I am trying to do the following integration $$\int_0^\infty d\rho \int_0^\pi \frac{\cos(\rho \cos\theta)-1}{\rho^{1+\alpha}}\sin^{d-2} \theta d\theta$$ where $\alpha\in(0,2), d\geq 2, d\in \mathbb{N}$.
The solution should be something consisting of Gamma functions $\Gamma(\cdot)$ but I don't know how to compute it.
To evaluate the integral \begin{equation} I=2\int_0^\infty d\rho \int_0^{\pi/2 }\frac{\cos(\rho \cos\theta)-1}{\rho^{1+\alpha}}\sin^{d-2} \theta \,d\theta \end{equation} one may notice that \begin{equation} \cos(\rho \cos\theta)-1=-2\sin^2\frac{\rho}{2} \cos\theta \end{equation} Then, changing $x=\frac{\rho}{2} \cos\theta$, the integral can be written as a product of two integrals: \begin{equation} I=-2^{2-\alpha}\int_0^\infty\frac{\sin^2 x}{x^{1+\alpha}}\, dx\int_0^{\pi/2}\sin^{d-2}\theta\cos^\alpha\theta\,d\theta \end{equation} From the integral representation of the Beta function (DLMF), \begin{equation} \int_0^{\pi/2}\sin^{d-2}\theta\cos^\alpha\theta\,d\theta=\frac{1}{2}\operatorname{B}\left( \frac{d-1}{2},\frac{1+\alpha}{2} \right) \end{equation} We define \begin{equation} J(t)=\int_0^\infty\frac{\sin^2 xt}{x^{1+\alpha}} \,dx \end{equation} then the first integral is $J(1)$. A differentiation wrt the auxiliary variable $t$ gives \begin{align} J'(t)&=\int_0^\infty\frac{\sin 2xt}{x^{\alpha}}\,dx\\ &=2^{\alpha-1} t^{\alpha-1}\int_0^\infty u^{-\alpha}\sin u\,du\\ &=2^{\alpha-1} t^{\alpha-1}\Gamma\left( 1-\alpha \right)\sin\left( \frac{\pi}{2}(1-\alpha) \right) \end{align} where the integral representation (DLMF) was used. Then, noticing that $J(0)=0$, \begin{align} J(1)&=2^{\alpha-1} \Gamma\left( 1-\alpha \right)\sin\left( \frac{\pi}{2}(1-\alpha) \right)\int_0^1t^{\alpha-1}\,dt\\ &=2^{\alpha-1} \frac{\Gamma\left( 1-\alpha \right)}{\alpha}\cos\left( \frac{\pi}{2}\alpha \right)\\ &=-2^{\alpha-1}\Gamma\left( -\alpha \right) \cos\left( \frac{\pi}{2}\alpha \right) \end{align} Finally, \begin{align} I&=\operatorname{B}\left( \frac{d-1}{2},\frac{1+\alpha}{2} \right)\Gamma\left( -\alpha \right)\cos\left( \frac{\pi}{2}\alpha \right)\\ &=\frac{\Gamma\left( -\alpha \right)\Gamma\left( \frac{d-1}{2} \right)\Gamma\left( \frac{1+\alpha}{2} \right)}{\Gamma\left( \frac{\alpha+d}{2} \right)}\cos\left( \frac{\pi}{2}\alpha \right) \end{align} as noticed in the comment by @Mariusz Iwaniuk