0-t-a-1-1-t-dt-pi-sin-pia-please-prove-that-

Question Number 45075 by arvinddayama01@gmail.com last updated on 08/Oct/18

\int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} dt = \frac{π}{\sin (π a)}

please prove that

Commented by abdo.msup.com last updated on 08/Oct/18

Commented by maxmathsup by imad last updated on 08/Oct/18

l e t n t e g r a t e t h e c o m p l e x f u n t i o n f (z) = \frac{z^{a - 1}}{1 + z} o n t h i s c o n t o u r s h o w n

w i t h 0 < a < 1 r e s i d u s t h e o r e m g i v e

\int_{Γ} f (z) d z + \int_{A B} f (z) d z + \int_{γ} f (z) d z + \int_{A^{^{'}} B^{^{'}}} f (z) d z = 2 i π R e s (f, - 1)

o n Γ z = R e^{i θ} \Rightarrow \int_{Γ} f (z) d z = \int_{0}^{2 π} f (R e^{i θ}) R i e^{i θ} d θ

= \int_{0}^{2 π} \frac{{(R e^{i θ})}^{a - 1}}{1 + {R e}^{i θ}} R i e^{i θ} d θ \Rightarrow \int_{Γ} ∣ f (z) ∣ {d z}_{R \to + \infty} \to 0

o n A B z = t e^{i 2 π} \Rightarrow \int_{A B} f (z) d z = \int_{R}^{r} \frac{{(t e^{i 2 π})}^{a - 1}}{1 + t} {d t}_{r \to 0 a n d R \to + \infty} \to - \int_{0}^{+ \infty} \frac{t^{a - 1}}{1 + t} e^{i 2 π (a - 1)} d t

o n γ z = r e^{i θ} a n d \int_{γ} f (z) d z = \int_{0}^{2 π} f (r e^{i θ}) d θ \to 0 w h e n r \to 0

o n A^{^{'}} B^{^{'}} z = t a n d \int_{A^{^{'}} B^{^{'}}} f (z) d z = \int_{r}^{R} \frac{t^{a - 1}}{1 + t} d t \to \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t (r \to 0 a n d R \to + \infty)

s o w e g e t \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t - e^{i 2 π (a - 1)} \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = 2 i π R e s (f, - 1) \Rightarrow

(1 - e^{i 2 π a}) \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = 2 i π {(- 1)}^{a - 1} \Rightarrow \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{2 i π e^{i π (a - 1)}}{1 - e^{i 2 π a}}

= \frac{2 i π e^{- i π a} e^{i π (a - 1)}}{e^{- i π a} (1 - e^{i 2 π a})} = \frac{2 i π (- 1)}{e^{i π a} - e^{i π a}} = \frac{2 i π}{e^{i π a} - e^{- i π a}} = \frac{2 i π}{2 i s i n (π a)} = \frac{π}{s i n (π a)} \Rightarrow

w e h a v e p r o v e d t h a t f o r 0 < a < 1 \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)} .

Answered by tanmay.chaudhury50@gmail.com last updated on 08/Oct/18

t = {t a n}^{2} θ d t = 2 t a n θ {s e c}^{2} θ d θ

\int_{0}^{\frac{π}{2}} \frac{{({t a n}^{2} θ)}^{a - 1}}{{s e c}^{2} θ} \times 2 t a n θ {s e c}^{2} θ d θ

\int_{0}^{\frac{π}{2}} \frac{{s i n}^{2 a - 2} θ}{{c o s}^{2 a - 2} θ} \times 2 \frac{s i n θ}{c o s θ} d θ

\int_{0}^{\frac{π}{2}} 2 {s i n}^{2 a - 1} \times {c o s}^{- 2 a + 1} θ

2 a - 1 = 2 p - 1 p = a

2 q - 1 = - 2 a + 1

2 q = - 2 a + 2 q = - a + 1

\int_{0}^{\frac{π}{2}} 2 {s i n}^{2 a - 1} {c o s}^{2 (1 - a) - 1} d θ

f o r m u l a \int_{0}^{\frac{π}{2}} 2 {s i n}^{2 p - 1} θ {c o s}^{2 q - 1} θ d θ

= \frac{⌈ (p) ⌈ (q)}{⌈ (p + q)}

\frac{⌈ a) ⌈ 1 - a)}{⌈ (a + 1 - a)} = ⌈ (a) ⌈ (1 - a) = \frac{π}{s i n (π a)} p r o v e d

f o r m u l a ⌈ (p) ⌈ (1 - p) = \frac{π}{s i n p π}

Commented by arvinddayama01@gmail.com last updated on 09/Oct/18

thanks sir

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Oct/18

m o s t w e l c o m e \dots