Convergence of $\int_0 ^{\infty}\frac{\cos t}{t^{\alpha}} dt$ related to $\Gamma$ function 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 $\alpha \in (0,1)$.
 A: 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
$$
\begin{align}
\int_{0}^{\infty} t^{-\alpha} \cos t \, dt & = \sin \left(\frac{\pi \alpha}{2}\right)\Gamma(1-\alpha), \quad 0<\alpha<1. \tag2 
\end{align}
$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[2]{\,\mathrm{Li}_{#1}\left(\,{#2}\,\right)}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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 $\color{#f00}{\alpha < 1}$.
$$
\begin{array}{|c|}\hline\mbox{}\\
\quad\mbox{Then, the original integral}\ 
\ds{\color{#f00}{\int_{0}^{\infty}{\cos\pars{t} \over t^{\alpha}}\,\dd t}}\
\mbox{converges when}\ \color{#f00}{\ds{0 < \alpha < 1}}.\quad
\\ \mbox{}\\ \hline
\end{array}
$$

$$
\color{#f00}{\int_{0}^{\infty}{\cos\pars{t} \over t^{\alpha}}\,\dd t} =
\sin\pars{\pi\alpha \over 2}\int_{0}^{\infty}y^{-\alpha}\expo{-y}\,\dd y =
\color{#f00}{\sin\pars{\pi\alpha \over 2}\Gamma\pars{1 - \alpha}}\,,\qquad
0 < \alpha < 1
$$
