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Question Number 129064 by Eric002 last updated on 12/Jan/21

$$valuatethefollowingintegral$$$$I={\int }_1^{\mathrm{\infty }}\frac{dt}{{\left(t\right)}^{2k+v+\frac{3}{2}}\sqrt{t-1}}$$$$andprovethat:$$$$I=\sqrt{\pi }\frac{\mathrm{\Gamma }(v+1)}{\mathrm{\Gamma }(v+\frac{3}{2})}\left(\frac{{\left(\frac{v+1}{2}\right)}_k{(1+\frac{v}{2})}_k}{{(\frac{v}{2}+\frac{3}{4})}_k{(\frac{v}{2}+\frac{5}{4})}_k}\right)$$

Answered by Dwaipayan Shikari last updated on 12/Jan/21

$${\int }_1^{\mathrm{\infty }}\frac{1}{t^z\sqrt{t-1}}dtz=2k+v+\frac{3}{2}t-1=u$$$$={\int }_0^{\mathrm{\infty }}{u}^{-\frac{1}{2}}.\frac{1}{{(u+1)}^z}dz={\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{2}-1}}{{(u+1)}^{(z-\frac{1}{2})+\frac{1}{2}}}du$$$$=\beta (\frac{1}{2},z-\frac{1}{2})=\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(z-\frac{1}{2})}{\mathrm{\Gamma }\left(z\right)}=\frac{\sqrt{\pi }}{\mathrm{\Gamma }(2k+v+\frac{3}{2})}\mathrm{\Gamma }(2k+v+1)$$$$=\sqrt{\pi }\frac{\mathrm{\Gamma }(2k+v+1)}{\mathrm{\Gamma }(2k+v+\frac{3}{2})}$$

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