calculate-0-log-2-x-1-x-2-dx-

Question Number 142430 by Mathspace last updated on 31/May/21

c a l c u l a t e \int_{0}^{\infty} \frac{{l o g}^{2} x}{1 + x^{2}} d x

Answered by Dwaipayan Shikari last updated on 31/May/21

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

\int_{0}^{\infty} \frac{x^{a} {l o g}^{2} (x)}{1 + x^{2}} d x = \frac{\partial^{2}}{\partial a^{2}} (\frac{π}{2} c o s e c (\frac{π}{2} (1 - a)))

\int_{0}^{\infty} \frac{{l o g}^{2} (x)}{1 + x^{2}} d x = \frac{\partial^{2}}{\partial a^{2}} ∣_{a = 0} \frac{π}{2} c o s e c (\frac{π}{2} (1 - a))

Answered by mathmax by abdo last updated on 01/Jun/21

Φ = \int_{0}^{\infty} \frac{\log^{2} x}{1 + x^{2}} dx changement x = t^{\frac{1}{2}} [give

Φ = \frac{1}{4} \int_{0}^{\infty} \frac{\log^{2} t}{1 + t} \frac{1}{2} t^{- \frac{1}{2}} dt = \frac{1}{8} \int_{0}^{\infty} \frac{t^{- \frac{1}{2}} \log^{2} t}{1 + t} dt

let f (a) = \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} dt \Rightarrow f^{^{'}} (a) = \int_{0}^{\infty} \frac{\partial}{\partial a} (\frac{e^{(a - 1) logt}}{1 + t}) dt

= \int_{0}^{\infty} \frac{t^{a - 1} logt}{1 + t} \Rightarrow f^{(2)} (a) = \int_{0}^{\infty} \frac{t^{a - 1} \log^{2} t}{1 + t} dt \Rightarrow

f^{(2)} (\frac{1}{2}) = \int_{0}^{\infty} \frac{t^{- \frac{1}{2}} \log^{2} t}{1 + t} dt = 8 Φ

we have f (a) = \frac{π}{\sin (π a)} \Rightarrow f^{^{'}} (a) = - \frac{π^{2} \cos (π a)}{\sin^{2} (π a)} (0 < a < 1)

and f^{(2)} (a) = - π^{2} . \frac{- π \sin (π a) \sin^{2} (π a) - \cos (π a) . 2 \sin (π a) π \cos (π a)}{\sin^{4} (π a)}

8 Φ = f^{(2)} (\frac{1}{2}) = - π^{2} \times \frac{- π}{1} = π^{3} \Rightarrow Φ = \frac{π^{3}}{8}