0-pi-4-log-1-tan-x-dx-

Question Number 151421 by peter frank last updated on 21/Aug/21

\int_{0}^{\frac{π}{4}} \log (1 + \tan x) dx

Commented by puissant last updated on 21/Aug/21

x = \frac{π}{4} - t \to d x = - d t

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

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

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

\Rightarrow Q = \int_{0}^{\frac{π}{4}} l n (\frac{2}{1 + t a n t}) d t

\Rightarrow Q = \int_{0}^{\frac{π}{4}} l n 2 d x - \int_{0}^{\frac{π}{4}} l n (1 + t a n t) d t

\Rightarrow 2 Q = \frac{π}{4} l n 2

∵∴ Q = \frac{π}{8} l n 2 \dots

Commented by peter frank last updated on 21/Aug/21

thank you

Answered by peter frank last updated on 21/Aug/21

x = \frac{π}{4} - θ

dx = - d θ

(0, \frac{π}{4}) \Leftrightarrow (\frac{π}{4}, 0)

\int_{0}^{\frac{π}{4}} \log [1 + \tan (\frac{π}{4} - θ)] d θ

\int_{0}^{\frac{π}{4}} \log [1 + \frac{\tan \frac{π}{4} - \tan θ}{1 + \tan \frac{π}{4} \tan θ}]

\int_{0}^{\frac{π}{4}} \log [\frac{2}{1 + \tan θ}] d θ

\int_{0}^{\frac{π}{4}} \log 2 d θ - \int_{0}^{\frac{π}{4}} \log (1 + \tan θ)

2 I = \log 2 {[θ]}_{0}^{\frac{π}{4}}

I = \frac{π}{8} \log 2

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

f (a) = \int_{0}^{\frac{π}{4}} \log (1 + atanx) dx (a > 0) \Rightarrow

f^{^{'}} (a) = \int_{0}^{\frac{π}{4}} \frac{tanx}{1 + atanx} = \frac{1}{a} \int_{0}^{\frac{π}{4}} \frac{1 + atanx - 1}{1 + atanx} dx

= \frac{π}{4 a} - \frac{1}{a} \int_{0}^{\frac{π}{4}} \frac{dx}{1 + atanx} [we have

\int_{0}^{\frac{π}{4}} \frac{dx}{1 + atanx} =_{tanx = t} \int_{0}^{1} \frac{dt}{(1 + t^{2}) (1 + at)}

F (t) = \frac{1}{(at + 1) (t^{2} + 1)} = \frac{α}{at + 1} + \frac{mt + n}{t^{2} + 1}

α = \frac{1}{\frac{1}{a^{2}} + 1} = \frac{a^{2}}{1 + a^{2}}

\lim_{t \to + \infty} tF (t) = \frac{α}{a} + m \Rightarrow m = - \frac{α}{a} = - \frac{a}{1 + a^{2}}

\Rightarrow F (t) = \frac{a^{2}}{(1 + a^{2}) (at + 1)} + \frac{- \frac{a}{1 + a^{2}} t + n}{t^{2} + 1}

F (0) = \frac{a^{2}}{1 + a^{2}} + n = 1 \Rightarrow n = 1 - \frac{a^{2}}{1 + a^{2}} = \frac{1}{1 + a^{2}} \Rightarrow

F (t) = \frac{a^{2}}{(a^{2} + 1) (at + 1)} + \frac{- \frac{a}{a^{2} + 1} t + \frac{1}{1 + a^{2}}}{t^{2} + 1}

\Rightarrow \int_{0}^{1} F (t) dt = \frac{a^{2}}{(a^{2} + 1)} \int_{0}^{1} \frac{dt}{at + 1} - \frac{1}{a^{2} + 1} \int_{0}^{1} \frac{at - 1}{t^{2} + 1} dt

= \frac{a}{a^{2} + 1} {[\ln (at + 1)]}_{0}^{1} - \frac{a}{2 (a^{2} + 1)} {[\ln (t^{2} + 1)]}_{0}^{1} + \frac{1}{a^{2} + 1} \frac{π}{4}

= \frac{aln (1 + a)}{a^{2} + 1} - \frac{\ln 2}{2} \times \frac{a}{a^{2} + 1} + \frac{π}{4 (a^{2} + 1)} \Rightarrow

f^{^{'}} (a) = \frac{π}{4 a} - \frac{\ln (1 + a)}{a^{2} + 1} + \frac{\ln 2}{2 (a^{2} + 1)} - \frac{π}{4 a (a^{2} + 1)}

f (1) = f (1) - f (0) = \int_{0}^{1} f^{^{'}} (a) da = \int_{0}^{\frac{π}{4}} \log (1 + tant) dt

= \int_{0}^{1} \frac{π}{4 a} (1 - \frac{1}{a^{2} + 1}) da - \int_{0}^{1} \frac{\ln (1 + a)}{a^{2} + 1} da + \frac{\ln 2}{2} \int_{0}^{1} \frac{da}{1 + a^{2}}

= \frac{π}{4} \int_{0}^{1} \frac{a}{a^{2} + 1} da - \int_{0}^{1} \frac{\ln (1 + a)}{1 + a^{2}} da + \frac{π}{4} \times \frac{\ln 2}{2}

= \frac{π}{8} {[\ln (a^{2} + 1)]}_{0}^{1} + \frac{π \ln 2}{8} - \int_{0}^{1} \frac{\ln (1 + a)}{1 + a^{2}} da (a = \tan θ)

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

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

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

Commented by peter frank last updated on 21/Aug/21

thank you