Prove that $\int_0^{\infty} \frac{1-\cos(at)}{t^{1+\alpha}} dt = \frac{\pi}{2 \Gamma(\alpha+1) \sin (\alpha \pi /2 )} |a|^{\alpha}$ $$\int_0^{\infty} \frac{1-\cos(at)}{t^{1+\alpha}} dt = \frac{\pi}{2 \Gamma(\alpha+1) \sin (\alpha \pi /2 )} |a|^{\alpha}$$
Is there a name for this integral? I saw someone refers to this integral as a well-known property. Is it from some probability distribution?
We might be integrated by parts like
$$\int \frac{\cos(at)}{t^{1+\alpha}} dt = \int \frac{d\sin(at)}{at^{1+\alpha}} $$
but it seems rather painful and I am going down a recursive rabbit hole.
 A: Let $I(a,\alpha)$, $0<\alpha<2$, be given by the integral
$$I(a,\alpha)=\int_0^\infty \frac{1-\cos(ax)}{x^{1+\alpha}}\,dx$$
Integrating by parts with $u=1-\cos(ax)$ and $v=\frac{x^{-\alpha}}{-\alpha}$ reveals
$$\begin{align}
I(a,\alpha)&=\frac{a}{\alpha}\int_0^\infty \frac{\sin(ax)}{x^\alpha}\,dx\\\\
&=\frac{|a|^\alpha}{\alpha}\int_0^\infty \frac{\sin(x)}{x^\alpha}\,dx\\\\
&=\frac{|a|^\alpha}{\alpha}\Gamma(1-\alpha)\sin(\pi(1-\alpha)/2)\tag1\\\\
&=\frac{\pi|a|^\alpha}{2\sin(\pi \alpha/2)\Gamma(1+\alpha)}\tag2
\end{align}$$
as was to be shown!
In going from $(1)$ to $(2)$, we applied the reflection formula along with the functional equation $\Gamma(1+x)=x\Gamma(x)$ for the Gamma function.


APPENDIX:
Here, we show that $\int_0^\infty \frac{\sin(x)}{x^\alpha}\,dx=\Gamma(1-\alpha)\sin(\pi(1-\alpha)/2)$.  To begin, we let $f(z)=\frac{e^{iz}}{z^\alpha}$ and  cut the plane along the negative real axis such that $\arg(z)\in (-\pi,\pi]$.
For $0<\varepsilon<R$, we apply Cauchy's Integral Theorem to write
$$\begin{align}
0&=\int_\varepsilon^R \frac{e^{ix}}{x^\alpha}\,dx+\int_0^{\pi/2}\frac{e^{iRe^{i\phi}}}{(Re^{i\phi})^\alpha}\,iRe^{i\phi}\,d\phi\\\\
&+e^{i(1-\alpha)\pi/2}\int_R^\varepsilon \frac{e^{-y}}{y^\alpha}\,dy+\int_0^{\pi/2}\frac{e^{i\varepsilon e^{i\phi}}}{(\varepsilon e^{i\phi})^\alpha}\,i\varepsilon e^{i\phi}\,d\phi\tag{A1}
\end{align}$$
Letting $R\to \infty$ in $(A1)$ and taking the imaginary part shows that
$$\begin{align}
\int_\varepsilon^\infty \frac{\sin(x)}{x^\alpha}&=\sin(\pi (1-\alpha)/2)\int_\varepsilon^\infty \frac{e^{-y}}{y^\alpha}\,dy\\\\
&+\frac{\sin(\pi(1-\alpha)/2)}{1-\alpha}\,\varepsilon^{1-\varepsilon}+O(\varepsilon^{2-\alpha})\tag{A2}
\end{align}$$
For $0<\alpha<1$, we let $\varepsilon\to 0^+$ in $(A2)$ to obtain
$$\int_0^\infty \frac{\sin(x)}{x^\alpha}\,dx=\sin(\pi (1-\alpha)/2)\Gamma(1-\alpha)\tag{A3}$$
For $1<\alpha<2$, we integrate by parts the integral on the right-hand side of $(A2)$ with $u=e^{-y}$ and $v=\frac{y^{1-\alpha}}{1-\alpha}$, and take the limit as $\varepsilon\to 0^+$ to obtain
$$\begin{align}
\int_0^\infty \frac{\sin(x)}{x^\alpha}\,dx&=\sin(\pi (1-\alpha)/2)\frac{\Gamma(2-\alpha)}{1-\alpha}\\\\
&=\sin(\pi (1-\alpha)/2)\Gamma(1-\alpha)\tag{A4}
\end{align}$$
Putting $(A3)$ and $(A4)$ together we find that for $\alpha\in (0,2)$
$$\int_0^\infty \frac{\sin(x)}{x^\alpha}\,dx=\sin(\pi (1-\alpha)/2)\Gamma(1-\alpha)$$
where $\alpha=1$ is a removeable discontinuity.
A: By the Tonelli's Theorem,
\begin{align*}
\int_{0}^{\infty} \frac{1-\cos(at)}{t^{1+\alpha}} \, \mathrm{d}t
&= \int_{0}^{\infty} (1-\cos(at))\int_{0}^{\infty} \frac{x^{\alpha}e^{-tx}}{\Gamma(\alpha+1)} \, \mathrm{d}x \mathrm{d}t \\
&= \int_{0}^{\infty} \frac{x^{\alpha}}{\Gamma(\alpha+1)} \int_{0}^{\infty} (1-\cos(at))e^{-xt} \, \mathrm{d}t \mathrm{d}x \\
&= \frac{1}{\Gamma(\alpha+1)} \int_{0}^{\infty} \frac{a^2 x^{\alpha-1}}{x^2 + a^2} \, \mathrm{d}x \\
&= \frac{|a|^{\alpha}}{\Gamma(\alpha+1)} \int_{0}^{\infty} \frac{s^{\alpha-1}}{s^2 + 1} \, \mathrm{d}s \tag{$x=|a|s$}
\end{align*}
Now the last integral can be computed in various ways to give the desired result.
