Fourier transform of a polynomial of fractional degree

Question

I'm trying to prove which polynomials $t^\alpha$ with $\alpha>0$ belong to $\hat{H}^{1+\beta}([0,T])$, $0<\beta\leq 1$, given by $$\hat{H}^{s}([0,T])=\left\{u \in L^2([0,T]): \int_{0}^T \left(1+|\xi|^{2}\right)^s|\hat{u}(\xi)|^2d\xi<\infty\right\}.$$ I get $$\hat{u}(\xi)=\dfrac{e^{i\xi}}{\sqrt{2\pi}}(-T)^{-\alpha}T^\alpha\left[\Gamma(\alpha+1,-T)-\Gamma(\alpha+1)\right]$$ so $$|\hat{u}(\xi)|^2=\dfrac{e^{2\xi}}{2\pi}(-T)^{-2\alpha}T^{2\alpha}\left[\Gamma(\alpha+1,-T)-\Gamma(\alpha+1)\right]^2.$$ Then, I can't move forward. I need to compute the following integral $$\dfrac{1}{2\pi}(-T)^{-2\alpha}T^{2\alpha}\left[\Gamma(\alpha+1,-T)-\Gamma(\alpha+1)\right]^2\int_{0}^T \left(1+|\xi|^{2}\right)^se^{2\xi}d\xi.$$ Any help? Thanks in advance!

Answer 1

If I am not wandering, we have: $$ \int_0^T \left(1 + \xi^2\right)^s\,e^{2\xi}\,\text{d}\xi = \int_0^T \sum_{k = 0}^{\infty} \binom{s}{k}\,\xi^{2k}\,e^{2\xi}\,\text{d}\xi = \sum_{k = 0}^{\infty} \binom{s}{k} \int_0^T \xi^{2k}\,e^{2\xi}\,\text{d}\xi $$ ie: $$ \int_0^T \left(1 + \xi^2\right)^s\,e^{2\xi}\,\text{d}\xi = \sum_{k = 0}^{\infty} \binom{s}{k}2^{-2k-1}\left(\Gamma(2k+1,\,-2T) - (2k)!\right) $$ and besides this I could not go.