valuate-the-following-integral-I-1-dt-t-2k-v-3-2-t-1-and-prove-that-I-pi-v-1-v-3-2-v-1-2-k-1-v-2-k-v-2-3-4-k-v-2-5-4-k-

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

v a l u a t e t h e f o l l o w i n g i n t e g r a l

I = \int_{1}^{\infty} \frac{d t}{{(t)}^{2 k + v + \frac{3}{2}} \sqrt{t - 1}}

a n d p r o v e t h a t :

I = \sqrt{π} \frac{Γ (v + 1)}{Γ (v + \frac{3}{2})} (\frac{{(\frac{v + 1}{2})}_{k} {(1 + \frac{v}{2})}_{k}}{{(\frac{v}{2} + \frac{3}{4})}_{k} {(\frac{v}{2} + \frac{5}{4})}_{k}})

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

\int_{1}^{\infty} \frac{1}{t^{z} \sqrt{t - 1}} d t z = 2 k + v + \frac{3}{2} t - 1 = u

= \int_{0}^{\infty} u^{- \frac{1}{2}} . \frac{1}{{(u + 1)}^{z}} d z = \int_{0}^{\infty} \frac{u^{\frac{1}{2} - 1}}{{(u + 1)}^{(z - \frac{1}{2}) + \frac{1}{2}}} d u

= β (\frac{1}{2}, z - \frac{1}{2}) = \frac{Γ (\frac{1}{2}) Γ (z - \frac{1}{2})}{Γ (z)} = \frac{\sqrt{π}}{Γ (2 k + v + \frac{3}{2})} Γ (2 k + v + 1)

= \sqrt{π} \frac{Γ (2 k + v + 1)}{Γ (2 k + v + \frac{3}{2})}