pi-4-pi-2-ln-ln-tan-x-dx-

Question Number 103190 by bemath last updated on 13/Jul/20

\underset{π / 4}{\int^{π / 2}} \ln (\ln (\tan x)) d x

Answered by abdomathmax last updated on 13/Jul/20

\ln (tanx) = t \Rightarrow I = \int_{\frac{π}{4}}^{\frac{π}{2}} \ln (\ln (tanx)) dx

tanx = e^{t} \Rightarrow x = \arctan (e^{t}) \Rightarrow

I = \int_{0}^{+ \infty} \ln (t) \times \frac{e^{t}}{1 + e^{2 t}} dt = \int_{0}^{\infty} \frac{e^{t} \ln (t)}{1 + e^{2 t}} dt

= \int_{0}^{\infty} \frac{e^{- t} \ln (t)}{1 + e^{- 2 x}} dt = \int_{0}^{\infty} e^{- t} \ln (t) (\sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- (nt}) dt

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} e^{- (n + 1) t} \ln (t) dt

A_{n} = \int_{0}^{\infty} e^{- (n + 1) t} \ln (t) dt =_{(n + 1) t = u} \int_{0}^{\infty} e^{- u} \ln (\frac{u}{n + 1}) \frac{d u}{n + 1}

= \frac{1}{n + 1} \int_{0}^{\infty} {e^{- u} l n (u) - e^{- u} l n (n + 1)) d u

= - \frac{γ}{n + 1} - \frac{\ln (n + 1)}{n + 1} \int_{0}^{\infty} e^{- u} du

= - \frac{γ}{n + 1} - \frac{l n (n + 1)}{n + 1} \Rightarrow

I = - γ \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} - \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} l n (n + 1)

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = \ln (2)

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} \ln (n + 1) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \ln (n)

\Rightarrow I = - γ \ln (2) + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} lnn}{n}

\dots be continued \dots

Commented by abdomathmax last updated on 13/Jul/20

i dont remember the value of \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n} \ln (n)}{n}

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