calculate-0-lnx-x-1-6-dx-

Question Number 130135 by mathmax by abdo last updated on 22/Jan/21

calculate \int_{0}^{\infty} \frac{lnx}{{(x + 1)}^{6}} dx

Answered by Dwaipayan Shikari last updated on 22/Jan/21

I (a) = \int_{0}^{\infty} \frac{x^{a + 1 - 1}}{{(1 + x)}^{6 + a + 1 - a - 1}} d x \Rightarrow I (a) = \frac{Γ (a + 1) Γ (5 - a)}{Γ (6)}

I^{'} (a) = \frac{1}{120} (Γ^{'} (a + 1) Γ (5 - a) - Γ^{'} (5 - a) Γ (a + 1))

I^{'} (0) = - \frac{γ}{5} - \frac{1}{120} (- 24 γ + 24 (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}))

= - \frac{5}{12}

M e a n s i n G e n e r a l \int_{0}^{\infty} \frac{l o g x}{{(1 + x)}^{n}} d x = - \frac{1}{n} \underset{n = 1}{\sum^{n - 2}} \frac{1}{n}