If-gt-0-and-gt-0-prove-0-ln-x-2-x-2-dx-pi-2-ln-

Question Number 136425 by Ar Brandon last updated on 21/Mar/21

If α > 0 and β > 0, prove

\int_{0}^{\infty} \frac{\ln (α x)}{β^{2} + x^{2}} dx = \frac{π}{2 β} \ln (α β)

Commented by Dwaipayan Shikari last updated on 21/Mar/21

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

= \frac{π l o g (α)}{2 β} + τ^{'} (0)

τ (η) = \int_{0}^{\infty} \frac{x^{η}}{β^{2} + x^{2}} d x = β^{η - 1} \int_{0}^{\infty} \frac{u^{η}}{u^{2} + 1} d u x = β u

= \frac{1}{2} β^{η - 1} \int_{0}^{\infty} \frac{t^{\frac{η + 1}{2} - 1}}{{(t + 1)}^{\frac{η}{2} + \frac{1}{2} - \frac{η}{2} + \frac{1}{2}}} d u = \frac{β^{η - 1}}{2} . \frac{Γ (\frac{η + 1}{2}) Γ (\frac{1}{2} - \frac{η}{2})}{Γ (2)}

= \frac{β^{η - 1}}{2} . \frac{π}{c o s (\frac{π η}{2})}

τ^{'} (η) = \frac{β^{η - 1} l o g (β)}{2} . \frac{π}{c o s (\frac{π η}{2})} + \frac{β^{n - 1} π^{2}}{4} s e c (\frac{π η}{2}) t a n (\frac{π η}{2})

τ^{'} (0) = \frac{l o g (β) π}{2 β}

Φ = \frac{π l o g (α)}{2 β} + \frac{π l o g (β)}{2 β} = \frac{π l o g (α β)}{2 β}

Commented by Ar Brandon last updated on 21/Mar/21

Thanks

Commented by Dwaipayan Shikari last updated on 21/Mar/21

Answered by mathmax by abdo last updated on 21/Mar/21

Φ = \int_{0}^{\infty} \frac{\log (α x)}{x^{2} + β^{2}} dx \Rightarrow Φ = \int_{0}^{\infty} \frac{\log α + logx}{x^{2} + β^{2}} dx

= \log α \int_{0}^{\infty} \frac{dx}{x^{2} + β^{2}} (\to x = β u) + \int_{0}^{\infty} \frac{logx}{x^{2} + β^{2}} dx (\to x = β u)

= \log α \int_{0}^{\infty} \frac{β du}{β^{2} (1 + u^{2})} + \int_{0}^{\infty} \frac{\log (β u)}{β^{2} (1 + u^{2})} β du

= \frac{\log α}{β} . \frac{π}{2} + \frac{1}{β} \int_{0}^{\infty} \frac{\log β + logu}{1 + u^{2}} du

= \frac{π \log α}{2 β} + \frac{\log β}{β} . \frac{π}{2} + \frac{1}{β} \int_{0}^{\infty} \frac{logu}{1 + u^{2}} du (\to 0)

= \frac{π (\log α + \log β)}{2 β} = \frac{π}{2 β} \log (α β)

Commented by Ar Brandon last updated on 21/Mar/21

Thanks