∫_( 0) ^( (π/2)) log^2 (tan(x))dx


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Question Number 123159 by Lordose last updated on 23/Nov/20

$\int_{0}^{\frac{π}{2}} \log^{2} (\tan (x)) dx$
	Answered by mnjuly1970 last updated on 23/Nov/20

	$a n s := \frac{π^{3}}{8}$
	Commented by Lordose last updated on 23/Nov/20

	${It}^{'} s correct sir$
	Commented by mnjuly1970 last updated on 23/Nov/20

	$t h a n k y o u$ $s o l u t i o n$ $I = \int_{0}^{\frac{π}{2}} {l o g}^{2} (t a n (x)) d x \overset{t a n (x) = y}{=}$ $= \int_{0}^{\infty} {l o g}^{2} (y) \frac{d y}{1 + y^{2}} \overset{y^{2} = t}{=} \frac{1}{8} \int_{0}^{\infty} \frac{t^{\frac{- 1}{2}} {l o g}^{2} (t)}{1 + t} d t$ $I (a) = \int_{0}^{\infty} \frac{t^{\frac{- 1}{2} + a}}{1 + t} d t$ $g o a l :: I = \frac{1}{8} I^{″} (0)$ $b u t : I (a) = β (\frac{1}{2} + a, \frac{1}{2} - a)$ $= Γ (\frac{1}{2} + a) Γ (\frac{1}{2} - a)$ $\underset{r e f l e c t i o n f o r m u l a}{\overset{e u l e r}{=}} \frac{π}{s i n (\frac{π}{2} + π a)} = \frac{π}{c o s (π a)}$ $I^{'} (a) = \frac{π^{2} s i n (π a)}{{c o s}^{2} (π a)}$ $I^{″} (a) = π^{2} (\frac{π {c o s}^{3} (π a) + 2 π {s i n}^{2} (π a) c o s (π a)}{{c o s}^{4} (π a)}) ∣_{a = 0}$ $I^{″} (0) = π^{3} \Rightarrow I = \frac{π^{3}}{8} ✓ ✓$
	Commented by Dwaipayan Shikari last updated on 23/Nov/20

	$G r e a t s i r! i {h a v e n}^{'} t t h o u g h t o f t h i s$
	Commented by mnjuly1970 last updated on 23/Nov/20

	$t h a n k y o u s o m u c h s i r p a y a n$ $t o m e y o u r s o l u t i o n i s v e r y v e r y$ $n i c e a n d u n i q u e . . .$
	Answered by Dwaipayan Shikari last updated on 23/Nov/20

	$\int_{0}^{\frac{π}{2}} {l o g}^{2} (t a n x) d x$ $= \int_{0}^{\infty} \frac{{l o g}^{2} t}{t^{2} + 1} d t$ $= \underset{n ⩾ 0}{\sum^{\infty}} \int_{0}^{\infty} {l o g}^{2} t {(- 1)}^{n} t^{2 n} d t l o g t = u \Rightarrow \frac{1}{t} = \frac{d u}{d t}$ $= \underset{n ⩾ 0}{\sum^{\infty}} {(- 1)}^{n} \int_{- \infty}^{\infty} u^{2} e^{2 n u + u} d t u (2 n + 1) = Λ$ $= 2 \underset{n ⩾ 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} \int_{0}^{\infty} Λ^{2} e^{- Λ} d Λ$ $= 2 \underset{n ⩾ 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} Γ (3)$ $= 4 \underset{n ⩾ 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} = 4 . \frac{π^{3}}{32} = \frac{π^{3}}{8}$
	Commented by mnjuly1970 last updated on 23/Nov/20

	$v e r y g o o d . . .$
	Commented by Lordose last updated on 23/Nov/20

	$Can you explain from 4 th line$
	Commented by Dwaipayan Shikari last updated on 24/Nov/20

	$\underset{n ⩾ 0}{\sum^{\infty}} {(- 1)}^{n} \int_{- \infty}^{\infty} \frac{Λ^{2}}{{(2 n + 1)}^{2} (2 n + 1)} e^{Λ} d Λ u (2 n + 1) = Λ$ $= \underset{n ⩾ 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} \int_{\infty}^{- \infty} - Φ^{2} e^{- Φ} d Φ Λ = - Φ$ $= 2 \underset{n ⩾ 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} \int_{0}^{\infty} Φ^{2} e^{- Φ} d Φ = 2 . Γ (3) \frac{π^{3}}{32} = \frac{π^{3}}{8}$
	Answered by mathmax by abdo last updated on 23/Nov/20
	$A =∫_0 ^(π/2) ln^2 (tanx)dx changementtanx=t give A =∫_0 ^∞ ((ln^2 (t))/(1+t^2 ))dt =∫_0 ^1 ((ln^2 (t))/(1+t^2 ))dt +∫_1 ^∞ ((ln^2 (t))/(1+t^2 ))dt (→t=(1/x)) =∫_0 ^1 ((ln^2 (t))/(1+t^2 ))dt −∫_0 ^1 ((ln^2 (x))/(1+(1/x^2 )))(−(dx/x^2 )) =2∫_0 ^1 ((ln^2 (x))/(1+x^2 ))dx =2∫_0 ^1 ln^2 (x)(Σ_(n=0) ^∞ (−1)^n x^(2n) )dx =2Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(2n) ln^2 (x)dx u_n =∫_0 ^(1 ) x^(2n) ln^2 (x)dx =[(x^(2n+1) /(2n+1))ln^2 (x)]_0 ^1 −∫_0 ^1 (x^(2n+1) /(2n+1))((2lnx)/x)dx =−(2/(2n+1))∫_0 ^1 x^(2n) ln(x)dx =−(2/(2n+1)){[(x^(2n+1) /(2n+1))lnx]_0 ^1 −∫_0 ^1 (x^(2n) /((2n+1)))dx} =(2/((2n+1)^3 )) ⇒ A =4 Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^3 )) rest to find tbe value of this serie by fourier....$
	$A = \int_{0}^{\frac{π}{2}} \ln^{2} (tanx) dx changementtanx = t give$ $A = \int_{0}^{\infty} \frac{\ln^{2} (t)}{1 + t^{2}} dt = \int_{0}^{1} \frac{\ln^{2} (t)}{1 + t^{2}} dt + \int_{1}^{\infty} \frac{\ln^{2} (t)}{1 + t^{2}} dt (\to t = \frac{1}{x})$ $= \int_{0}^{1} \frac{\ln^{2} (t)}{1 + t^{2}} dt - \int_{0}^{1} \frac{\ln^{2} (x)}{1 + \frac{1}{x^{2}}} (- \frac{dx}{x^{2}}) = 2 \int_{0}^{1} \frac{\ln^{2} (x)}{1 + x^{2}} dx$ $= 2 \int_{0}^{1} \ln^{2} (x) (\sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n}) dx = 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{2 n} \ln^{2} (x) dx$ $u_{n} = \int_{0}^{1} x^{2 n} \ln^{2} (x) dx = {[\frac{x^{2 n + 1}}{2 n + 1} \ln^{2} (x)]}_{0}^{1} - \int_{0}^{1} \frac{x^{2 n + 1}}{2 n + 1} \frac{2 lnx}{x} dx$ $= - \frac{2}{2 n + 1} \int_{0}^{1} x^{2 n} \ln (x) dx = - \frac{2}{2 n + 1} {{[\frac{x^{2 n + 1}}{2 n + 1} lnx]}_{0}^{1} - \int_{0}^{1} \frac{x^{2 n}}{(2 n + 1)} dx}$ $= \frac{2}{{(2 n + 1)}^{3}} \Rightarrow A = 4 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} rest to find tbe value of this$ $serie by fourier . . . .$
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