nice-calculus-please-prove-that-0-1-Arctan-x-1-x-dx-pi-8-log-2-

Question Number 129806 by mnjuly1970 last updated on 19/Jan/21

\dots n i c e c a l c u l u s \dots

p l e a s e p r o v e t h a t ::

Φ = \int_{0}^{1} \frac{Arctan (x)}{1 + x} d x = \frac{π}{8} l o g (2) \dots

Answered by Dwaipayan Shikari last updated on 19/Jan/21

{[{t a n}^{- 1} (x) l o g (x + 1)]}_{0}^{1} - \int_{0}^{1} \frac{l o g (x + 1)}{x^{2} + 1} d x

= \frac{π}{4} l o g (2) - \int_{0}^{\frac{π}{4}} l o g (1 + t a n θ) d θ

= \frac{π}{4} l o g (2) - \frac{π}{8} l o g (2)

= \frac{π}{8} l o g (2)

Commented by mnjuly1970 last updated on 19/Jan/21

n i c e v e r y n i c e . .

Answered by mnjuly1970 last updated on 19/Jan/21

Answered by mathmax by abdo last updated on 19/Jan/21

Φ = \int_{0}^{1} \frac{arctanx}{1 + x} dx by parts Φ = {[\ln (1 + x) arctanx]}_{0}^{1} - \int_{0}^{1} \ln (1 + x) \times \frac{dx}{1 + x^{2}}

= \frac{π}{4} \ln (2) - \int_{0}^{1} \frac{\ln (1 + x)}{1 + x^{2}} dx we hsve

\int_{0}^{1} \frac{\ln (1 + x)}{1 + x^{2}} dx =_{x = \tan θ} \int_{0}^{\frac{π}{4}} \frac{\ln (1 + \tan θ)}{1 + \tan^{2} θ} (1 + \tan^{2} θ) d θ

= \int_{0}^{\frac{π}{4}} \ln (1 + \tan θ) d θ =_{θ = \frac{π}{4} - u} \int_{0}^{\frac{π}{4}} \ln (1 + \frac{1 - tanu}{1 + tanu}) du

= \int_{0}^{\frac{π}{4}} \ln (\frac{2}{1 + tanu}) du = \frac{π}{4} \ln (2) - \int_{0}^{\frac{π}{4}} \ln (1 + tanu) du \Rightarrow

\int_{0}^{\frac{π}{4}} \ln (1 + \tan θ) d θ = \frac{π}{8} \ln (2) \Rightarrow

Φ = \frac{π}{4} \ln (2) - \frac{π}{8} \ln (2) = \frac{π}{8} \ln (2)