calculate-0-log-2-x-8-x-4-2-dx-

Question Number 143546 by mathmax by abdo last updated on 15/Jun/21

calculate \int_{0}^{\infty} \frac{\log^{2} x}{{(8 + x^{4})}^{2}} dx

Answered by mathmax by abdo last updated on 16/Jun/21

f (a) = \int_{0}^{\infty} \frac{\log^{2} x}{a^{4} + x^{4}} dx \Rightarrow f^{^{'}} (a) = - \int_{0}^{\infty} \frac{{4 a}^{3} \log^{2} x}{{(a^{4} + x^{4})}^{2}} dx \Rightarrow

f^{^{'}} (2^{\frac{3}{4}}) = - 4 {(2)}^{\frac{3}{4}} \int_{0}^{\infty} \frac{\log^{2} x}{{(8 + x^{4})}^{2}} dx = - 2^{2 + \frac{3}{4}} \int_{0}^{\infty} \frac{\log^{2} x}{{(8 + x^{4})}^{2}} dx

\Rightarrow \int_{0}^{\infty} \frac{\log^{2} x}{{(8 + x^{4})}^{2}} dx = - \frac{1}{2^{\frac{11}{4}}} f^{^{'}} (2^{\frac{3}{4}})

f (a) =_{x = at} \int_{0}^{\infty} \frac{\log^{2} (at)}{a^{4} (1 + t^{4})} adt = \frac{1}{a^{3}} \int_{0}^{\infty} \frac{\log^{2} a + 2 logalogt + \log^{2} t}{1 + t^{4}} dt

= \frac{\log^{2} a}{a^{3}} \int_{0}^{\infty} \frac{dt}{1 + t^{4}} + \frac{2 loga}{a^{3}} \int_{0}^{\infty} \frac{logt}{1 + t^{4}} dt + \frac{1}{a^{3}} \int_{0}^{\infty} \frac{\log^{2} (t)}{1 + t^{4}} dt

\int_{0}^{\infty} \frac{dt}{1 + t^{4}} =_{t^{4} = y} \frac{1}{4} \int_{0}^{\infty} \frac{y^{\frac{1}{4} - 1}}{1 + y} dy = \frac{1}{4} \times \frac{π}{\sin (\frac{π}{4})} = \frac{π}{4 \times \frac{\sqrt{2}}{2}}

= \frac{π}{2 \sqrt{2}}

\int_{0}^{\infty} \frac{logt}{1 + t^{4}} dt =_{t^{4} = y} \frac{1}{4} \int_{0}^{\infty} \frac{logy}{1 + y} \times \frac{1}{4} y^{\frac{1}{4} - 1} dy

= \frac{1}{16} \int_{0}^{\infty} \frac{y^{\frac{1}{4} - 1} logy}{1 + y} dy = \frac{1}{16} w^{^{'}} (\frac{1}{4}) with

w (λ) = \int_{0}^{\infty} \frac{y^{λ - 1}}{1 + y} dy \Rightarrow w^{^{'}} (λ) = \int_{0}^{\infty} \frac{y^{λ - 1} logy}{1 + y} dy

w (λ) = \frac{π}{\sin (π λ)} \Rightarrow w^{^{'}} (λ) = - \frac{π^{2} \cos (π λ)}{\sin^{2} (π λ)} \Rightarrow

w^{^{'}} (\frac{1}{4}) = - π^{2} \times \frac{\frac{1}{\sqrt{2}}}{{(\frac{1}{\sqrt{2}})}^{2}} = - 2 π^{2} . \frac{1}{\sqrt{2}} = - π^{2} \sqrt{2} \Rightarrow

\int_{0}^{\infty} \frac{logt}{1 + t^{4}} dt = - \frac{\sqrt{2}}{16} π^{2}

\int_{0}^{\infty} \frac{\log^{2} t}{1 + t^{4}} dt =_{t^{4} = y} \frac{1}{16} \int_{0}^{\infty} \frac{\log^{2} y}{1 + y} \times \frac{1}{4} y^{\frac{1}{4} - 1} dt

= \frac{1}{64} \int_{0}^{\infty} \frac{y^{\frac{1}{4} - 1} \log^{2} y}{1 + y} dy = \frac{1}{64} w^{(2)} (\frac{1}{4})

w^{(2)} (λ) = - π^{2} \times \frac{- π \sin (π λ) \sin^{2} (π λ) - 2 π \sin (π λ) \cos (π λ)}{\sin^{4} (π λ)}

= - π^{3} \times \frac{\sin^{2} (π λ) - 2 \cos (π λ)}{\sin^{3} (π λ)} \Rightarrow

w^{(2)} (\frac{1}{4}) = - π^{3} \times \frac{\frac{1}{2} - 2 \times \frac{1}{\sqrt{2}}}{{(\frac{1}{\sqrt{2}})}^{3}} = - π^{3} \times \frac{\frac{1}{2} - \sqrt{2}}{\frac{1}{2 \sqrt{2}}}

= - 2 \sqrt{2} π^{3} (\frac{1 - 2 \sqrt{2}}{2}) = \sqrt{2} π^{3} (2 \sqrt{2} - 1) \Rightarrow

f (a) = \frac{\log^{2} a}{a^{3}} \times \frac{π}{2 \sqrt{2}} + \frac{2 loga}{a^{3}} (- \frac{\sqrt{2}}{16}) π^{2} + \frac{1}{a^{3}} \times \frac{1}{64} \sqrt{2} π^{3} (2 \sqrt{2} - 1)

Φ = - \frac{1}{2^{\frac{11}{4}}} f^{^{'}} (2^{\frac{3}{4}}) \dots .