Question Number 129064 by Eric002 last updated on 12/Jan/21
$${valuate}\:{the}\:{following}\:{integral} \\ $$$${I}=\int_{\mathrm{1}} ^{\infty} \frac{{dt}}{\left({t}\right)^{\mathrm{2}{k}+{v}+\frac{\mathrm{3}}{\mathrm{2}}} \sqrt{{t}−\mathrm{1}}} \\ $$$${and}\:{prove}\:{that}: \\ $$$${I}=\sqrt{\pi}\frac{\Gamma\left({v}+\mathrm{1}\right)}{\Gamma\left({v}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\:\left(\frac{\left(\frac{{v}+\mathrm{1}}{\mathrm{2}}\right)_{{k}} \left(\mathrm{1}+\frac{{v}}{\mathrm{2}}\right)_{{k}} }{\left(\frac{{v}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}\right)_{{k}} \left(\frac{{v}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{4}}\right)_{{k}} }\right) \\ $$
Answered by Dwaipayan Shikari last updated on 12/Jan/21
$$\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{t}^{{z}} \sqrt{{t}−\mathrm{1}}}{dt}\:\:\:\:\:\:\:\:\:{z}=\mathrm{2}{k}+{v}+\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:{t}−\mathrm{1}={u} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {u}^{−\frac{\mathrm{1}}{\mathrm{2}}} .\frac{\mathrm{1}}{\left({u}+\mathrm{1}\right)^{{z}} }{dz}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\left({u}+\mathrm{1}\right)^{\left({z}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}} }{du} \\ $$$$=\beta\left(\frac{\mathrm{1}}{\mathrm{2}},{z}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({z}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({z}\right)}=\frac{\sqrt{\pi}}{\Gamma\left(\mathrm{2}{k}+{v}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\Gamma\left(\mathrm{2}{k}+{v}+\mathrm{1}\right) \\ $$$$=\sqrt{\pi}\frac{\Gamma\left(\mathrm{2}{k}+{v}+\mathrm{1}\right)}{\Gamma\left(\mathrm{2}{k}+{v}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$