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I would like to show that the integral

$$\int_0 ^\infty \frac{\cos t}{t^{\alpha}} dt$$ converges for $0<\alpha <1$. I already showed that it does not converge for $\alpha\leq 0$ or $\alpha \geq 1$. Do you have any hints for me? I know that I can use the gamme function (using $\cos t=\frac{e^{it}+e^{-it}}{2}$) but I don't see why I can use the gamma function just for $\alpha \in (0,1)$.

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  • $\begingroup$ Are you sure it does not converge for $\alpha \geq 1$? $\endgroup$ – Michael Feb 23 '15 at 12:01
  • $\begingroup$ @slinshady Are you OK with my answer? Thanks. $\endgroup$ – Olivier Oloa Feb 24 '15 at 16:34
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Hint. Recall that, from the definition of the Euler $\Gamma$ function, we have $$ \begin{align} \int_{0}^{\infty} e^{-bt} \, t^{-\alpha} \, dt = \frac{\Gamma(1-\alpha)}{b^{1-\alpha}}, \quad 0<\alpha<1, \Re b>0. \tag1 \end{align} $$ Then put $b:=b_\epsilon:=\epsilon+i,\, \epsilon>0$, in $(1)$, let $\epsilon \to 0^+$ and take the real part to get $$ \begin{align} \int_{0}^{\infty} t^{-\alpha} \cos t \, dt & = \sin \left(\frac{\pi \alpha}{2}\right)\Gamma(1-\alpha), \quad 0<\alpha<1. \tag2 \end{align} $$

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  • $\begingroup$ you have to justify why $(1)$ still works with $b$ complex $\endgroup$ – reuns Jul 1 '16 at 3:55
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[2]{\,\mathrm{Li}_{#1}\left(\,{#2}\,\right)} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\int_{0}^{\infty}{\cos\pars{t} \over t^{\alpha}}\,\dd t} & = \Re\int_{0}^{\infty}{\expo{\ic t} \over t^{\alpha}}\,\dd t = -\,\Re\lim_{R \to \infty}\int_{0}^{\pi/2}{\expo{\ic R\cos\pars{\theta}} \expo{-R\sin\pars{\theta}} \over R^{\alpha}\expo{\ic\alpha\theta}} \,R\expo{\ic\theta}\ic\,\dd\theta\tag{1} \\[3mm] & - \Re\lim_{R \to \infty}\int_{R}^{0} {\expo{-y} \over y^{\alpha}\expo{\pi\alpha\ic/2}}\,\ic\tag{2} \,\dd y \end{align}


Above, we closed the contour in the first quadrant and takes the $z^{\alpha}$ branch-cut 'along the negative real axis' with $-\pi < \,\mathrm{arg}\pars{z} < \pi$. It turns out that \begin{align} 0 &< \verts{\int_{0}^{\pi/2}{\expo{\ic R\cos\pars{\theta}} \expo{-R\sin\pars{\theta}} \over R^{\alpha}\expo{\ic\alpha\theta}} \,R\expo{\ic\theta}\ic\,\dd\theta} \leq {1 \over R^{\alpha - 1}}\verts{\int_{0}^{\pi/2} \expo{-R\sin\pars{\theta}}\,\dd\theta} \\[3mm] & < {1 \over R^{\alpha - 1}}\int_{0}^{\pi/2}\expo{-2R\theta/\pi}\,\dd\theta = {1 \over R^{\alpha - 1}}\,{1 - \expo{-R} \over 2R/\pi} = {\pi \over 2}\pars{{1 \over R^{\alpha}} - {\expo{-R} \over R^{\alpha}}} \\[1mm] & \mbox{The RHS}\ \to 0\ \mbox{whenever}\ \color{#f00}{\alpha > 0}.\ \mbox{In such case, the}\ \mbox{'}\theta\mbox{-integral' in}\ \pars{1} \to\ \stackrel{R\ \to\ \infty}{\color{#f00}{\large 0}} \end{align}
Moreover, the '$y$-integral', in $\pars{2}$, converges in the $R \to \infty$ limit whenever $\color{#f00}{\alpha < 1}$. $$ \begin{array}{|c|}\hline\mbox{}\\ \quad\mbox{Then, the original integral}\ \ds{\color{#f00}{\int_{0}^{\infty}{\cos\pars{t} \over t^{\alpha}}\,\dd t}}\ \mbox{converges when}\ \color{#f00}{\ds{0 < \alpha < 1}}.\quad \\ \mbox{}\\ \hline \end{array} $$
$$ \color{#f00}{\int_{0}^{\infty}{\cos\pars{t} \over t^{\alpha}}\,\dd t} = \sin\pars{\pi\alpha \over 2}\int_{0}^{\infty}y^{-\alpha}\expo{-y}\,\dd y = \color{#f00}{\sin\pars{\pi\alpha \over 2}\Gamma\pars{1 - \alpha}}\,,\qquad 0 < \alpha < 1 $$

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