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Question Number 129064 by Eric002 last updated on 12/Jan/21
valuate the following integral I=∫_1 ^∞ (dt/((t)^(2k+v+(3/2)) (√(t−1)))) and prove that: I=(√π)((Γ(v+1))/(Γ(v+(3/2)))) ((((((v+1)/2))_k (1+(v/2))_k )/(((v/2)+(3/4))_k ((v/2)+(5/4))_k )))
$${valuate}\:{the}\:{following}\:{integral} \\ $$$${I}=\int_{\mathrm{1}} ^{\infty} \frac{{dt}}{\left({t}\right)^{\mathrm{2}{k}+{v}+\frac{\mathrm{3}}{\mathrm{2}}} \sqrt{{t}−\mathrm{1}}} \\ $$$${and}\:{prove}\:{that}: \\ $$$${I}=\sqrt{\pi}\frac{\Gamma\left({v}+\mathrm{1}\right)}{\Gamma\left({v}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\:\left(\frac{\left(\frac{{v}+\mathrm{1}}{\mathrm{2}}\right)_{{k}} \left(\mathrm{1}+\frac{{v}}{\mathrm{2}}\right)_{{k}} }{\left(\frac{{v}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}\right)_{{k}} \left(\frac{{v}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{4}}\right)_{{k}} }\right) \\ $$
Answered by Dwaipayan Shikari last updated on 12/Jan/21
∫_1 ^∞ (1/(t^z (√(t−1))))dt z=2k+v+(3/2) t−1=u =∫_0 ^∞ u^(−(1/2)) .(1/((u+1)^z ))dz = ∫_0 ^∞ (u^((1/2)−1) /((u+1)^((z−(1/2))+(1/2)) ))du =β((1/2),z−(1/2))=((Γ((1/2))Γ(z−(1/2)))/(Γ(z)))=((√π)/(Γ(2k+v+(3/2))))Γ(2k+v+1) =(√π)((Γ(2k+v+1))/(Γ(2k+v+(3/2))))
$$\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{t}^{{z}} \sqrt{{t}−\mathrm{1}}}{dt}\:\:\:\:\:\:\:\:\:{z}=\mathrm{2}{k}+{v}+\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:{t}−\mathrm{1}={u} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {u}^{−\frac{\mathrm{1}}{\mathrm{2}}} .\frac{\mathrm{1}}{\left({u}+\mathrm{1}\right)^{{z}} }{dz}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} }{\left({u}+\mathrm{1}\right)^{\left({z}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}} }{du} \\ $$$$=\beta\left(\frac{\mathrm{1}}{\mathrm{2}},{z}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({z}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({z}\right)}=\frac{\sqrt{\pi}}{\Gamma\left(\mathrm{2}{k}+{v}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\Gamma\left(\mathrm{2}{k}+{v}+\mathrm{1}\right) \\ $$$$=\sqrt{\pi}\frac{\Gamma\left(\mathrm{2}{k}+{v}+\mathrm{1}\right)}{\Gamma\left(\mathrm{2}{k}+{v}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$

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