Evaluate $\int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy$ How do I evaluate the following integral?
$$\int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy=\frac{\pi}{\Gamma(1+\alpha)\sin(\frac{\pi\alpha}{2})}$$
Thank you in advance.

I found the answer. Thank you all! 
 A: Let $I(\alpha)$, $0<\alpha<2$, be the integral given by
$$\begin{align}
I(\alpha)&=\int_{-\infty}^\infty \frac{1-\cos ( y  )}{|y|^{1+\alpha}}\,dy\\\\
&=2\int_0^\infty \frac{1-\cos ( y  )}{y^{1+\alpha}}\,dy \tag 1
\end{align}$$
Integrating by parts $(1)$ with $u=1-\cos(y)$ and $v=-\frac{1}{\alpha y^{\alpha}}$ reveals
$$\begin{align}
I(\alpha)&=\frac{2}{\alpha}\int_0^\infty \frac{\sin(y)}{y^\alpha}\,dy \tag 2\\\\
\end{align}$$
For $1<\alpha <2$, integrating by parts $(2)$ with $u=\sin(y)$ and $v=\frac{1}{1-\alpha}y^{1-\alpha}$ yields
$$\begin{align}
I(\alpha)&=\frac{2}{\alpha(\alpha-1)}\int_0^\infty \frac{\cos(y)}{y^{\alpha-1}}\,dy \tag 3\\\\
\end{align}$$


CASE $1$: $0<\alpha<1$ 

Note that for $0<\alpha<1$, we can write $(2)$ as
$$I(\alpha)=\frac{2}{\alpha}\text{Im}\left(\int_0^\infty \frac{e^{iy}}{y^\alpha}\,dy\right) \tag 4$$
Enforcing the substitution $y\to iy$ in the integral on the right-hand side of $(4)$ yields
$$\int_0^\infty \frac{e^{iy}}{y^\alpha}\,dy=e^{i\pi (1-\alpha)/2}\int_0^{-i\infty} \frac{e^{-y}}{y^\alpha}\,dy \tag 5$$
Using Cauchy's Integral Theorem, we can deform the contour back to the real line and write $(5)$ as 
$$\begin{align}
\int_0^\infty \frac{e^{iy}}{y^\alpha}\,dy&=e^{i\pi (1-\alpha)/2}\int_0^\infty y^{-\alpha}e^{-y}\,dy\\\\&=e^{i\pi (1-\alpha)/2}\Gamma(1-\alpha) \tag 6
\end{align}$$
Substituting $(6)$ into $(4)$, we obtain 
$$\begin{align}
\int_{-\infty}^\infty \frac{1-\cos ( y  )}{|y|^{1+\alpha}}\,dy&=\frac{2}{\alpha}\sin((1-\alpha)\pi/2)\Gamma(1-\alpha)\\\\
&=\frac{2}{\alpha}\cos(\pi \alpha/2)\Gamma(1-\alpha) \tag 7\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{\pi}{\Gamma(1+\alpha)\sin(\pi \alpha/2)}} \tag 8
\end{align}$$
where in going from $(7)$ to $(8)$ we used the functional relationship $\Gamma(1+z)=z\Gamma(z)$ along with Euler's Reflection formula $\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}$


CASE $2$: $1<\alpha<2$ 

Note that for $1<\alpha<2$, we can write $(3)$ as
$$I(\alpha)=\frac{2}{\alpha(\alpha -1)}\text{Re}\left(\int_0^\infty \frac{e^{iy}}{y^{\alpha-1}}\,dy\right) \tag 9$$
Enforcing the substitution $y\to iy$ in the integral on the right-hand side of $(9)$ yields
$$\int_0^\infty \frac{e^{iy}}{y^{\alpha-1}}\,dy=-e^{-i\pi \alpha/2}\int_0^{-i\infty} \frac{e^{-y}}{y^{\alpha-1}}\,dy \tag {10}$$
Using Cauchy's Integral Theorem, we can deform the contour back to the real line and write $(10)$ as 
$$\begin{align}
\int_0^\infty \frac{e^{iy}}{y^{\alpha-1}}\,dy&=-e^{i\pi \alpha/2}\int_0^\infty y^{1-\alpha}e^{-y}\,dy\\\\&=-e^{i\pi \alpha/2}\Gamma(2-\alpha) \tag{11}
\end{align}$$
Substituting $(11)$ into $(9)$, we obtain 
$$\begin{align}
\int_{-\infty}^\infty \frac{1-\cos ( y  )}{|y|^{1+\alpha}}\,dy&=\frac{2}{\alpha(\alpha-1)}\cos(\alpha\pi/2)\Gamma(2-\alpha) \tag {12}\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{\pi}{\Gamma(1+\alpha)\sin(\pi \alpha/2)}}\tag {13}
\end{align}$$
where in going from $(12)$ to $(13)$ we used the functional relationship $\Gamma(1+z)=z\Gamma(z)$ along with Euler's Reflection formula $\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}$.


PUTTING IT ALL TOGETHER:
Using $(8)$ and $(13)$ along with the well-known result $I(1)=\pi/2$, we find that for all $0<\alpha<2$ we have
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{1-\cos ( y  )}{|y|^{1+\alpha}}\,dy=\frac{\pi}{\Gamma(1+\alpha)\sin(\pi \alpha/2)}}$$

as was to be shown!
A: Just an idea:
$$  \int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy = 2\int_0^{\infty}{\sum_{k=0}^{\infty}{ \frac{{\left(y\right)}^{2k+1-\alpha}}{(2k)!} } dy}$$
Let me know if it helps any.
Its complex integration according to one your tags....do you care what method is used for it?
A: Thank you Dr.MV for your answer. Since you found the solution first, I voted your answer as the correct one.
I was also able to find the solution, though in a different way. So for the sake of completeness, I´ll live it here.
First, let´s rewrite  the integral as  $$\int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy=2\int_{0}^{\infty} \frac{(1-\cos { y } )}{y^{1+\alpha}}dy$$
Then use the folowing integral representation:
$$\frac{1}{y^{1+\alpha}}=\frac{1}{\Gamma(1+\alpha)}\int_{0}^{\infty}u^{\alpha}e^{-yu}
du$$ 
to rewrite the original integral as a double integral
$$\int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy=\frac{2}{\Gamma(1+\alpha)}\int_{0}^{\infty}\int_{0}^{\infty} u^{\alpha}e^{-yu}(1- \cos{y}) dy du$$
Swap the order of integration and then concentrate in the inner integral first, which can be splited into two intergals
$$\int_{0}^{\infty}e^{-yu}dy - \int_{0}^{\infty} e^{-yu}\cos{y}dy$$
the result is:
$$\frac{1}{u}-\frac{u}{u^{2}+1} $$
plugging it back into the double integral leads us to:
$$\int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy=\frac{2}{\Gamma(1+\alpha)}\int_{0}^{\infty} u^{\alpha} \left[\frac{1}{u}-\frac{u}{u^{2}+1}\right]du$$ 
$$=\frac{2}{\Gamma(1+\alpha)}\int_{0}^{\infty}  \frac{u^{\alpha}}{u(u^{2}+1)}du =\frac{2}{\Gamma(1+\alpha)}\int_{0}^{\infty}  \frac{u^{\alpha-1}}{(u^{2}+1)}du$$
now, make the substitution $$u=v^{2} \Rightarrow du=\frac{1}{2}v^{-\frac{1}{2}}dv$$
$$\frac{2}{\Gamma(1+\alpha)}\int_{0}^{\infty}  \frac{v^{\frac{\alpha-2}{2}}}{(v+1)}\frac{dv}{2}$$
using the complex analysis result:
$$\int_{0}^{\infty}  \frac{x^{\alpha-1}}{(x+1)}dx=\frac{\pi}{\sin{\pi\alpha}}$$
Leads us to the final answer
$$\int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy=\frac{\pi}{\Gamma(1+\alpha)\sin(\frac{\pi\alpha}{2})}$$
