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Someone could help me to solve the following integral?

$$\int_{\mathbb{R}} \dfrac{1-\cos(z)}{|z|^{1+\alpha}}dz.$$

I think that the result should be $$\dfrac{\pi^{\frac{1}{2}} \Gamma(1-\alpha/2)}{\alpha 2^{\alpha-1} \Gamma((1+\alpha)/2)},$$ but I don't know how to see it.

Thanks a lot.

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    $\begingroup$ If $z \in \mathbb R^d$, what is $\cos (z)$ ?? $\endgroup$
    – Fred
    Nov 25 at 10:44
  • $\begingroup$ Well, in this moment I don't know how, so I change the question for only $d=1$. $\endgroup$
    – Ivan
    Nov 25 at 11:20
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Too long for a comment

For $d=1$. The integrand is even, therefore $$I=\int_{\mathbb{R}} \dfrac{1-\cos z}{|z|^{1+\alpha}}dz=2\int_0^\infty\frac{1-\cos z}{z^{1+\alpha}}dz$$ Integrating by part $$I=\frac{2}{\alpha}\int_0^\infty x^{-\alpha}\sin x dx=\frac{2}{\alpha}\Im\int_0^\infty x^{-\alpha}e^{ix} dx$$ The last integral is known; we can also evaluate it integrating in the complex plane along the following contour: from $r$ to $R$ along the axis $X$, then along the big quarter-circle (of the radius $R$) counter-clockwise, then along the axis $Y$ from $R$ to $r$, and finally along the small quarter-circle of radius $r$ clockwise - to the destination point. There are no singularities inside the contour, and it is easy to show that the integrals along quarter-circles $\to0$ as $R\to\infty$ and $r\to0$.

Therefore, $\oint=0$, and $$I=-\frac{2}{\alpha}\Im\int_\infty^0 \Big(xe^{\frac{\pi i}{2}}\Big)^{-\alpha}e^{-x}idx=\frac{2}{\alpha}\Re \Big(\,i\,\Gamma(1-\alpha)e^{-\frac{\pi i\alpha}{2}}\Big)=\frac{2}{\alpha}\Gamma(1-\alpha)\cos\frac{\pi \alpha}{2}$$

On the other hand, using the reflection formula $\Gamma(1-\frac{\alpha}{2})=\frac{\pi}{\sin\frac{\pi \alpha}{2}}\frac{1}{\Gamma(\frac{\alpha}{2})}$ and duplication formula $\Gamma(\frac{\alpha}{2})\Gamma(\frac{\alpha}{2}+\frac{1}{2})=2^{1-\alpha}\sqrt\pi\Gamma(\alpha)$ $$\dfrac{\pi^{\frac{1}{2}} \Gamma(1-\alpha/2)}{\alpha 2^{\alpha-1} \Gamma((1+\alpha)/2)}=\frac{\sqrt\pi}{\alpha2^{\alpha-1}}\frac{\pi}{\sin\frac{\pi \alpha}{2}\Gamma(\frac{\alpha}{2})}\frac{1}{\Gamma(\frac{\alpha}{2}+\frac{1}{2})}=\frac{\pi}{\alpha}\frac{1}{\Gamma(\alpha)\sin\frac{\pi \alpha}{2}}$$ $$=\frac{\pi}{\alpha}\frac{2\cos\frac{\pi \alpha}{2}}{\Gamma(\alpha)2\sin\frac{\pi \alpha}{2}\cos\frac{\pi \alpha}{2}}=\frac{2\cos\frac{\pi \alpha}{2}}{\alpha}\frac{\pi}{\Gamma(\alpha)\sin\pi\alpha}=\frac{2}{\alpha}\Gamma(1-\alpha)\cos\frac{\pi \alpha}{2}$$

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