find-the-value-of-0-t-a-1-1-t-2-dt-with-0-lt-a-lt-1-

Question Number 62330 by maxmathsup by imad last updated on 19/Jun/19

f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{t^{a - 1}}{{(1 + t)}^{2}} d t w i t h 0 < a < 1

Commented by mathmax by abdo last updated on 04/Jul/19

w e h a v e p r o v e d t h a t B (x, y) = \int_{0}^{\infty} \frac{t^{x - 1}}{{(1 + t)}^{x + y}} d t l e t t a k e x = a a n d x + y = 2 \Rightarrow

x = a a n d y = 2 - a \Rightarrow

\int_{0}^{\infty} \frac{t^{a - 1}}{{(1 + t)}^{2}} d t = B (a, 2 - a) = \frac{Γ (a) Γ (2 - a)}{Γ (a + 2 - a)} = Γ (a) Γ (2 - a) (Γ (2) = 1! = 1) \Rightarrow

\int_{0}^{\infty} \frac{t^{a - 1}}{{(1 + t)}^{2}} d t = Γ (a) Γ (2 - a)

Answered by tanmay last updated on 19/Jun/19

t = {t a n}^{2} θ \to 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} \times 2 t a n θ {s e c}^{2} θ d θ}{{s e c}^{4} θ}

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

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

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

= \frac{⌈ (p) ⌈ (q)}{⌈ (p + q)} h e r e 2 p - 1 = 2 a - 1 p = a

2 q - 1 = 3 - 2 a

2 q = 4 - 2 a \to q = 2 - a \leftarrow l o o k h e r e

a n s w e r = \frac{⌈ (p) ⌈ (q)}{⌈ (p + q)} = \frac{⌈ (a) \times ⌈ (2 - a)}{⌈ (a + 2 - a)}

= \frac{⌈ (a) ⌈ (2 - a)}{1}

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