Convergence $I=\int_0^\infty \frac{\sin x}{x^s}dx$

This is a good problem to analyze. We can solve it by just series methods and careful thought.

Given the following integral \begin{equation} \int_{0}^\infty \frac{\sin x}{x^s}dx, \tag1 \end{equation} for what values of the real parameter s is the integral convergent and absolutely convergent.

(a) In order to solve this problem we break (1) into two pieces \begin{equation} \int_{0}^\infty \frac{\sin x}{x^s}dx=\int_{0}^1 \frac{\sin x}{x^s}dx + \int_{1}^\infty \frac{\sin x}{x^s}dx \tag2 \end{equation} We can analyze each term separately. It is easy to see that the term $$ \int_{1}^\infty \frac{\sin x}{x^s}dx $$ is divergent for $ s \leq 0$ since integral is proportional to $x^s$ which diverges as $x \to \infty$. For $ s > 0$, the series is convergent since $x^{-s} \downarrow 0 \ \text{as}\ x \to \infty$. We now consider the other term in (2) and write it explicitly in terms of a sum $$ \int_{0}^1 \frac{\sin x}{x^s}dx=\int_{0}^1 \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}{x^{-s}}dx= \int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1-s}}{(2n+1)!}dx. $$ We can evaluate if this integral is convergent by analyzing the series inside which is \begin{equation} \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1-s}}{(2n+1)!}\equiv \xi \end{equation} Using the ratio test on $\xi$, we have $$ \lim_{n\to \infty}\bigg| \frac{(-1)^{n+1} x^{2n+3-s} \cdot (2n+1)}{(2n+3)! \cdot (-1)^n x^{2n+1-s}} \bigg|=\lim_{n\to \infty} \frac{x^2}{4n^2+10n+6}=0. $$ By the definition of the ratio test, this series is absolutely convergent since $$ \lim_{n\to \infty} \bigg|\frac{\xi_{n+1}}{\xi_n}\bigg| =0 <1. $$ We now check for uniform convergence by swapping the order of summation and integration, that is doing the integral first which yields $$ \sum_{n=0}^{\infty} \frac{(-1)^n} {(2n+1)!}\int_{0}^{1} x^{2n+1-s} dx=\sum_{n=0}^{\infty} \frac{(-1)^n} {(2n+1)! \cdot (2n+2-s)}. $$ Note, the $(2n+2-s) >0$ to be defined. Computing the sum for $n=0$ we have the condition $2 -s > 0$, or $ 2>s$. Evaluating the integral at $n=0, s=2$ we have $$ \int_{0}^{1} x^{2n+1-s} dx=\int_{0}^{1} {x^{-1}} dx $$ which diverges as the logarithm.

We can conclude that (1) is convergent for $s \in (0,2)$.

(b):For absolute convergence we check the convergence of $$ \int_{0}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx. $$ Once again, we break the integral into two parts $$ \int_{0}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx=\int_{0}^{1} \bigg|\frac{\sin x}{x^s}\bigg| dx + \int_{1}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx. $$ The second term on the right converges for $s > 1$ and is seen easily since $$ \int_{1}^{\infty} \bigg|\frac{\sin x}{x^s}\bigg| dx < \int_{1}^{\infty} \bigg|\frac{1}{x^s}\bigg| dx $$ which is convergent for $s > 1$. We check the other term for convergence by noting that $$ \bigg|\frac{\sin x}{x^s}\bigg|=\frac{\sin x}{x^s} $$ for $ x \in [0,1]$. Thus we conclude that $$ \int_0^1 \frac{\sin x}{x^s} $$ is absolutely convergent for $s \in (0,2)$.

Therefore, the integral in (1) is absolutely convergent for $s \in (1,2)$.

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I \equiv \int_{0}^{\infty}{\sin\pars{x} \over x^{s}}\,\dd x:\ {\large ?}}$

\begin{align} I&=\lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty} {1 \over x^{s - 1}}{\sin\pars{x} \over x}\,\dd x =\lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty}{1 \over x^{s - 1}}\bracks{% \half\Re\int_{-1}^{1}\expo{\ic\verts{k}x}\,\dd k}\,\dd x \\[3mm]&=\half\Re\int_{-1}^{1}\bracks{\color{blue}{% \lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty} {\expo{\ic\verts{k}x} \over x^{s - 1}}\,\dd x}}\,\dd k\tag{1} \end{align}

\begin{align} &\overbrace{\color{blue}{% \lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty}{\expo{\ic\verts{k}x} \over x^{s - 1}} \,\dd x}}^{\ds{\ic\verts{k}x = -t\ \imp\ x = {\ic \over \verts{k}}\,t}} =\lim_{\epsilon \to 0^{+}}\int_{-\epsilon\ic}^{-\infty\ic}\pars{\expo{\ic\pi/2}t \over \verts{k}}^{1 - s} \expo{-t}\,{\ic \over \verts{k}}\,\dd t \\[3mm]&=-\,{\expo{-\pi s\ic/2} \over \verts{k}^{2 - s}} \lim_{\epsilon \to 0^{+}}\int_{-\epsilon\ic}^{-\infty\ic}t^{1 - s}\expo{-t}\,\dd t \\[3mm]&=-\,{\expo{-\pi s\ic/2} \over \verts{k}^{2 - s}}\times \\[3mm]&\lim_{\epsilon \to 0^{+}}\bracks{% -\int^{\epsilon}_{\infty}t^{1 - s}\expo{-t}\,\dd t -\lim_{R \to \infty}\int_{-\pi/2}^{0} R^{1 - s}\expo{\ic\pars{1 - s}\theta}\exp\pars{-R\expo{\ic\theta}}R \expo{\ic\theta}\ic\,\dd\theta}\qquad\pars{2} \end{align}

$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \mbox{When}\quad\epsilon \to 0^{+},\ \mbox{the first integral converges when}\ \Re\pars{1 - s} > -1\ \imp\ \Re\pars{s} < 2\tag{3} $$

Let's study the second integral in the limit $\ds{R \to \infty}$: \begin{align} &\verts{\int_{-\pi/2}^{0} R^{1 - s}\expo{\ic\pars{1 - s}\theta}\exp\pars{-R\expo{\ic\theta}}R \expo{\ic\theta}\ic\,\dd\theta} \leq R^{2 - s}\int_{-\pi/2}^{0}\exp\pars{-R\cos\pars{\theta}}\,\dd\theta \\[3mm]&=R^{2 - s}\int_{0}^{\pi/2}\exp\pars{-R\sin\pars{\theta}}\,\dd\theta <R^{2 - s}\int_{0}^{\pi/2}\exp\pars{-R\,{2\theta \over \pi}}\,\dd\theta \\[3mm]&={\pi \over 2}\pars{R^{1 - s} - R^{1 - s}\expo{-R}} \to 0\ \mbox{when}\ \Re\pars{1 - s} < 0\ \imp\ \Re\pars{s} > 1\tag{4} \end{align}

$\pars{3}$ and $\pars{4}$ show that both terms in $\pars{2}$ converge whenever $\ds{1 < \Re\pars{s} < 2}$: $$ \color{blue}{\lim_{\epsilon \to 0^{+}} \int_{\epsilon}^{\infty}{\expo{\ic\verts{k}x} \over x^{s - 1}}\,\dd x} =-\,{\expo{-\pi s\ic/2} \over \verts{k}^{2 - s}}\,\Gamma\pars{2 - s}\,,\qquad\qquad 1 < \Re\pars{s} < 2 $$ where $\ds{\Gamma\pars{z}}$ is the Gamma Function. This result is replaced in $\pars{1}$ to find: \begin{align} I&=-\,\half\,\cos\pars{\pi s \over 2}\Gamma\pars{2 - s} \int_{-1}^{1}\verts{k}^{s - 2}\,\dd k =-\,\half\,\cos\pars{\pi s \over 2}\Gamma\pars{2 - s}\,{2 \over s - 1} \end{align}

$$\color{#00f}{\large% I = \lim_{\epsilon \to 0^{+}}\int_{\epsilon}^{\infty} {\sin\pars{x} \over x^{s}}\,\dd x = \cos\pars{\pi s \over 2}\Gamma\pars{1 - s}}\,,\qquad 1 < \Re\pars{s} < 2 $$ where we used the Gamma Recurrence Formula ${\bf\mbox{6.1.15}}$.