Evaluate $\int_{0}^{\infty} \frac {\cos x - 1}{x^{1+\alpha}}\ dx$, $0 < \alpha < 1$ I am interested in finding a formula for the evaluation of
\begin{equation}
\int_0^\infty \frac{\cos(x) - 1}{x^{1+\alpha}}dx, \quad 0<\alpha<1.
\end{equation}
I believe the integral exists as an improper integral, however the calculation becomes tricky due to our choice of $\alpha$.  Given the similar look to integrating $\sin x /x$, I decided to try using the Laplace transform.  So I let
\begin{equation}
A(t) = \int_0^\infty \frac{\cos(tx) - 1}{x^{1+\alpha}}dx
\end{equation}
and then do
\begin{align}
\mathcal{L}(A)(t) ={}& \int_0^\infty \int_0^\infty \frac{\cos(tx) - 1}{x^{1+\alpha}} e^{-st}dx dt\\
={}& \int_0^\infty \frac{1}{x^{1+\alpha}} \int_0^\infty (\cos(tx) - 1)e^{-st}dtdx\\
={}& \int_0^\infty \frac{1}{x^{1+\alpha}} \mathcal{L}(\cos(tx) - 1)(t)dx\\
={}& \int_0^\infty \frac{1}{x^{1+\alpha}} \left( \frac{s}{s^2+x^2} - \frac{1}{s} \right)dx\\
={}& - \frac{1}{s} \int_0^\infty \frac{x^{1-\alpha}}{s^2+x^2}dx.
\end{align}
Unfortunately I've found myself unable to solve this improper integral.  Given that our only condition on $\alpha$ is $0 < \alpha < 1$, it doesn't seem to lend itself to trigonometric substitution.  Integration by parts didn't appear to clean anything up either.
I greatly appreciate any clarity or ideas.  To be honest, even if I can solve the last improper integral, I can't be sure the inverse Laplace transform will be clean either; so perhaps my initial approach is where I need work.  Thank you.
 A: I thought it might be instructive to present an approach that circumvents use of complex analysis and relies on real analysis only.  To that end, we now proceed.

Let $I(\alpha)$, $0<\alpha<2$, be given by the integral
$$I(\alpha)=\int_0^\infty \frac{1-\cos(x)}{x^{1+\alpha}}\,dx$$
Let $f(t)=1-\cos(t)$ and $G(s) = \frac1{s^{1+\alpha}}$.  Then appealing to this property of the Laplace Transform, we have
$$\begin{align}
I(\alpha)&=\int_0^\infty \color{blue}{\mathscr{L}\{f\}(x)}
\color{red}{ \mathscr{L^{-1}}\{G\}(x)}\,dx\\\\
&=\int_0^\infty \color{blue}{\frac{1}{x(x^2+1)}}\,\,\,\color{red}{\frac{x^{\alpha}}{\Gamma(1+\alpha)}}\,dx\\\\
&=\frac{1}{\Gamma(1+\alpha)}\int_0^\infty \frac{x^{\alpha-1}}{x^2+1}\,dx\tag1
\end{align}$$
In THIS ANSWER, I showed using real analysis only that the integral on the right-hand side of $(1)$ is given by
$$\int_0^\infty \frac{x^{\alpha-1}}{x^2+1}\,dx=\frac{\pi}{2}\sec\left(\frac{\pi}{2}(\alpha-1)\right)\tag2$$
Using $(2)$ in $(1)$ yields for $\alpha\in (0,2)$
$$\bbox[5px,border:2px solid #C0A000]{I(\alpha)=\frac{\pi}{2\Gamma(1+\alpha)\sin(\pi \alpha/2)}}\tag3\\$$


NOTE:  ALTERNATIVE FORM
Using the reflection formula $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ for the Gamma function, we can write $(3)$ alternatively as for $\alpha\in (0,2)\setminus { 1}$
$$\bbox[5px,border:2px solid #C0A000]{I(\alpha)=-\cos(\pi \alpha)\Gamma(-\alpha)}\tag4$$
which agrees with previously reported results.
For $\alpha\to 1$, the right-hand side of $(4)$ approaches $\pi/2$.  So, if we define $(4)$ as a function with a removeable discontinuity at $\alpha=1$, the the result holds for $\alpha \in (0,2)$.
A: Integrating by parts gives that
$$ \int_0^{+\infty}\frac{1-\cos(x)}{x^{1+\alpha}}dx=\frac{1}{\alpha}\int_0^{+\infty}\frac{\sin(x)}{x^{\alpha}}dx $$
Let $f(z):=\frac{e^{iz}}{z^{\alpha}}$ and let $\gamma_R$ be the contour integration $[0,R]\cup\left\{Re^{i\vartheta},\vartheta\in\left[0,\frac{\pi}{2}\right]\right\}\cup[iR,0]$. Then because $f$ has no singularities inside the contour, we have
$$ \int_{\gamma_R}f(z)dz=0 $$
However,
$$ \int_{\gamma_R}f(z)dz=\int_0^R f(t)dt+iR\int_0^{\frac{\pi}{2}}f(Re^{i\vartheta})e^{i\vartheta}d\vartheta-i\int_0^R f(it)dt $$
And,
$$ \left|\int_0^{\frac{\pi}{2}}f(Re^{i\vartheta})e^{i\vartheta}d\vartheta\right|\leqslant\frac{1}{R^{\alpha}}\int_0^{\frac{\pi}{2}}e^{-R\sin\vartheta}d\vartheta\leqslant\frac{1}{R^{\alpha}}\int_0^{\frac{\pi}{2}}e^{-R\frac{2}{\pi}\vartheta}d\vartheta\ll\frac{1}{R^{1+\alpha}} $$
Therefore
$$ \lim\limits_{R\rightarrow +\infty}iR\int_0^{\frac{\pi}{2}}f(Re^{i\vartheta})e^{i\vartheta}d\vartheta=0 $$
Taking the limit as $R\rightarrow +\infty$ gives that
$$ \int_0^{+\infty}f(t)dt=i\int_0^{+\infty}f(it)dt $$
That is
$$ \int_0^{+\infty}\frac{e^{it}}{t^{\alpha}}dt=e^{i\frac{\pi}{2}(1-\alpha)}\Gamma(1-\alpha) $$
Taking the imaginary part gives that
$$ \int_0^{+\infty}\frac{\sin(t)}{t^{\alpha}}dt=\sin\left(\frac{\pi}{2}(1-\alpha)\right)\Gamma(1-\alpha)=-\cos\left(\frac{\pi\alpha}{2}\right)\alpha\Gamma(-\alpha) $$
We can therefore conclude that
$$ \int_0^{+\infty}\frac{1-\cos(x)}{x^{1+\alpha}}dx=-\cos\left(\frac{\pi\alpha}{2}\right)\Gamma(-\alpha) $$
which is the formula given by Hans Olo in the comments section.
