How do you integrate $\int_0^\infty \exp(it^k)\,\mathrm dt$ for $k \in \Bbb N$?

Question

My problem is with the integral $$\int^\infty_0 e^{it^k}\,\mathrm dt$$ with $k\in\mathbb{N}$.

Somehow it can be evaluated by use of Cauchy's theorem. But I don't see how. The best thing I can imagine is drawing a fourth part of a pizza with radius $r$ in for instance the north east quadrant of the complex plane and then do a contour integral around the boundary of the pizza slice and finally see what happens when $r$ is send to infinity.

What seems to be especially nasty is the complex line integral around the circular curved part of the contour of the pizza slice since it is going to contain some difficult "$\leq k$"-powers of angle functions. But to be honest I am also not sure about the rest of the contour integral.

Kindly appreciated,

Aris

Answer 1

Make the angle of your pizza slice equal to $\pi/(2 k)$. Then we have

$$\oint_C dz \, e^{i z^k} = \int_0^R dx \, e^{i x^k} + i R \int_0^{\pi/(2 k)} d\phi \, e^{i \phi} \, e^{-R^k \sin{k \phi}} e^{i R^k \cos{k \phi}} + e^{i \pi/(2 k)} \int_{\infty}^0 dt \, e^{-t^k} $$

which is zero due to Cauchy's theorem. Use the fact that $\sin{k \phi} \ge (2/\pi) k \phi$ over $\phi \in [0,\pi/(2 k)]$, so that

$$\left | i R \int_0^{\pi/(2 k)} d\phi \, e^{i \phi} \, e^{-R^k \sin{k \phi}} e^{i R^k \cos{k \phi}}\right| \le R \int_0^{\pi/(2 k)} d\phi \, e^{-(2 k R^k/\pi) \phi}\le \frac{\pi}{2 k R^{k-1}} $$

which vanishes as $R \to \infty$. In this limit, then,

$$\int_0^{\infty} dx \, e^{i x^k} = e^{i \pi/(2 k)} \int_0^{\infty} dx \, e^{-x^k} = e^{i \pi/(2 k)} \Gamma{\left ( 1+ \frac{1}{k}\right)} $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}\expo{\ic t^{k}} \,\dd t} \,\,\,\stackrel{t^{k}\ \mapsto\ t}{=}\,\,\, {1 \over k}\int_{0}^{\infty}t^{\color{red}{1/k} - 1}\,\,\expo{\ic t}\,\dd t \end{align} The integral will be evaluated with Ramanujan Master Theorem. Note that $\ds{\expo{\ic t} = \sum_{n = 0}^{\infty}{\pars{\ic t}^{n} \over n!} = \sum_{n = 0}^{\infty}\color{red}{\expo{-\ic n\pi/2}} \,\,{\pars{-t}^{n} \over n!}}$.

Then, \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}\expo{\ic t^{k}} \,\dd t} = {1 \over k}\,\Gamma\pars{\color{red}{1 \over k}} \expo{-\ic\pars{-\color{red}{1/k}}\pi/2} \\[5mm] = &\ \bbx{{1 \over k}\,\Gamma\pars{1 \over k} \exp\pars{{\pi \over 2k}\,\ic}} \\ & \end{align}