find-0-lnx-x-2-a-2-dx-2-find-the-value-of-0-lnx-x-2-a-2-3-

Question Number 31102 by abdo imad last updated on 02/Mar/18

f i n d \int_{0}^{+ \infty} \frac{l n x}{x^{2} + a^{2}} d x

2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{3}} .

Commented by abdo imad last updated on 05/Mar/18

l e t p u t f (a) = \int_{0}^{\infty} \frac{l n x}{x^{2} + a^{2}} d x c h . x = a t w i t h a > 0 g i v e

f (a) = \int_{0}^{\infty} \frac{l n a + l n t}{a^{2} (1 + t^{2})} d t = \frac{l n a}{a^{2}} \int_{0}^{\infty} \frac{d t}{1 + t^{2}} + \frac{1}{a^{2}} \int_{0}^{\infty} \frac{l n t}{1 + t^{2}} d t b u t

w e h a v e p r o v e d t h a t \int_{0}^{\infty} \frac{l n t}{1 + t^{2}} d t = 0 \Rightarrow

f (a) = \frac{π l n a}{2 a^{2}}

2) w e h a v e f^{^{'}} (a) = - 2 a \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{2}} d x \Rightarrow

\int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{2}} = \frac{- 1}{2 a} f^{^{'}} (a) = \frac{- π l n a}{4 a^{3}} .

Commented by abdo imad last updated on 05/Mar/18

\frac{d}{d a} (\frac{- f^{^{'}} (a)}{2 a}) = \int_{0}^{\infty} \frac{- 2 2 a (x^{2} + a^{2})}{{(x^{2} + a^{2})}^{4}} l n x d x = - 4 a \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{3}} d x

\Rightarrow \int_{0}^{\infty} \frac{l n x}{{(x^{2} + a^{2})}^{3}} d x = \frac{- 1}{4 a} \frac{\partial}{\partial a} (\frac{- π l n a}{4 a^{3}})

= \frac{π}{16 a} {(\frac{l n a}{a^{3}})}^{^{'}} = \frac{π}{16 a} (\frac{a^{2} - 3 a^{2} l n a}{a^{6}}) = \frac{π a^{2} (1 - 3 l n a)}{16 a^{7}}

= \frac{π (1 - 3 l n a)}{16 a^{4}} .