work for all functions -- when I tried combining a few of the above powers into a polynomial, it wasn't 1 anymore.

Does anyone know why this would work for any real $$n \gt 1$$? And is there any chance I screwed it up and it actually does work for any function (as long as both integrals converge), not just $$x^n$$? That would be so cool...

You can relate the numerator to the denominator by contour integration. The idea is to relate the integral in the numerator to the integral of the same function along a ray from the origin in the $$e^{i \pi/(2n)}$$ direction. So you take a contour going from $$0$$ to $$R$$ along a straight line, then from $$R$$ to $$R e^{i\pi/(2n)}$$ along a circular arc, then from $$R e^{i \pi/(2n)}$$ back to $$0$$ along a straight line.
After you've shown that the integral along the arc goes to $$0$$ as $$R \to \infty$$, the residue theorem tells you that the integrals along the two straight lines sum to zero. So you get $$\int_0^\infty e^{ix^n} dx = \int_0^\infty e^{-x^n} e^{i\pi/(2n)} dx$$.
To go this way, you definitely need $$n>1$$, so that the LHS converges in improper Riemann sense. To go exactly the way I described it, you also need $$n$$ to be an integer, so that $$e^{i x^n}$$ is analytic everywhere. But if $$n$$ is not an integer, I think you can select a logarithm so that $$e^{i x^n}$$ is analytic on this wedge domain whenever $$n>1$$.
You can actually evaluate this in closed form as well: $$\int_0^\infty e^{-x^n} dx = \int_0^\infty e^{-u} \frac{dx}{du} du$$ where $$u(x)=x^n,\frac{du}{dx}=n x^{n-1},\frac{dx}{du}=\frac{1}{n} x^{1-n}=\frac{1}{n} u^{1/n-1}$$. So you have $$\frac{1}{n} \Gamma(1/n)$$, which you can also write as $$\Gamma(1+1/n)$$.
• I think you can define $e^{iz^n}$ appropriately so that it's analytic in the region you consider, since that region is simply connected. Commented Sep 17, 2022 at 16:57