please-0-xlogx-1-x-2-2-dx-

Question Number 11420 by ainstain last updated on 25/Mar/17

p l e a s e

\int_{0}^{\infty} \frac{x l o g x}{{(1 + x^{2})}^{2}} d x =

Commented by FilupS last updated on 26/Mar/17

is \log (x) in base 10 or base e ?

Answered by sm3l2996 last updated on 25/Mar/17

{_{v^{'} = \frac{x}{{(1 + x^{2})}^{2}}}^{u = \log (x)} \Rightarrow {_{v = \frac{- 1}{2 (1 + x^{2})}}^{u^{^{'}} = \frac{1}{x}}

I = \int_{0}^{\infty} \frac{xlog (x)}{{(1 + x^{2})}^{2}} dx = {[\frac{- \log (x)}{2 (1 + x^{2})}]}_{0}^{\infty} + \int_{0}^{\infty} \frac{dx}{2 x (1 + x^{2})}

I = \lim_{x \to \infty} (- \frac{\log (x)}{2 (1 + x^{2})}) - \lim_{x \to 0} (- \frac{\log (x)}{2 (1 + x^{2})}) + \frac{1}{2} \int_{0}^{\infty} \frac{dx}{x (1 + x^{2})}

\frac{1}{x (1 + x^{2})} = \frac{a}{x} + \frac{bx + c}{1 + x^{2}}

a = 1;

{_{\frac{c - b}{2} - 1 = \frac{- 1}{2}}^{\frac{b + c}{2} + 1 = \frac{1}{2}} \Leftrightarrow {_{c - b = 1}^{b + c = - 1}

b = - 1; c = 0

so : \frac{1}{x (1 + x^{2})} = \frac{1}{x} - \frac{x}{1 + x^{2}}

\int_{0}^{\infty} \frac{dx}{x (1 + x^{2})} = \int_{0}^{\infty} \frac{dx}{x} - \int_{0}^{\infty} \frac{xdx}{1 + x^{2}}

= {[\log ∣ x ∣ - \frac{1}{2} \log ∣ 1 + x^{2} ∣]}_{0}^{\infty}

= \underset{x \to + \infty}{\lim l o g} (\frac{x}{\sqrt{1 + x^{2}}}) - \underset{x \to 0}{\lim l o g} (\frac{x}{\sqrt{1 + x^{2}}})

I = \underset{x \to + \infty}{\lim l o g} (\frac{x}{\sqrt{1 + x^{2}}}) - \lim_{x \to 0} (\log (\frac{x}{\sqrt{1 + x^{2}}}) + \frac{\log (x)}{2 (1 + x^{2})})

Commented by sm3l2996 last updated on 25/Mar/17

working

Answered by ajfour last updated on 26/Mar/17

zero

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