Trouble in complex integral while calculating 2-point function in AdS-CFT correspondence [closed]

Question

Upon calculating the 2-point function of a scalar, I ran in a problem in the final calculations. For reference, I am basiccaly following the methodology of https://www.sissa.it/tpp/phdsection/OnlineResources/16/SISSA_AdSCFT_course2022.pdf. My problem appears at eq 6.40 (in the footnote). I am not quite certain how to properly solve the integral in eq 6.38. I used a few tricks my professor told me and arrived in the integral below.

The integral in question is $$2 \int_0^{+\infty}z^{2b-1} sinz \cdot dz \quad \quad b>0$$ I have tried 2 different but similar approaches, one calculating the integral $\int_{-\infty}^{+\infty}z^{2b-1} e^{iz} \cdot dz$ and then take the imaginary part, but using Jordan's lemma and since the integral has no poles, it vanishes!

My second approach uses the fact that $sinz = \dfrac{e^{iz} - e^{-iz}}{2i}$. I use that identity to break the integral in 2 parts and once again integrate in closed contours and use Jordan's lemma. I get $$ \int_0^{+\infty}z^{2b-1} e^{iz} \cdot dz = \int_0^{+\infty}z^{2b-1} e^{-iz} \cdot dz = (-1)^{b-1} \Gamma(2b-1)$$ But, since I want to subtract one integral from the other, the final result once again vanishes! I am sure the result cannot be zero, and I know it must contain Gamma functions, but I cannot figure out my mistake

Answer 1

By integrating $$ \int_C e^{iz} z^{\alpha-1}dz $$ about the boundary of the first quadrant and taking the imaginary part, one can show that $$ \int_0^\infty x^{\alpha-1} \sin x dx=\Gamma(\alpha)\sin (\pi \alpha/2), \quad 0<\alpha<1. $$ The real part gives the cosine version. The limits on $\alpha$ are required to ensure that the large circle at infinity, and a small circle avoiding the origin, make no contribution. The total contour integral is zero and the gamma functions come from the integral down the $y$ axis from infinity back to zero.