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Question Number 108920 by mathdave last updated on 20/Aug/20

	Answered by 1549442205PVT last updated on 20/Aug/20

	$I = \int_{0}^{\frac{π}{12}} \ln (tanx) dx . Put tanx = t$ $\Rightarrow dt = (1 + t^{2}) dx \Rightarrow I = \int_{0}^{2 - \sqrt{3}} \frac{\ln (t)}{1 + t^{2}} dt$ $= \int_{0}^{2 - \sqrt{3}} \ln (t) (\underset{k = 0}{\overset{\infty}{Σ}} {(- 1)}^{k} x^{2 k}) dt =$ $= \underset{k = 0}{\overset{\infty}{Σ}} {(- 1)}^{k} \int_{0}^{2 - \sqrt{3}} x^{2 k} \ln (t) dt = \underset{k = 0}{\overset{\infty}{Σ}} {(- 1)}^{k} A_{k}$ $Extra \left or missing \right$ $= \frac{\ln (2 - \sqrt{3}) {(2 - \sqrt{3})}^{2 k + 1}}{2 k + 1} - {[\frac{1}{2 k + 1} . \frac{x^{2 k + 1}}{2 k + 1}]}_{0}^{2 - \sqrt{3}}$ $= \frac{\ln (2 - \sqrt{3}) {(2 - \sqrt{3})}^{2 k + 1}}{2 k + 1} - \frac{1}{{(2 k + 1)}^{2}} \times {(2 - \sqrt{3})}^{2 k + 1}$ $I = \underset{k = 0}{\overset{\infty}{Σ}} {(- 1)}^{k} A_{k} = \underset{k = 0}{\overset{\infty}{Σ}} {(- 1)}^{k} [\frac{\ln (2 - \sqrt{3}) {(2 - \sqrt{3})}^{2 k + 1}}{2 k + 1} - \frac{1}{{(2 k + 1)}^{2}} \times {(2 - \sqrt{3})}^{2 k + 1}]$ $= \underset{k = 0}{\overset{\infty}{Σ}} {(- 1)}^{k} {[2 - \sqrt{3})}^{2 k + 1} (\frac{(2 k + 1) \ln (2 - \sqrt{3}) - 1}{{(2 k + 1)}^{2}})$ $Since G = \underset{k = 1}{\overset{\infty}{Σ}} {(- 1)}^{k} \frac{1}{{(2 k + 1)}^{2}}, we need$ $to prove that :$ ${[2 - \sqrt{3})}^{2 k + 1} (\frac{(2 k + 1) \ln (2 - \sqrt{3}) - 1}{{(2 k + 1)}^{2}}) = \frac{2}{3} \times \frac{1}{{(2 k + 1)}^{2}}$ $\Leftrightarrow [(2 k + 1) \ln (2 - \sqrt{3}) - 1] \times {(2 - \sqrt{3})}^{2 k + 1}$ $= \frac{- 2}{3} ? . . . . .$
	Commented by mathdave last updated on 20/Aug/20

	$u t r y i w i l l s e n d m y s o l u t i o n a s w e l l$
	Answered by mathdave last updated on 20/Aug/20

	$s o l u t i o n$ $t o s h o w t h a t \int_{0}^{\frac{π}{12}} \ln (\tan x) d x = - \frac{2}{3} G$ $r e c a l l t o l e m m a (2) t h e o r e m w h i c h s t a t e$ $\int_{0}^{\frac{π}{4}} \ln (\tan x) d x = 2 \int_{0}^{\frac{π}{12}} \ln (\tan 3 x) d y$ $b y l e t y = 3 x a n d d x = \frac{1}{3} d y$ $\frac{2}{3} \int_{0}^{\frac{π}{4}} \ln (\tan y) d y = \int_{0}^{\frac{π}{4}} \ln (\tan x) d x$ $l e t t = \tan y a n d d y = \frac{d t}{1 + t^{2}}$ $∵ \int_{0}^{\frac{π}{4}} \ln (\tan x) d x = \frac{2}{3} \int_{0}^{1} \frac{\ln t}{1 + t^{2}} d t$ $b u t t h e s e r i e s o f \frac{1}{1 + t^{2}} = {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} t^{2 n}$ $= \frac{2}{3} \int_{0}^{1} {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} t^{2 n} \ln t$ $u s i n g f e y n m a n n t r i c k$ $\frac{d}{d a} ∣_{a = 1} I = \frac{2}{3} ∙ {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} t^{2 n} . t^{a - 1} d t = \frac{2}{3} ∙ {(- 1)}^{n} \underset{n = 0}{\sum^{\infty}} \frac{1}{(2 n + a)}$ $\frac{d}{d a} ∣_{a = 1} I = - \frac{2}{3} ∙ \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + a)}^{2}} = - \frac{2}{3} ∙ \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}}$ $b u t \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}} = G (c a l l c a t a l a n c o n s t a n t)$ $∵ \int_{0}^{\frac{π}{12}} \ln (\tan x) d x = - \frac{2}{3} G Q . E . D$ $b y m a t h d a v e (20 / 08 / 2020)$
	Commented by 1549442205PVT last updated on 20/Aug/20

	$Thank you Sir . I think that just integrating$ $by parts obtain also above result$ $but {don}^{'} t need to use {feynmann}^{'} s trick,$ $it seems becomes trouble more$ $However, {Sir}^{'} s idea can using to solve$ $other problems$
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