let-f-x-0-t-a-1-x-t-dt-with-x-gt-0-and-0-lt-a-lt-1-1-calculate-f-x-2-calculate-g-x-0-t-a-1-x-t-2-dt-3-find-the-value-of-0-t-a-1-1-t-2-dt-

Question Number 63510 by turbo msup by abdo last updated on 05/Jul/19

l e t f (x) = \int_{0}^{\infty} \frac{t^{a - 1}}{x + t} d t w i t h x > 0

a n d 0 < a < 1

1) c a l c u l a t e f (x)

2) c a l c u l a t e g (x) = \int_{0}^{\infty} \frac{t^{a - 1}}{{(x + t)}^{2}} d t

3) 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

Commented by mathmax by abdo last updated on 06/Jul/19

1) c h a n g e m e n t t = x u g i v e f (x) = \int_{0}^{\infty} \frac{{(x u)}^{a - 1}}{x + x u} x d u = x^{a - 1} \int_{0}^{\infty} \frac{u^{a - 1}}{1 + u} d u

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

2) w e h a v e f^{^{'}} (x) = \int_{0}^{\infty} \frac{\partial}{\partial x} (\frac{t^{a - 1}}{x + t}) d t = - \int_{0}^{\infty} \frac{t^{a - 1}}{{(x + t)}^{2}} d t = - g (x) \Rightarrow

g (x) = - f^{^{'}} (x) b u t f (x) = \frac{π}{s i n (π a)} e^{(a - 1) l n (x)} \Rightarrow

f^{^{'}} (x) = \frac{π (a - 1)}{x s i n (π a)} x^{a - 1} = \frac{π (a - 1) x^{a - 2}}{s i n (π a)} \Rightarrow g (x) = \frac{π (1 - a)}{s i n (π a)} x^{a - 2}

3) \int_{0}^{\infty} \frac{t^{a - 1}}{{(1 + t)}^{2}} d t = g (1) = \frac{π (1 - a)}{s i n (π a)} (0 < a < 1) .

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

f^{(n)} (x) = \int_{0}^{\infty} \frac{{(- 1)}^{n} n! t^{a - 1}}{{(x + t)}^{n + 1}} d t \Rightarrow \int_{0}^{\infty} \frac{t^{a - 1}}{{(x + t)}^{n + 1}} d t = \frac{{(- 1)}^{n}}{n!} f^{(n)} (x)

w e h a v e f (x) = \frac{π}{s i n (π a)} e^{(a - 1) l n (x)} l e t f i n d {(e^{λ l n (x)})}^{(n)}

{(e^{λ l n x})}^{(1)} = \frac{λ}{x} e^{λ l n (x)} \Rightarrow {(e^{λ l n x})}^{(2)} = \frac{λ^{2}}{x^{2}} e^{λ l n (x)} \Rightarrow {(e^{λ l n (x)})}^{(n)} = \frac{λ^{n}}{x^{n}} e^{λ l n x} \Rightarrow

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

\int_{0}^{\infty} \frac{t^{a - 1}}{{(x + t)}^{n + 1}} d t = \frac{π {(1 - a)}^{n}}{n! s i n (π a)} x^{a - n - 1}

s p e c i a l c a s e x = 1 \Rightarrow \int_{0}^{\infty} \frac{t^{a - 1}}{{(1 + t)}^{n + 1}} d t = \frac{π {(1 - a)}^{n}}{n! s i n (π a)} .