calculate-0-ln-x-1-x-3-dx-

Question Number 79096 by mathmax by abdo last updated on 22/Jan/20

c a l c u l a t e \int_{0}^{\infty} \frac{l n (x)}{{(1 + x)}^{3}} d x

Commented by mind is power last updated on 22/Jan/20

s a m i d e e w i l l w o r c k

f o r

\int_{0}^{+ \infty} \frac{l n {(x)}^{p}}{{(x + 1)}^{r}} d x

w i t h e r > 1, p \in N

l n {(x)}^{p} = \frac{\partial^{p}}{\partial t^{p}} x^{t} ∣_{t = 0} a n d s a m i d e a l i k e p r e v i o u s o n e

b u t w e w i l l h a v e m o r c o m p l i c a t e d f u n c t i o n Ψ_{p} (x) \dots .

Answered by mind is power last updated on 22/Jan/20

l n (x) = \frac{d}{d x} x^{t} ∣_{t = 0}

f (t) = \int_{0}^{+ \infty} \frac{x^{t}}{{(1 + x)}^{3}} d x, w e w a n t f^{'} (0)

x = {t g}^{2} (y) \Rightarrow

f (t) = \int_{0}^{\frac{π}{2}} \frac{{t g}^{2 t} (y)}{{(1 + {t g}^{2} (y))}^{3}} . 2 (1 + {t g}^{2} (y)) t g (y) d y

= 2 \int_{0}^{\frac{π}{2}} \frac{{t g}^{2 t + 1} (y)}{{(1 + {t g}^{2} (y))}^{2}} d y

= 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 t + 1} (y) {c o s}^{- 2 t + 3} (y) d y

= β (t + 1, - t + 2) = \frac{Γ (t + 1) Γ (2 - t)}{Γ (3)} = \frac{Γ (t + 1) Γ (2 - t)}{2}

f (t) = \frac{Γ (t + 1) Γ (2 - t)}{2}

f^{'} (t) = \frac{1}{2} {Γ^{'} (t + 1) Γ (2 - t) - Γ^{'} (2 - t) Γ (t + 1)}

= \frac{1}{2} {Ψ (t + 1) Γ (t + 1) Γ (2 - t) - Ψ (2 - t) Γ (2 - t) Γ (t + 1)}

f^{'} (0) = \frac{1}{2} {Ψ (1) - Ψ (2)} = \frac{1}{2} {- γ - (1 - γ)} = - \frac{1}{2} = - 0.5

Commented by msup trace by abdo last updated on 23/Jan/20

t h a n k y o u s i r n o w t r y t o s o l v e i t

c o m p l e x a n a l y s i s \dots