# Integral of $d$ dimension Calculation

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.

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

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 $$\Gamma(\frac{\alpha}{2})\Gamma(\frac{\alpha}{2}+\frac{1}{2})=2^{1-\alpha}\sqrt\pi\Gamma(\alpha)$$ $$\dfrac{\pi^{\frac{1}{2}} \Gamma(1-\alpha/2)}{\alpha 2^{\alpha-1} \Gamma((1+\alpha)/2)}=\frac{\sqrt\pi}{\alpha2^{\alpha-1}}\frac{\pi}{\sin\frac{\pi \alpha}{2}\Gamma(\frac{\alpha}{2})}\frac{1}{\Gamma(\frac{\alpha}{2}+\frac{1}{2})}=\frac{\pi}{\alpha}\frac{1}{\Gamma(\alpha)\sin\frac{\pi \alpha}{2}}$$ $$=\frac{\pi}{\alpha}\frac{2\cos\frac{\pi \alpha}{2}}{\Gamma(\alpha)2\sin\frac{\pi \alpha}{2}\cos\frac{\pi \alpha}{2}}=\frac{2\cos\frac{\pi \alpha}{2}}{\alpha}\frac{\pi}{\Gamma(\alpha)\sin\pi\alpha}=\frac{2}{\alpha}\Gamma(1-\alpha)\cos\frac{\pi \alpha}{2}$$