0-1-log-tan-d-

Question Number 105940 by Dwaipayan Shikari last updated on 01/Aug/20

\int_{0}^{1} l o g (t a n θ) d θ

Answered by mathmax by abdo last updated on 01/Aug/20

A = \int_{0}^{1} \ln (\tan θ) d θ by parts A = {[θ \ln (\tan θ)]}_{0}^{1} - \int_{0}^{1} θ \times \frac{1 + \tan^{2} θ}{\tan θ} d θ

= \ln (\frac{π}{4}) - \int_{0}^{1} θ {\frac{1}{\tan θ} + \tan θ} d θ changement \tan θ = x give

\int_{0}^{1} θ {\frac{1}{\tan θ} + \tan θ} d θ = \int_{0}^{\tan (1)} arctanx {\frac{1}{x} + x} \frac{dx}{1 + x^{2}}

= \int_{0}^{\tan (1)} \frac{1 + x^{2}}{x} \times \frac{arctanx}{1 + x^{2}} dx = \int_{0}^{\tan (1)} \frac{\arctan (x)}{x} dx

let f (a) = \int_{0}^{\tan (1)} \frac{\arctan (α x)}{x} dx \Rightarrow f^{^{'}} (a) = \int_{0}^{\tan (1)} \frac{x}{(1 + α^{2} x^{2}) x} dx

= \int_{0}^{\tan (1)} \frac{dx}{1 + α^{2} x^{2}} =_{α x = z} \int_{0}^{α \tan (1)} \frac{dz}{α (1 + z^{2})} = \frac{1}{α} {[arctanz]}_{0}^{α \tan (1)}

= \frac{\arctan (α \tan (1))}{α} \Rightarrow f (α) = \int_{0}^{α} \frac{\arctan (utan (1))}{u} du + C

and \int_{0}^{\tan (1)} \frac{arctanx}{x} dx = f (1) = \int_{0}^{1} \frac{\arctan (utan (1))}{u} du + C

. . be continued \dots