.... calculus .... evaluate ::: i::∫_0 ^( 1) t^2 ln(t)ln(1−t)dt=??? ii::: ψ^′ ((1/4))=??? iii::: ∫


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 114302 by mnjuly1970 last updated on 18/Sep/20

$. . . . c a l c u l u s . . . .$ $e v a l u a t e :::$ $i :: \int_{0}^{1} t^{2} l n (t) l n (1 - t) d t = ? ? ?$ $i i ::: ψ^{^{'}} (\frac{1}{4}) = ? ? ?$ $i i i ::: \int_{0}^{\frac{π}{8}} l n (t a n (x)) d x = ? ? ?$
Commented by MJS_new last updated on 18/Sep/20

$for \int \ln \sin x d x use the same path as I used$ $for \int \ln \cos x d x in question 113634$
Commented by MJS_new last updated on 18/Sep/20

$. . . you changed it from \sin to \tan after I$ $commented . . .$
	Answered by Olaf last updated on 18/Sep/20

	$I_{n} = \int_{0}^{1} x^{n} \ln x d x$ $I_{n} = {[\frac{x^{n + 1}}{n + 1} \ln x]}_{0}^{1} - \int_{0}^{1} \frac{x^{n + 1}}{n + 1} . \frac{d x}{x}$ $I_{n} = - {[\frac{x^{n + 1}}{{(n + 1)}^{2}}]}_{0}^{1} = - \frac{1}{{(n + 1)}^{2}}$ $\frac{1}{1 - t} = \underset{n = 0}{\sum^{\infty}} t^{n}$ $\ln ∣ 1 - t ∣ = - \underset{n = 0}{\sum^{\infty}} \frac{t^{n + 1}}{n + 1} if ∣ t ∣< 1$ $I = \int_{0}^{1} t^{2} \ln (t) \ln (1 - t) d t$ $I = - \int_{0}^{1} t^{2} \ln (t) \underset{n = 0}{\sum^{\infty}} \frac{t^{n + 1}}{n + 1} d t$ $I = - \underset{n = 0}{\sum^{\infty}} \frac{1}{n + 1} \int_{0}^{1} t^{n + 3} \ln t d t$ $I = - \underset{n = 0}{\sum^{\infty}} \frac{I_{n + 3}}{n + 1} = \underset{n = 0}{\sum^{\infty}} \frac{1}{(n + 1) {(n + 4)}^{2}}$ $I = \underset{n = 1}{\sum^{\infty}} \frac{1}{n {(n + 3)}^{2}}$ $\frac{1}{n {(n + 3)}^{2}} = \frac{1}{9 n} - \frac{1}{9 (n + 3)} - \frac{1}{3 {(n + 3)}^{2}}$ $I = \frac{1}{9} (\frac{1}{1} + \frac{1}{2} + \frac{1}{3}) - \frac{1}{3} \underset{n = 1 (}{\sum^{\infty}} \frac{1}{{n + 3)}^{2}}$ $I = \frac{11}{54} - \frac{1}{3} \underset{n = 4}{\sum^{\infty}} \frac{1}{n^{2}}$ $With \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} = ξ (2) = \frac{π^{2}}{6}$ $I = \frac{11}{54} - \frac{1}{3} (\frac{π^{2}}{6} - \frac{1}{1^{2}} - \frac{1}{2^{2}} - \frac{1}{3^{2}})$ $I = \frac{71}{108} - \frac{π^{2}}{18} \approx 0, 109096051$
	Commented by mnjuly1970 last updated on 19/Sep/20

	$v e r y g o o d . t h a n k y o u s i r$ $y o u r w o r k i s a d m i r a b l e . . .$
	Answered by mathmax by abdo last updated on 19/Sep/20
	$3) I =∫_0 ^(π/8) ln(tanx)dx by parts I= [xln(tanx)]_0 ^(π/8) −∫_0 ^(π/8) x×((1+tan^2 x)/(tanx))dx =(π/8)ln((√2)−1)−∫_0 ^(π/8) (x/(tanx))dx−∫_0 ^(π/8) tanx dx ∫_0 ^(π/8) tanx dx =[−ln∣cosx∣]_0 ^(π/8) =−ln(((√2)/2)) =−ln((1/(√2))) =((ln2)/2) A =∫_0 ^(π/8) (x/(tanx))dx ⇒ A =_(tanx=t) ∫_0 ^((√2)−1) ((arctant)/t)(dt/(1+t^2 )) =∫_0 ^((√2)−1) arctan(t){(1/t)−(t/(1+t^2 ))}dt =∫_0 ^((√2)−1) ((arctant)/t)dt −∫_0 ^((√2)−1) (t/(1+t^2 )) arctan(t)dt by parts ∫_0 ^((√2)−1) (t/(1+t^2 )) arctan(t)dt =[(1/2)ln(1+t^2 )arctant]_0 ^((√2)−1) −∫_0 ^((√2)−1) (1/2)ln(1+t^2 )×(dt/(1+t^2 )) =(1/2)ln(4−2(√2)) arctan((√2)−1) −(1/2) ∫_0 ^((√2)−1) ((ln(1+t^2 ))/(1+t^2 )) dt we considere f(a) =∫_0 ^((√2)−1) ((ln(1+at^2 ))/(1+t^2 ))dt with a>0 f^′ (a) =∫_0 ^((√2)−1) (t^2 /((1+t^2 )(1+at^2 ))) dt let decompose F(t) =(t^2 /((t^2 +1)(at^2 +1))) ⇒F(t)=((αt +β)/(t^2 +1)) +((mt+n)/(at^2 +1)) ⇒ F(−t)=F(t) ⇒((−αt +β)/(t^2 +1)) +((−mt +n)/(at^2 +1)) =F(t)⇒α=m=0 ⇒ F(t)=(β/(t^2 +1)) +(n/(at^2 +1)) we haveF(0) =0 =β+n ⇒n=−β lim_(t→+∞) t^2 F(t) =(1/a) =β +(n/a) ⇒1=aβ +n ⇒1=aβ−β =(a−1)β ⇒β =(1/(a−1)) ⇒F(t) =(1/((a−1)(t^2 +1)))−(1/((a−1)(at^2 +1))) ⇒ f^′ (a) =(1/(a−1)) ∫_0 ^((√2)−1) (dt/(t^2 +1))−(1/(a−1))∫_0 ^((√2)−1 ) (dt/(at^2 +1))(→(√a)t =u) =((arctan((√2)−1))/(a−1))−(1/(a−1)) ∫_0 ^(((√2)−1)(√a)) (du/((√a)(u^2 +1))) =((arctan((√2)−1))/(a−1))−(1/((√a)(a−1)))arctan(((√2)−1)(√a)) ⇒ f(a) =arctan((√2)−1)ln∣a−1∣−∫ ((arctan(((√2)−1)(√a)))/((√a)(a−1))) da +C ....be continued...$
	$3) I = \int_{0}^{\frac{π}{8}} \ln (tanx) dx by parts I = {[xln (tanx)]}_{0}^{\frac{π}{8}} - \int_{0}^{\frac{π}{8}} x \times \frac{1 + \tan^{2} x}{tanx} dx$ $= \frac{π}{8} \ln (\sqrt{2} - 1) - \int_{0}^{\frac{π}{8}} \frac{x}{tanx} dx - \int_{0}^{\frac{π}{8}} tanx dx$ $\int_{0}^{\frac{π}{8}} tanx dx = {[- \ln ∣ cosx ∣]}_{0}^{\frac{π}{8}} = - \ln (\frac{\sqrt{2}}{2}) = - \ln (\frac{1}{\sqrt{2}}) = \frac{\ln 2}{2}$ $A = \int_{0}^{\frac{π}{8}} \frac{x}{tanx} dx \Rightarrow A =_{tanx = t} \int_{0}^{\sqrt{2} - 1} \frac{arctant}{t} \frac{dt}{1 + t^{2}}$ $= \int_{0}^{\sqrt{2} - 1} \arctan (t) {\frac{1}{t} - \frac{t}{1 + t^{2}}} dt = \int_{0}^{\sqrt{2} - 1} \frac{arctant}{t} dt - \int_{0}^{\sqrt{2} - 1} \frac{t}{1 + t^{2}} \arctan (t) dt$ $by parts \int_{0}^{\sqrt{2} - 1} \frac{t}{1 + t^{2}} \arctan (t) dt = {[\frac{1}{2} \ln (1 + t^{2}) arctant]}_{0}^{\sqrt{2} - 1}$ $- \int_{0}^{\sqrt{2} - 1} \frac{1}{2} \ln (1 + t^{2}) \times \frac{dt}{1 + t^{2}} = \frac{1}{2} \ln (4 - 2 \sqrt{2}) \arctan (\sqrt{2} - 1)$ $- \frac{1}{2} \int_{0}^{\sqrt{2} - 1} \frac{\ln (1 + t^{2})}{1 + t^{2}} dt we considere f (a) = \int_{0}^{\sqrt{2} - 1} \frac{\ln (1 + {at}^{2})}{1 + t^{2}} dt$ $with a > 0$ $f^{^{'}} (a) = \int_{0}^{\sqrt{2} - 1} \frac{t^{2}}{(1 + t^{2}) (1 + {at}^{2})} dt let decompose$ $F (t) = \frac{t^{2}}{(t^{2} + 1) ({at}^{2} + 1)} \Rightarrow F (t) = \frac{α t + β}{t^{2} + 1} + \frac{mt + n}{{at}^{2} + 1} \Rightarrow$ $F (- t) = F (t) \Rightarrow \frac{- α t + β}{t^{2} + 1} + \frac{- mt + n}{{at}^{2} + 1} = F (t) \Rightarrow α = m = 0 \Rightarrow$ $F (t) = \frac{β}{t^{2} + 1} + \frac{n}{{at}^{2} + 1} we haveF (0) = 0 = β + n \Rightarrow n = - β$ $\lim_{t \to + \infty} t^{2} F (t) = \frac{1}{a} = β + \frac{n}{a} \Rightarrow 1 = a β + n \Rightarrow 1 = a β - β = (a - 1) β$ $\Rightarrow β = \frac{1}{a - 1} \Rightarrow F (t) = \frac{1}{(a - 1) (t^{2} + 1)} - \frac{1}{(a - 1) ({at}^{2} + 1)} \Rightarrow$ $f^{^{'}} (a) = \frac{1}{a - 1} \int_{0}^{\sqrt{2} - 1} \frac{dt}{t^{2} + 1} - \frac{1}{a - 1} \int_{0}^{\sqrt{2} - 1} \frac{dt}{{at}^{2} + 1} (\to \sqrt{a} t = u)$ $= \frac{\arctan (\sqrt{2} - 1)}{a - 1} - \frac{1}{a - 1} \int_{0}^{(\sqrt{2} - 1) \sqrt{a}} \frac{du}{\sqrt{a} (u^{2} + 1)}$ $= \frac{\arctan (\sqrt{2} - 1)}{a - 1} - \frac{1}{\sqrt{a} (a - 1)} \arctan ((\sqrt{2} - 1) \sqrt{a}) \Rightarrow$ $f (a) = \arctan (\sqrt{2} - 1) \ln ∣ a - 1 ∣ - \int \frac{\arctan ((\sqrt{2} - 1) \sqrt{a})}{\sqrt{a} (a - 1)} da + C$ $. . . . be continued . . .$
	Answered by maths mind last updated on 19/Sep/20
	$∫_0 ^(π/8) ln(tg(x))dx ln(tg(x))=−2Σ_(k≥0) ((cos(2(2k+1)x))/(2k+1)) =∫_0 ^(π/8) (−2Σ_(k≥0) ((cos(2(2k+1)x))/(2k+1)))dx =Σ_(k≥0) −(1/((2k+1)^2 ))sin((((2k+1)π)/4)) =−Σ_(k≥0) (1/((2k+1)^2 ))sin(((kπ)/2)+(π/4)) =−Σ_(k≥0) ((sin((π/4)))/((8k+1)^2 ))−Σ_(k≥0) ((sin((π/4)))/((8k+3)^2 ))+Σ((sin((π/4)))/((8k+5)^2 ))+Σ((sin((π/4)))/((8k+7)^2 )) =((sin((π/4)))/(64)){−Σ(1/((k+(1/8))^2 ))−Σ(1/((k+(3/8))^2 ))+Σ(1/((k+(5/8))^2 ))+Σ(1/((k+(7/8))^2 ))} Σ_(n≥0) (1/((n+a)^p ))=ζ(a,p) hurwitz zeta function =((sin((π/4)))/(64)){−ζ((1/8),2)−ζ((3/8),2)+ζ((5/8),2)+ζ((7/8),2)}$
	$\int_{0}^{\frac{π}{8}} l n (t g (x)) d x$ $l n (t g (x)) = - 2 \sum_{k ⩾ 0} \frac{c o s (2 (2 k + 1) x)}{2 k + 1}$ $= \int_{0}^{\frac{π}{8}} (- 2 \sum_{k ⩾ 0} \frac{c o s (2 (2 k + 1) x)}{2 k + 1}) d x$ $= \sum_{k ⩾ 0} - \frac{1}{{(2 k + 1)}^{2}} s i n (\frac{(2 k + 1) π}{4})$ $= - \sum_{k ⩾ 0} \frac{1}{{(2 k + 1)}^{2}} s i n (\frac{k π}{2} + \frac{π}{4})$ $= - \sum_{k ⩾ 0} \frac{s i n (\frac{π}{4})}{{(8 k + 1)}^{2}} - \sum_{k ⩾ 0} \frac{s i n (\frac{π}{4})}{{(8 k + 3)}^{2}} + Σ \frac{s i n (\frac{π}{4})}{{(8 k + 5)}^{2}} + Σ \frac{s i n (\frac{π}{4})}{{(8 k + 7)}^{2}}$ $= \frac{s i n (\frac{π}{4})}{64} {- Σ \frac{1}{{(k + \frac{1}{8})}^{2}} - Σ \frac{1}{{(k + \frac{3}{8})}^{2}} + Σ \frac{1}{{(k + \frac{5}{8})}^{2}} + Σ \frac{1}{{(k + \frac{7}{8})}^{2}}}$ $\sum_{n ⩾ 0} \frac{1}{{(n + a)}^{p}} = ζ (a, p) h u r w i t z z e t a f u n c t i o n$ $= \frac{s i n (\frac{π}{4})}{64} {- ζ (\frac{1}{8}, 2) - ζ (\frac{3}{8}, 2) + ζ (\frac{5}{8}, 2) + ζ (\frac{7}{8}, 2)}$
	Commented by mnjuly1970 last updated on 19/Sep/20

	$v e r y n i c e s i r . t h a n k s a l o t . .$
	Answered by Olaf last updated on 19/Sep/20

	$i i i . . .$ $I = \int_{0}^{\frac{π}{8}} \ln (\tan x) d x$ $\tan 2 θ = \frac{2 \tan θ}{1 - \tan^{2} θ}$ $with θ = \frac{π}{8}, \tan \frac{π}{4} = \frac{2 \tan \frac{π}{8}}{1 - \tan^{2} \frac{π}{8}} = 1$ $\tan^{2} \frac{π}{8} + 2 \tan \frac{π}{8} - 1 = 0$ $\Rightarrow \tan \frac{π}{8} = \sqrt{2} - 1$ $Now u = \tan x$ $d u = (1 + \tan^{2} x) d x = (1 + u^{2}) d x$ $I = \int_{0}^{\sqrt{2} - 1} \frac{\ln u}{1 + u^{2}} d u$ $I = \int_{0}^{\sqrt{2} - 1} \ln u \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} u^{2 n} d u$ $I = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{\sqrt{2} - 1} u^{2 n} \ln u d u$ $I = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} ({[\frac{u^{2 n + 1}}{2 n + 1} \ln u]}_{0}^{\sqrt{2} - 1} - \int_{0}^{\sqrt{2} - 1} \frac{u^{2 n + 1}}{2 n + 1} . \frac{d u}{u})$ $I = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} (\ln (\sqrt{2} - 1) [\frac{{(\sqrt{2} - 1)}^{2 n + 1}}{2 n + 1}] - {[\frac{u^{2 n + 1}}{{(2 n + 1)}^{2}}]}_{0}^{\sqrt{2} - 1})$ $I = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} (\ln (\sqrt{2} - 1) [\frac{{(\sqrt{2} - 1)}^{2 n + 1}}{2 n + 1}] - [\frac{{(\sqrt{2} - 1)}^{2 n + 1}}{{(2 n + 1)}^{2}}])$ $\arctan x = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{x^{2 n + 1}}{2 n + 1}$ $I = \ln (\sqrt{2} - 1) \arctan (\sqrt{2} - 1) - \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{{(\sqrt{2} - 1)}^{2 n + 1}}{{(2 n + 1)}^{2}}$ $I = \frac{π}{8} \ln (\sqrt{2} - 1) - \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{{(\sqrt{2} - 1)}^{2 n + 1}}{{(2 n + 1)}^{2}}$ $I think we cannot simplify$
	Answered by mathmax by abdo last updated on 20/Sep/20

	$A = \int_{0}^{1} x^{2} \ln (x) \ln (1 - x) dx we have \frac{d}{dx} \ln (1 - x) = \frac{- 1}{1 - x} = - \sum_{n = 0}^{\infty} x^{n}$ $\Rightarrow \ln (1 - x) = - \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1} + c (c = 0) = - \sum_{n = 1}^{\infty} \frac{x^{n}}{n} \Rightarrow$ $A = - \int_{0}^{1} x^{2} lnx (\sum_{n = 1}^{\infty} \frac{x^{n}}{n}) dx = - \sum_{n = 1}^{\infty} \frac{1}{n} \int_{0}^{1} x^{n + 2} \ln (x) dx$ $u_{n} = \int_{0}^{1} x^{n + 2} \ln (x) dx = {[\frac{x^{n + 3}}{n + 3} \ln (x)]}_{0}^{1} - \int_{0}^{1} \frac{x^{n + 2}}{n + 3} dx$ $= - \frac{1}{{(n + 3)}^{2}} \Rightarrow A = \sum_{n = 1}^{\infty} \frac{1}{n {(n + 3)}^{2}} let decompose$ $F (x) = \frac{1}{x {(x + 3)}^{2}} \Rightarrow F (x) = \frac{a}{x} + \frac{b}{x + 3} + \frac{c}{{(x + 3)}^{2}}$ $a = \frac{1}{9}, c = - \frac{1}{3} \Rightarrow F (x) = \frac{1}{9 x} + \frac{b}{x + 3} - \frac{1}{3 {(x + 3)}^{2}}$ $\lim_{x \to + \infty} xF (x) = 0 = \frac{1}{9} + b \Rightarrow b = - \frac{1}{9} \Rightarrow$ $A = \lim_{n \to + \infty} A_{n} / A_{n} = \sum_{k = 1}^{n} \frac{1}{k {(k + 3)}^{2}}$ $= \frac{1}{9} \sum_{k = 1}^{n} \frac{1}{k} - \frac{1}{9} \sum_{k = 1}^{n} \frac{1}{k + 3} - \frac{1}{3} \sum_{k = 1}^{n} \frac{1}{{(k + 3)}^{2}}$ $= \frac{1}{9} \sum_{k = 1}^{n} \frac{1}{k} - \frac{1}{9} \sum_{k = 4}^{n + 3} - \frac{1}{3} \sum_{k = 4}^{n + 3} \frac{1}{k^{2}}$ $= \frac{1}{9} (1 + \frac{1}{2} + \frac{1}{3}) - \frac{1}{9} (\frac{1}{n + 1} + \frac{1}{n + 2} + \frac{1}{n + 3}) - \frac{1}{3} (\sum_{k = 1}^{n + 3} \frac{1}{k^{2}} - 1 - \frac{1}{2^{2}} - \frac{1}{3^{2}})$ $\to \frac{1}{9} (\frac{3}{2} + \frac{1}{3}) - \frac{1}{3} \times \frac{π^{2}}{6} + \frac{1}{3} (1 + \frac{1}{4} + \frac{1}{9})$ $= \frac{1}{9} (\frac{11}{6}) - \frac{π^{2}}{18} + \frac{1}{3} (\frac{5}{4} + \frac{1}{9}) = \frac{11}{54} - \frac{π^{2}}{18} + \frac{1}{3} (\frac{49}{36}) = \frac{11}{54} + \frac{49}{108} - \frac{π^{2}}{18}$ $\Rightarrow \int_{0}^{1} x^{2} lnx \ln (1 - x) dx = \frac{11}{54} + \frac{49}{108} - \frac{π^{2}}{18}$
	Commented by mathmax by abdo last updated on 20/Sep/20

	$\Rightarrow \int_{0}^{1} x^{2} lnxln (1 - x) dx = \frac{71}{108} - \frac{π^{2}}{18}$
	Answered by maths mind last updated on 20/Sep/20

	$Ψ_{1} (\frac{1}{4}) = \int_{0}^{\infty} \frac{l n (x) x^{- \frac{3}{4}}}{x - 1} d x$ $x^{\frac{1}{4}} = t \Rightarrow d t = \frac{x^{- \frac{3}{4}} d x}{4}$ $= \int_{0}^{\infty} \frac{4 l n (t^{4})}{t^{4} - 1} d t$ $= 16 \int_{0}^{1} \frac{l n (t)}{(t^{2} - 1) (t^{2} + 1)} d t$ $= 8 \int_{0}^{1} \frac{l n (t)}{t^{2} - 1} - \int_{0}^{1} \frac{l n (t)}{t^{2} + 1}$ $= - 8 \int_{0}^{1} \sum_{k ⩾ 0} t^{2 k} l n (t) d t - 8 \sum_{m ⩾ 0} \int_{0}^{1} {(- t^{2})}^{m} l n (t) d t$ $= - 8 \sum_{k ⩾ 0} \int_{0}^{1} t^{2 k} l n (t) d t - 8 Σ {(- 1)}^{m} \int_{0}^{1} t^{2 m} l n (t) d t$ $\int_{0}^{1} t^{n} l n (t) = {[\frac{t^{n + 1} l n (t)}{n + 1}]}_{0}^{1} - \int \frac{t^{n}}{n + 1} d t = - \frac{1}{{(n + 1)}^{2}}$ $= 8 \sum_{k ⩾ 0} \frac{1}{{(2 k + 1)}^{2}} + 8 . Σ \frac{{(- 1)}^{k}}{{(2 k + 1)}^{2}}$ $= 8 . \frac{3}{4} ζ (2) + 8 . G$ $= π^{2} + 8 G$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum