Solving $\int_{0}^{\infty}{\frac{\cos(tx^n)}{x^n+a}\, dx}$ via residues

Question

I was trying to evaluate $\int_{0}^{\infty}{\frac{\cos(tx^n)}{x^n+a}\, dx}$, with a semicircle in the upper half plane we have :

$$\oint{\frac{e^{itz^n}}{z^n+a}\, dz}=\int_{-\infty}^{\infty}{\frac{e^{itx^n}}{x^n+a}\, dx}+\int_{\Gamma}{f(z)\, dz=2\pi i\sum{\mathrm{Res}f(z)}}$$

I think $\int_{\Gamma}\to0$ as the radius goes to infinity. Then the residue when $z^n+a=0$ :

$$\begin{align}\lim_{z\to a^{1/n}e^{\frac{i\pi}{n}(2k+1)}}{(z-a^{1/n}e^{\frac{i\pi}{n}(2k+1)})}\frac{e^{itz^n}}{z^n+a}=\frac{e^{itae^{i\pi(2k+1)}}}{n(a^{1/n}e^{\frac{i\pi}{n}(2k+1)})^{n-1}}=-a^{1/n-1}n^{-1}e^{-ita}e^{\frac{i\pi}{n}(2k+1)}\end{align}$$

But after summing, it doesn't seem like this arrives at the answer given here. Specifically the $\csc(\frac{\pi}{n})$.

EDIT : Since part of the numerator of $\sum_{k=0}^{n-1}{e^{\frac{i\pi}{n}(2k+1)}}$ is $1-e^{2\pi i}=0$

Answer 1

The linked answer is alleged to be incorrect on numerical grounds (see comments). It's also quite unrigorous; it uses non-convergent integrals in at least one place. I suspect it is simply wrong.

I don't yet have an evaluation for the integral, but I can explain why your method probably doesn't work. At the very least, to make sure it works you'd need to do a lot of hard work convincing me that the arc integral vanishes.

---

A serious issue with your approach is that it's not obvious that $\int_{\Gamma}$ goes to zero, and it probably doesn't.

You're probably used to $\int_{\Gamma}$ vanishing when we have functions of the form: $$z\mapsto e^{iz}f(z)$$Because in the upper half plane, all the imaginary parts are positive; thus, the real parts of $iz$ are all negative (and large) and you can expect the exponential term to decay sufficiently quickly.

But with $z\mapsto e^{iz^n}f(z)$, it's not so simple. Visualise $z\to z^n$ as a stretch on the unit circle; each sector of angle $2\pi/n$ is blown up to the entire circle. In particular, the semicircle (of angle $\pi$) is blown up and wraps around the unit circle $\lfloor n/2\rfloor$ times. This is $\ge1$ when $n\ge2$, so we wrap around the circle at least once.

Long story short, the imaginary parts of $z^n$ are not always positive if you take $z$ in the upper half plane, so $e^{iz^n}$ does not experience exponential decay. Rather, it will decay on some subsectors of $\Gamma$ and blow up on the others; so, the usual argument wouldn't show $\int_{\Gamma}\to0$ as the radius tends to infinity.

In some cases, the use instead of a wedge contour with angle $\le\pi/n$ (so that $e^{iz^n}$ decays on the arc) works as an alternative. I don't think it works here.

Answer 2

This is not an answer since it does not use residues. Just for the pleasure of working an interesting integral. $$I=\int_{0}^{\infty}{\frac{\cos(tx^n)}{x^n+a}\, dx}=t\int_{0}^{\infty}{\frac{\cos(tx^n)}{tx^n+at}\, dx}$$ $$t x^n=y \quad \implies \quad I=\frac{t^{\frac{n-1}{n}}}{n}\int_{0}^{\infty}\frac{y^{\frac{1}{n}-1} \cos (y)}{y+a t}\,dy$$ Using $b= a t$

$$\int_{0}^{\infty}\frac{y^{\frac{1}{n}-1} \cos (y)}{y+b}\,dy=\pi \, b^{\frac{1}{n}-1} \cos (b) \csc \left(\frac{\pi }{n}\right)+\frac{\Gamma \left(\frac{1}{n}-2\right)}{n}\,\Big[\cdots\Big]$$ where $$\Big[\cdots\Big]=b \,n \cos \left(\frac{\pi }{2 n}\right) \, _1F_2\left(1;\frac{3n-2}{2n},\frac{4n-1}{2 n};-\frac{b^2}{4}\right)-$$ $$(2 n-1) \sin \left(\frac{\pi }{2 n}\right) \, _1F_2\left(1;\frac{2n-1}{2 n},\frac{3n-1}{2n};-\frac{b^2}{4}\right)$$