calculate-0-lnx-1-x-2-2-dx-

Question Number 82755 by mathmax by abdo last updated on 23/Feb/20

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

Commented by Ajao yinka last updated on 24/Feb/20

Commented by mathmax by abdo last updated on 24/Feb/20

l e t f (a) = \int_{0}^{\infty} \frac{l n x}{a + x^{2}} d x w i t h a > 0 w e c a n v e r i f y e a z l y t h a t f i s

d e r i v a b l e o n] 0, + \infty [a n d f^{^{'}} (a) = - \int_{0}^{\infty} \frac{l n x}{{(a + x^{2})}^{2}} d x \Rightarrow

\int_{0}^{\infty} \frac{l n x}{{(a + x^{2})}^{2}} d x = - f^{^{'}} (a) l e t e x p l i c i t f c h a n g e m e n t x = \sqrt{a} t g i v e

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

= \frac{1}{2} \frac{l n a}{\sqrt{a}} \times \frac{π}{2} + \frac{1}{\sqrt{a}} \int_{0}^{\infty} \frac{l n t}{1 + t^{2}} d t b u t \int_{0}^{\infty} \frac{l n (t)}{1 + t^{2}} d t = 0 ((p u t t = \frac{1}{u}) \Rightarrow

f (a) = \frac{π l n a}{4 \sqrt{a}} \Rightarrow f^{^{'}} (a) = \frac{π}{4} {\frac{\frac{\sqrt{a}}{a} - l n a \times \frac{1}{2 \sqrt{a}}}{a}}

= \frac{π}{4 a \sqrt{a}} - \frac{π}{4} \times \frac{l n a}{2 a \sqrt{a}} \Rightarrow \int_{0}^{\infty} \frac{l n x}{{(a + x^{2})}^{2}} d x = - \frac{π}{4 a \sqrt{a}} + \frac{π l n a}{8 a \sqrt{a}}

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

Answered by mind is power last updated on 24/Feb/20

\int_{0}^{+ \infty} \frac{l n (t)}{{(1 + t^{2})}^{n}} d t = A (n)

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

A = \frac{\partial}{\partial x} \int_{0}^{+ \infty} \frac{t^{x}}{{(1 + t^{2})}^{n}} d t ∣_{x = 0}

= \frac{\partial}{\partial x} \int_{0}^{\frac{π}{2}} \frac{{t g}^{x} (t)}{{(1 + {t g}^{2} (t))}^{n}} (1 + {t g}^{2} (t)) d t ∣_{x = 0}

= \frac{\partial}{\partial x} \int_{0}^{\frac{π}{2}} \frac{{t g}^{x} (t)}{{(1 + {t g}^{2} (t))}^{n}} d t ∣_{x = 0}

= \frac{1}{2} \frac{\partial}{\partial x} . 2 \int_{0}^{\frac{π}{2}} {s i n}^{x} (t) {c o s}^{- x + 2 n} (t) d t ∣_{x = 0}

= \frac{1}{2} \frac{\partial}{\partial x} β (\frac{x}{2} + \frac{1}{2}; - \frac{x}{2} + n + \frac{1}{2}) ∣_{x = 0}

= \frac{1}{2} \frac{\partial}{\partial x} \frac{Γ (\frac{x}{2} + \frac{1}{2}) Γ (n + \frac{1}{2} - \frac{x}{2})}{Γ (n + 1)} ∣_{x = 0} = \frac{1}{2} \frac{\partial}{\partial x} \frac{Γ (\frac{x}{2} + \frac{1}{2}) Γ (n + \frac{1}{2} - \frac{x}{2})}{n!} ∣_{x = 0}

Γ (n + \frac{1}{2} - \frac{x}{2}) = Γ (\frac{1}{2} - \frac{x}{2}) \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2} - \frac{x}{2})

= \frac{1}{2} . \frac{\partial}{\partial x} \frac{Γ (\frac{x}{2} + \frac{1}{2}) Γ (\frac{1}{2} - \frac{x}{2})}{n!} . \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2} - \frac{x}{2}) ∣_{x = 0}

= \frac{1}{2 (n!)} \frac{\partial}{\partial x} . \frac{π}{s i n (\frac{π}{2} (x + 1))} . \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2} - \frac{x}{2}) ∣_{x = 0}

f (x) = \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2} - \frac{x}{2})

l n (f (x)) = \underset{k = 0}{\sum^{n - 1}} l n (k + \frac{1}{2} - \frac{x}{2})

\Rightarrow \frac{f^{'} (0)}{f (0)} = - \underset{k = 0}{\sum^{n - 1}} \frac{1}{2 (k + \frac{1}{2})} = - \underset{k = 0}{\sum^{n - 1}} \frac{1}{2 k + 1}

= \underset{k = 1}{\sum^{n}} \frac{1}{2 k} - \underset{k = 1}{\sum^{2 n}} \frac{1}{k} = \frac{H_{n}}{2} - H_{2 n}

f (0) = \underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2}) = \frac{(2 n - 1)!!}{2^{n}} = \frac{(2 n)!}{2^{2 n} . n!}

\Rightarrow f^{'} (0) = \frac{(2 n)!}{2^{2 n} n!} . (\frac{H_{n}}{2} - H_{2 n})

A (n) = \frac{(2 n)!}{2^{2 n + 1} {(n!)}^{2}} (\frac{H_{n}}{2} - H_{2 n}) π = \int_{0}^{+ \infty} \frac{l n (x)}{{(1 + x^{2})}^{n}} d x

n = 2 \Rightarrow \int_{0}^{+ \infty} \frac{l n (x)}{{(1 + x^{2})}^{2}} d x = A (2)

A (2) = \frac{4!}{2^{5} {(2!)}^{2}} (- 1 - \frac{1}{3}) π = \frac{24}{32.4} . \frac{- 4}{3} π = \frac{- 8}{32} . π = - \frac{π}{4}