Question Number 114302 by mnjuly1970 last updated on 18/Sep/20
$$\:\:\:\:\:\:\:\:….\:{calculus}\:…. \\ $$$$\:\:\:\:{evaluate}\:::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${i}::\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{\mathrm{2}} {ln}\left({t}\right){ln}\left(\mathrm{1}−{t}\right){dt}=??? \\ $$$${ii}:::\:\psi^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=??? \\ $$$${iii}:::\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} {ln}\left({tan}\left({x}\right)\right){dx}\:=??? \\ $$$$ \\ $$
Commented by MJS_new last updated on 18/Sep/20
$$\mathrm{for}\:\int\mathrm{ln}\:\mathrm{sin}\:{x}\:{dx}\:\mathrm{use}\:\mathrm{the}\:\mathrm{same}\:\mathrm{path}\:\mathrm{as}\:\mathrm{I}\:\mathrm{used} \\ $$$$\mathrm{for}\:\int\mathrm{ln}\:\mathrm{cos}\:{x}\:{dx}\:\mathrm{in}\:\mathrm{question}\:\mathrm{113634} \\ $$
Commented by MJS_new last updated on 18/Sep/20
$$…\mathrm{you}\:\mathrm{changed}\:\mathrm{it}\:\mathrm{from}\:\mathrm{sin}\:\mathrm{to}\:\mathrm{tan}\:\mathrm{after}\:\mathrm{I} \\ $$$$\mathrm{commented}… \\ $$
Answered by Olaf last updated on 18/Sep/20
![I_n = ∫_0 ^1 x^n lnxdx I_n = [(x^(n+1) /(n+1))lnx]_0 ^1 − ∫_0 ^1 (x^(n+1) /(n+1)).(dx/x) I_n = −[(x^(n+1) /((n+1)^2 ))]_0 ^1 = −(1/((n+1)^2 )) (1/(1−t)) = Σ_(n=0) ^∞ t^n ln∣1−t∣ = −Σ_(n=0) ^∞ (t^(n+1) /(n+1)) if ∣t∣<1 I = ∫_0 ^1 t^2 ln(t)ln(1−t)dt I = −∫_0 ^1 t^2 ln(t)Σ_(n=0) ^∞ (t^(n+1) /(n+1))dt I = −Σ_(n=0) ^∞ (1/(n+1))∫_0 ^1 t^(n+3) lntdt I = −Σ_(n=0) ^∞ (I_(n+3) /(n+1)) = Σ_(n=0) ^∞ (1/((n+1)(n+4)^2 )) I = Σ_(n=1) ^∞ (1/(n(n+3)^2 )) (1/(n(n+3)^2 )) = (1/(9n))−(1/(9(n+3)))−(1/(3(n+3)^2 )) I = (1/9)((1/1)+(1/2)+(1/3))−(1/3)Σ_(n=1() ^∞ (1/(n+3)^2 )) I = ((11)/(54))−(1/3)Σ_(n=4) ^∞ (1/n^2 ) With Σ_(n=1) ^∞ (1/n^2 ) = ξ(2) = (π^2 /6) I = ((11)/(54))−(1/3)((π^2 /6)−(1/1^2 )−(1/2^2 )−(1/3^2 )) I = ((71)/(108))−(π^2 /(18)) ≈ 0,109096051](https://www.tinkutara.com/question/Q114382.png)
$$\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \mathrm{ln}{xdx} \\ $$$$\mathrm{I}_{{n}} \:=\:\left[\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\mathrm{ln}{x}\right]_{\mathrm{0}} ^{\mathrm{1}} −\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}.\frac{{dx}}{{x}} \\ $$$$\mathrm{I}_{{n}} \:=\:−\left[\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{t}}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{t}^{{n}} \\ $$$$\mathrm{ln}\mid\mathrm{1}−{t}\mid\:=\:−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:\mathrm{if}\:\mid{t}\mid<\mathrm{1} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}} \mathrm{ln}\left({t}\right)\mathrm{ln}\left(\mathrm{1}−{t}\right){dt} \\ $$$$\mathrm{I}\:=\:−\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}} \mathrm{ln}\left({t}\right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{dt} \\ $$$$\mathrm{I}\:=\:−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}+\mathrm{3}} \mathrm{ln}{tdt} \\ $$$$\mathrm{I}\:=\:−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{I}_{{n}+\mathrm{3}} }{{n}+\mathrm{1}}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\mathrm{I}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{n}\left({n}+\mathrm{3}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{9}{n}}−\frac{\mathrm{1}}{\mathrm{9}\left({n}+\mathrm{3}\right)}−\frac{\mathrm{1}}{\mathrm{3}\left({n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{9}}\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}\left(\right.} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left.{n}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{11}}{\mathrm{54}}−\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{4}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\mathrm{With}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\:\xi\left(\mathrm{2}\right)\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{11}}{\mathrm{54}}−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right) \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{71}}{\mathrm{108}}−\frac{\pi^{\mathrm{2}} }{\mathrm{18}}\:\approx\:\mathrm{0},\mathrm{109096051} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 19/Sep/20
$${very}\:{good}\:\:.{thank}\:{you}\:{sir} \\ $$$${your}\:{work}\:{is}\:{admirable}… \\ $$
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...](https://www.tinkutara.com/question/Q114396.png)
$$\left.\mathrm{3}\right)\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\mathrm{ln}\left(\mathrm{tanx}\right)\mathrm{dx}\:\:\:\mathrm{by}\:\mathrm{parts}\:\mathrm{I}=\:\left[\mathrm{xln}\left(\mathrm{tanx}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \mathrm{x}×\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}{\mathrm{tanx}}\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{8}}\mathrm{ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\frac{\mathrm{x}}{\mathrm{tanx}}\mathrm{dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\mathrm{tanx}\:\mathrm{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\mathrm{tanx}\:\mathrm{dx}\:=\left[−\mathrm{ln}\mid\mathrm{cosx}\mid\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:=−\mathrm{ln}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\:=−\mathrm{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:=\frac{\mathrm{ln2}}{\mathrm{2}} \\ $$$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:\frac{\mathrm{x}}{\mathrm{tanx}}\mathrm{dx}\:\Rightarrow\:\mathrm{A}\:=_{\mathrm{tanx}=\mathrm{t}} \:\:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{arctant}}{\mathrm{t}}\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \mathrm{arctan}\left(\mathrm{t}\right)\left\{\frac{\mathrm{1}}{\mathrm{t}}−\frac{\mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right\}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{arctant}}{\mathrm{t}}\mathrm{dt}\:−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\mathrm{arctan}\left(\mathrm{t}\right)\mathrm{dt} \\ $$$$\mathrm{by}\:\mathrm{parts}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\mathrm{arctan}\left(\mathrm{t}\right)\mathrm{dt}\:=\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{arctant}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \\ $$$$−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)×\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)\:\mathrm{arctan}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt}\:\:\:\mathrm{we}\:\mathrm{considere}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{at}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\mathrm{with}\:\mathrm{a}>\mathrm{0} \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{at}^{\mathrm{2}} \right)}\:\mathrm{dt}\:\mathrm{let}\:\mathrm{decompose} \\ $$$$\mathrm{F}\left(\mathrm{t}\right)\:=\frac{\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{at}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow\mathrm{F}\left(\mathrm{t}\right)=\frac{\alpha\mathrm{t}\:+\beta}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{\mathrm{mt}+\mathrm{n}}{\mathrm{at}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{F}\left(−\mathrm{t}\right)=\mathrm{F}\left(\mathrm{t}\right)\:\Rightarrow\frac{−\alpha\mathrm{t}\:+\beta}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−\mathrm{mt}\:+\mathrm{n}}{\mathrm{at}^{\mathrm{2}} \:+\mathrm{1}}\:=\mathrm{F}\left(\mathrm{t}\right)\Rightarrow\alpha=\mathrm{m}=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{\beta}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{\mathrm{n}}{\mathrm{at}^{\mathrm{2}} \:+\mathrm{1}}\:\mathrm{we}\:\mathrm{haveF}\left(\mathrm{0}\right)\:=\mathrm{0}\:=\beta+\mathrm{n}\:\Rightarrow\mathrm{n}=−\beta \\ $$$$\mathrm{lim}_{\mathrm{t}\rightarrow+\infty} \mathrm{t}^{\mathrm{2}} \mathrm{F}\left(\mathrm{t}\right)\:=\frac{\mathrm{1}}{\mathrm{a}}\:=\beta\:+\frac{\mathrm{n}}{\mathrm{a}}\:\Rightarrow\mathrm{1}=\mathrm{a}\beta\:+\mathrm{n}\:\Rightarrow\mathrm{1}=\mathrm{a}\beta−\beta\:=\left(\mathrm{a}−\mathrm{1}\right)\beta \\ $$$$\Rightarrow\beta\:=\frac{\mathrm{1}}{\mathrm{a}−\mathrm{1}}\:\Rightarrow\mathrm{F}\left(\mathrm{t}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{a}−\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left(\mathrm{a}−\mathrm{1}\right)\left(\mathrm{at}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\frac{\mathrm{1}}{\mathrm{a}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{a}−\mathrm{1}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}\:} \:\frac{\mathrm{dt}}{\mathrm{at}^{\mathrm{2}} \:+\mathrm{1}}\left(\rightarrow\sqrt{\mathrm{a}}\mathrm{t}\:=\mathrm{u}\right) \\ $$$$=\frac{\mathrm{arctan}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{a}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{a}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\sqrt{\mathrm{a}}} \:\frac{\mathrm{du}}{\:\sqrt{\mathrm{a}}\left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{arctan}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{a}−\mathrm{1}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}\left(\mathrm{a}−\mathrm{1}\right)}\mathrm{arctan}\left(\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\sqrt{\mathrm{a}}\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=\mathrm{arctan}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{ln}\mid\mathrm{a}−\mathrm{1}\mid−\int\:\:\frac{\mathrm{arctan}\left(\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\sqrt{\mathrm{a}}\right)}{\:\sqrt{\mathrm{a}}\left(\mathrm{a}−\mathrm{1}\right)}\:\mathrm{da}\:+\mathrm{C} \\ $$$$….\mathrm{be}\:\mathrm{continued}… \\ $$$$ \\ $$
Answered by maths mind last updated on 19/Sep/20
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} {ln}\left({tg}\left({x}\right)\right){dx} \\ $$$${ln}\left({tg}\left({x}\right)\right)=−\mathrm{2}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{cos}\left(\mathrm{2}\left(\mathrm{2}{k}+\mathrm{1}\right){x}\right)}{\mathrm{2}{k}+\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \left(−\mathrm{2}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{cos}\left(\mathrm{2}\left(\mathrm{2}{k}+\mathrm{1}\right){x}\right)}{\mathrm{2}{k}+\mathrm{1}}\right){dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}−\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }{sin}\left(\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{4}}\right) \\ $$$$=−\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }{sin}\left(\frac{{k}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right) \\ $$$$=−\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}{\left(\mathrm{8}{k}+\mathrm{1}\right)^{\mathrm{2}} }−\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}{\left(\mathrm{8}{k}+\mathrm{3}\right)^{\mathrm{2}} }+\Sigma\frac{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}{\left(\mathrm{8}{k}+\mathrm{5}\right)^{\mathrm{2}} }+\Sigma\frac{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}{\left(\mathrm{8}{k}+\mathrm{7}\right)^{\mathrm{2}} } \\ $$$$=\frac{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{64}}\left\{−\Sigma\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{1}}{\mathrm{8}}\right)^{\mathrm{2}} }−\Sigma\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{3}}{\mathrm{8}}\right)^{\mathrm{2}} }+\Sigma\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{5}}{\mathrm{8}}\right)^{\mathrm{2}} }+\Sigma\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{7}}{\mathrm{8}}\right)^{\mathrm{2}} }\right\} \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({n}+{a}\right)^{{p}} }=\zeta\left({a},{p}\right)\:{hurwitz}\:{zeta}\:{function} \\ $$$$ \\ $$$$=\frac{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{64}}\left\{−\zeta\left(\frac{\mathrm{1}}{\mathrm{8}},\mathrm{2}\right)−\zeta\left(\frac{\mathrm{3}}{\mathrm{8}},\mathrm{2}\right)+\zeta\left(\frac{\mathrm{5}}{\mathrm{8}},\mathrm{2}\right)+\zeta\left(\frac{\mathrm{7}}{\mathrm{8}},\mathrm{2}\right)\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 19/Sep/20
$${very}\:{nice}\:{sir}\:.{thanks}\:{alot}.. \\ $$
Answered by Olaf last updated on 19/Sep/20
![iii... I = ∫_0 ^(π/8) ln(tanx)dx tan2θ = ((2tanθ)/(1−tan^2 θ)) with θ = (π/8), tan(π/4) = ((2tan(π/8))/(1−tan^2 (π/8))) = 1 tan^2 (π/8)+2tan(π/8)−1 = 0 ⇒ tan(π/8) = (√2)−1 Now u = tanx du = (1+tan^2 x)dx = (1+u^2 )dx I = ∫_0 ^((√2)−1) ((lnu)/(1+u^2 ))du I = ∫_0 ^((√2)−1) lnuΣ_(n=0) ^∞ (−1)^n u^(2n) du I = Σ_(n=0) ^∞ (−1)^n ∫_0 ^((√2)−1) u^(2n) lnudu I = Σ_(n=0) ^∞ (−1)^n ([(u^(2n+1) /(2n+1))lnu]_0 ^((√2)−1) −∫_0 ^((√2)−1) (u^(2n+1) /(2n+1)).(du/u)) I = Σ_(n=0) ^∞ (−1)^n (ln((√2)−1)[((((√2)−1)^(2n+1) )/(2n+1))]−[(u^(2n+1) /((2n+1)^2 ))]_0 ^((√2)−1) ) I = Σ_(n=0) ^∞ (−1)^n (ln((√2)−1)[((((√2)−1)^(2n+1) )/(2n+1))]−[((((√2)−1)^(2n+1) )/((2n+1)^2 ))]) arctanx = Σ_(n=0) ^∞ (−1)^n (x^(2n+1) /(2n+1)) I = ln((√2)−1)arctan((√2)−1)−Σ_(n=0) ^∞ (−1)^n ((((√2)−1)^(2n+1) )/((2n+1)^2 )) I = (π/8)ln((√2)−1)−Σ_(n=0) ^∞ (−1)^n ((((√2)−1)^(2n+1) )/((2n+1)^2 )) I think we cannot simplify](https://www.tinkutara.com/question/Q114563.png)
$${iii}… \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \mathrm{ln}\left(\mathrm{tan}{x}\right){dx} \\ $$$$\mathrm{tan2}\theta\:=\:\frac{\mathrm{2tan}\theta}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \theta} \\ $$$$\mathrm{with}\:\theta\:=\:\frac{\pi}{\mathrm{8}},\:\mathrm{tan}\frac{\pi}{\mathrm{4}}\:=\:\frac{\mathrm{2tan}\frac{\pi}{\mathrm{8}}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}}\:=\:\mathrm{1} \\ $$$$\mathrm{tan}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}+\mathrm{2tan}\frac{\pi}{\mathrm{8}}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{tan}\frac{\pi}{\mathrm{8}}\:=\:\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\mathrm{Now}\:{u}\:=\:\mathrm{tan}{x} \\ $$$${du}\:=\:\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}\right){dx}\:=\:\left(\mathrm{1}+{u}^{\mathrm{2}} \right){dx} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \frac{\mathrm{ln}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \mathrm{ln}{u}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {u}^{\mathrm{2}{n}} {du} \\ $$$$\mathrm{I}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} {u}^{\mathrm{2}{n}} \mathrm{ln}{udu} \\ $$$$\mathrm{I}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\left[\frac{{u}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\mathrm{ln}{u}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} −\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \frac{{u}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}.\frac{{du}}{{u}}\right) \\ $$$$\mathrm{I}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left[\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\right]−\left[\frac{{u}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \right) \\ $$$$\mathrm{I}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left[\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\right]−\left[\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right]\right) \\ $$$$\mathrm{arctan}{x}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\mathrm{I}\:=\:\mathrm{ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{arctan}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{I}\:=\:\frac{\pi}{\mathrm{8}}\mathrm{ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{simplify} \\ $$
Answered by mathmax by abdo last updated on 20/Sep/20
![A =∫_0 ^1 x^2 ln(x)ln(1−x)dx we have (d/dx)ln(1−x) =((−1)/(1−x))=−Σ_(n=0) ^∞ x^n ⇒ln(1−x) =−Σ_(n=0) ^∞ (x^(n+1) /(n+1)) +c (c=0) =−Σ_(n=1) ^∞ (x^n /n) ⇒ A =−∫_0 ^1 x^2 lnx(Σ_(n=1) ^∞ (x^n /n))dx =−Σ_(n=1) ^∞ (1/n) ∫_0 ^1 x^(n+2) ln(x)dx u_n =∫_0 ^1 x^(n+2) ln(x)dx =[(x^(n+3) /(n+3))ln(x)]_0 ^1 −∫_0 ^1 (x^(n+2) /(n+3)) dx =−(1/((n+3)^2 )) ⇒ A =Σ_(n=1) ^∞ (1/(n(n+3)^2 )) let decompose F(x) =(1/(x(x+3)^2 )) ⇒F(x) =(a/x) +(b/(x+3)) +(c/((x+3)^2 )) a=(1/9) , c =−(1/3) ⇒F(x) =(1/(9x)) +(b/(x+3))−(1/(3(x+3)^2 )) lim_(x→+∞) xF(x) =0 =(1/9) +b ⇒b=−(1/9) ⇒ A =lim_(n→+∞) A_n / A_n =Σ_(k=1) ^n (1/(k(k+3)^2 )) =(1/9) Σ_(k=1) ^n (1/k)−(1/9)Σ_(k=1) ^n (1/(k+3))−(1/3) Σ_(k=1) ^n (1/((k+3)^2 )) =(1/9)Σ_(k=1) ^n (1/k)−(1/9) Σ_(k=4) ^(n+3) −(1/3)Σ_(k=4) ^(n+3) (1/k^2 ) =(1/9)(1 +(1/2)+(1/3))−(1/9)((1/(n+1))+(1/(n+2))+(1/(n+3)))−(1/3)(Σ_(k=1) ^(n+3) (1/k^2 )−1−(1/2^2 )−(1/3^2 )) →(1/9)((3/2)+(1/3))−(1/3)×(π^2 /6) +(1/3)(1+(1/4)+(1/9)) =(1/9)(((11)/6))−(π^2 /(18)) +(1/3)((5/4)+(1/9)) =((11)/(54))−(π^2 /(18))+(1/3)(((49)/(36))) =((11)/(54))+((49)/(108))−(π^2 /(18)) ⇒∫_0 ^1 x^2 lnx ln(1−x)dx =((11)/(54))+((49)/(108))−(π^2 /(18))](https://www.tinkutara.com/question/Q114632.png)
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{2}} \mathrm{ln}\left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}\:\:\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\:=\frac{−\mathrm{1}}{\mathrm{1}−\mathrm{x}}=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{n}} \\ $$$$\Rightarrow\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\:=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}\:+\mathrm{c}\:\:\left(\mathrm{c}=\mathrm{0}\right)\:=−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}}\:\Rightarrow \\ $$$$\mathrm{A}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{2}} \mathrm{lnx}\left(\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}}\right)\mathrm{dx}\:=−\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{n}+\mathrm{2}} \:\mathrm{ln}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\mathrm{u}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{n}+\mathrm{2}} \mathrm{ln}\left(\mathrm{x}\right)\mathrm{dx}\:=\left[\frac{\mathrm{x}^{\mathrm{n}+\mathrm{3}} }{\mathrm{n}+\mathrm{3}}\mathrm{ln}\left(\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{n}+\mathrm{2}} }{\mathrm{n}+\mathrm{3}}\:\mathrm{dx} \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{3}\right)^{\mathrm{2}} }\:\Rightarrow\:\mathrm{A}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{3}\right)^{\mathrm{2}} }\:\mathrm{let}\:\mathrm{decompose} \\ $$$$\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{x}\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} }\:\Rightarrow\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{a}}{\mathrm{x}}\:+\frac{\mathrm{b}}{\mathrm{x}+\mathrm{3}}\:+\frac{\mathrm{c}}{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\mathrm{a}=\frac{\mathrm{1}}{\mathrm{9}}\:,\:\mathrm{c}\:=−\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{9x}}\:+\frac{\mathrm{b}}{\mathrm{x}+\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{xF}\left(\mathrm{x}\right)\:=\mathrm{0}\:=\frac{\mathrm{1}}{\mathrm{9}}\:+\mathrm{b}\:\Rightarrow\mathrm{b}=−\frac{\mathrm{1}}{\mathrm{9}}\:\Rightarrow \\ $$$$\mathrm{A}\:=\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{A}_{\mathrm{n}} /\:\:\:\:\:\:\:\:\mathrm{A}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{k}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{9}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}+\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{9}}\:\sum_{\mathrm{k}=\mathrm{4}} ^{\mathrm{n}+\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}}\sum_{\mathrm{k}=\mathrm{4}} ^{\mathrm{n}+\mathrm{3}} \:\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{9}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{n}+\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\left(\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}+\mathrm{3}} \:\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right) \\ $$$$\rightarrow\frac{\mathrm{1}}{\mathrm{9}}\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{3}}×\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\frac{\mathrm{11}}{\mathrm{6}}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{18}}\:+\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}\right)\:=\frac{\mathrm{11}}{\mathrm{54}}−\frac{\pi^{\mathrm{2}} }{\mathrm{18}}+\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{49}}{\mathrm{36}}\right)\:=\frac{\mathrm{11}}{\mathrm{54}}+\frac{\mathrm{49}}{\mathrm{108}}−\frac{\pi^{\mathrm{2}} }{\mathrm{18}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{2}} \mathrm{lnx}\:\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}\:=\frac{\mathrm{11}}{\mathrm{54}}+\frac{\mathrm{49}}{\mathrm{108}}−\frac{\pi^{\mathrm{2}} }{\mathrm{18}} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 20/Sep/20
$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{2}} \mathrm{lnxln}\left(\mathrm{1}−\mathrm{x}\right)\mathrm{dx}\:=\frac{\mathrm{71}}{\mathrm{108}}−\frac{\pi^{\mathrm{2}} }{\mathrm{18}} \\ $$$$ \\ $$
Answered by maths mind last updated on 20/Sep/20
![Ψ_1 ((1/4))=∫_0 ^∞ ((ln(x)x^(−(3/4)) )/(x−1))dx x^(1/4) =t⇒dt=((x^(−(3/4)) dx)/4) =∫_0 ^∞ ((4ln(t^4 ))/(t^4 −1))dt =16∫_0 ^1 ((ln(t))/((t^2 −1)(t^2 +1)))dt =8∫_0 ^1 ((ln(t))/(t^2 −1))−∫_0 ^1 ((ln(t))/(t^2 +1)) =−8∫_0 ^1 Σ_(k≥0) t^(2k) ln(t)dt−8Σ_(m≥0) ∫_0 ^1 (−t^2 )^m ln(t)dt =−8Σ_(k≥0) ∫_0 ^1 t^(2k) ln(t)dt−8Σ(−1)^m ∫_0 ^1 t^(2m) ln(t)dt ∫_0 ^1 t^n ln(t)=[((t^(n+1) ln(t))/(n+1))]_0 ^1 −∫(t^n /(n+1))dt=−(1/((n+1)^2 )) =8Σ_(k≥0) (1/((2k+1)^2 ))+8.Σ(((−1)^k )/((2k+1)^2 )) =8.(3/4)ζ(2)+8.G =π^2 +8G](https://www.tinkutara.com/question/Q114729.png)
$$\Psi_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right){x}^{−\frac{\mathrm{3}}{\mathrm{4}}} }{{x}−\mathrm{1}}{dx} \\ $$$${x}^{\frac{\mathrm{1}}{\mathrm{4}}} ={t}\Rightarrow{dt}=\frac{{x}^{−\frac{\mathrm{3}}{\mathrm{4}}} {dx}}{\mathrm{4}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{4}{ln}\left({t}^{\mathrm{4}} \right)}{{t}^{\mathrm{4}} −\mathrm{1}}{dt} \\ $$$$=\mathrm{16}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt} \\ $$$$=\mathrm{8}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} −\mathrm{1}}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=−\mathrm{8}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}\geqslant\mathrm{0}} {\sum}{t}^{\mathrm{2}{k}} {ln}\left({t}\right){dt}−\mathrm{8}\underset{{m}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−{t}^{\mathrm{2}} \right)^{{m}} {ln}\left({t}\right){dt} \\ $$$$=−\mathrm{8}\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}{k}} {ln}\left({t}\right){dt}−\mathrm{8}\Sigma\left(−\mathrm{1}\right)^{{m}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}{m}} {ln}\left({t}\right){dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}} {ln}\left({t}\right)=\left[\frac{{t}^{{n}+\mathrm{1}} {ln}\left({t}\right)}{{n}+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int\frac{{t}^{{n}} }{{n}+\mathrm{1}}{dt}=−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{8}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{8}.\Sigma\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{8}.\frac{\mathrm{3}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)+\mathrm{8}.{G} \\ $$$$=\pi^{\mathrm{2}} +\mathrm{8}{G} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$