... ◂nice integral▶... please prove : Ω=∫_0 ^( (π/2)) ln^2 (cot(x))dx =(π^3 /8) ...♠m.n.1070♠...


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Question Number 117953 by mnjuly1970 last updated on 14/Oct/20

$. . . ◂ n i c e i n t e g r a l ▸ . . .$ $p l e a s e p r o v e :$ $Ω = \int_{0}^{\frac{π}{2}} {l n}^{2} (c o t (x)) d x = \frac{π^{3}}{8}$ $. . . ♠ m . n . 1070 ♠ . . .$
	Answered by mindispower last updated on 14/Oct/20

	$Ω = \int_{0}^{\frac{π}{2}} {l n}^{2} (c o t (x)) d x; c o t (x) = s \Rightarrow$ $Ω = \int_{0}^{\infty} \frac{{l n}^{2} (s)}{1 + s^{2}} d s = 2 \int_{0}^{1} \frac{{l n}^{2} (s)}{1 + s^{2}} d s = 2 \int_{0}^{1} Σ {(- s^{2})}^{k} {l n}^{2} (s) d s$ $= 2 \sum_{n ⩾ 0} {(- 1)}^{n} \int_{0}^{1} s^{2 k} {l n}^{2} (s) d s$ $\int_{0}^{1} s^{n} {l n}^{2} (s) d s = - \frac{2}{n + 1} \int_{0}^{1} s^{n} l n (s) d s$ $= \frac{2}{{(n + 1)}^{2}} \int_{0}^{1} s^{n} d s = \frac{2}{{(n + 1)}^{3}}$ $w e g e t \sum_{n ⩾ 0} \frac{4 {(- 1)}^{n}}{{(2 n + 1)}^{3}} = \sum_{n ⩾ 0} \frac{4}{{(4 n + 1)}^{3}} - \sum_{n ⩾ 0} \frac{4}{{(4 n + 3)}^{3}}$ $= \frac{1}{16} [\sum_{n ⩾ 0} \frac{1}{{(n + \frac{1}{4})}^{3}} - \sum_{n ⩾ 0} \frac{1}{{(n + \frac{3}{4})}^{3}}]$ $= \frac{1}{16} [Ψ^{3} (\frac{1}{4}) - Ψ^{3} (\frac{3}{4})]$ $Ψ (1 - x) - Ψ (x) = π c o t (π x)$ $\Rightarrow Ψ^{3} (1 - x) - Ψ^{3} (x) = \frac{d^{2}}{{d x}^{2}} (π c o t (π x))$ $= \frac{d}{d x} (- π^{2} (1 + {c o t}^{2} (π x)) = 2 π^{3} (1 + {c o t}^{2} (π x)) c o t (π x))$ $\frac{1}{16} [Ψ^{3} (\frac{1}{4}) - Ψ^{3} (\frac{3}{4})] = \frac{1}{16} (2 π^{3} (1 + {c o t}^{2} (\frac{π}{4})) c o t (\frac{π}{4})$ $= \frac{π^{3}}{4} m a y b e e f o r g e t s o m t h i n g$
	Commented by mnjuly1970 last updated on 14/Oct/20

	$t h a n k y o u m a s t e r$ $y o u r w o r k i s v e r y n i c e a n d$ $a d m i r a b l e$ $f i n a l a n s w e r i s n o t v e r y$ $i m p o r t a n t . y o u r s o l u t i o n$ $i s f a b u l o u s . g r a t e f u l .$
	Commented by mindispower last updated on 19/Oct/20

	$t h a n k y o u w i t h e p l e s u r$ $s i r d o y o u h a v e s o m m e l e c t u r a b o u t p o l y g a r i t h m s$ $f u n c t i o n i d e n t i t i e s$ $t h a n k y o u f o r y o u a n s w e r$
	Answered by mathmax by abdo last updated on 14/Oct/20
	$A =∫_0 ^(π/2) ln^2 (cotanx)dx ⇒ A =∫_0 ^(π/2) ln^2 ((1/(tanx)))dx =∫_0 ^(π/2) ln^2 (tanx)dx =_(tanx=t) ∫_0 ^∞ ((ln^2 (t))/(1+t^2 ))dt =∫_0 ^1 ((ln^2 t)/(1+t^2 ))dt +∫_1 ^∞ ((ln^2 t)/(1+t^2 ))dt(→t=(1/u)) =∫_0 ^1 ((ln^2 t)/(1+t^2 ))dt +∫_0 ^1 ((ln^2 u)/((1+(1/u^2 ))))(du/u^2 ) =2∫_0 ^1 ((ln^2 x)/(1+x^2 ))dx =2 ∫_0 ^1 ln^2 x(Σ_(n=0) ^∞ (−1)^n x^(2n) )dx =2 Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(2n) ln^2 x dx =2Σ_(n=0) ^∞ (−1)^n A_n by parts A_n =∫_0 ^1 x^(2n) ln^2 xdx =[(x^(2n+1) /(2n+1))ln^2 x]_0 ^1 −∫_0 ^1 (x^(2n+1) /(2n+1))((2lnx)/x)dx =−(2/((2n+1)))∫_0 ^1 x^(2n) lnx dx =−(2/((2n+1))){ [(x^(2n+1) /(2n+1))lnx]_0 ^1 −∫_0 ^1 (x^(2n) /(2n+1))} =(2/((2n+1)^3 )) ⇒ A =4 Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^3 )) rest to find the value of this serie by sir fourier...be continued....$
	$A = \int_{0}^{\frac{π}{2}} \ln^{2} (cotanx) dx \Rightarrow A = \int_{0}^{\frac{π}{2}} \ln^{2} (\frac{1}{tanx}) dx$ $= \int_{0}^{\frac{π}{2}} \ln^{2} (tanx) dx =_{tanx = t} \int_{0}^{\infty} \frac{\ln^{2} (t)}{1 + t^{2}} dt$ $= \int_{0}^{1} \frac{\ln^{2} t}{1 + t^{2}} dt + \int_{1}^{\infty} \frac{\ln^{2} t}{1 + t^{2}} dt (\to t = \frac{1}{u})$ $= \int_{0}^{1} \frac{\ln^{2} t}{1 + t^{2}} dt + \int_{0}^{1} \frac{\ln^{2} u}{(1 + \frac{1}{u^{2}})} \frac{du}{u^{2}} = 2 \int_{0}^{1} \frac{\ln^{2} x}{1 + x^{2}} dx$ $= 2 \int_{0}^{1} \ln^{2} x (\sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n}) dx$ $= 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{2 n} \ln^{2} x dx = 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} A_{n} by parts$ $A_{n} = \int_{0}^{1} x^{2 n} \ln^{2} xdx = {[\frac{x^{2 n + 1}}{2 n + 1} \ln^{2} x]}_{0}^{1} - \int_{0}^{1} \frac{x^{2 n + 1}}{2 n + 1} \frac{2 lnx}{x} dx$ $= - \frac{2}{(2 n + 1)} \int_{0}^{1} x^{2 n} lnx dx = - \frac{2}{(2 n + 1)} {{[\frac{x^{2 n + 1}}{2 n + 1} lnx]}_{0}^{1} - \int_{0}^{1} \frac{x^{2 n}}{2 n + 1}}$ $= \frac{2}{{(2 n + 1)}^{3}} \Rightarrow A = 4 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} rest to find the value of$ $this serie by sir fourier . . . be continued . . . .$
	Commented by mnjuly1970 last updated on 15/Oct/20

	$t h a n k y o u s o m u c h s i r .$ $g r a t e f u l . . .$
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