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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=∫_0 ^(π/(12)) ln(tanx)dx.Put tanx =t ⇒dt=(1+t^2 )dx⇒I=∫_0 ^( 2−(√3)) ((ln(t))/(1+t^2 ))dt =∫_0 ^( 2−(√3)) ln(t)(Σ_(k=0) ^(∞) (−1)^k x^(2k) )dt= =Σ_(k=0) ^(∞) (−1)^k ∫_0 ^( 2−(√3)) x^(2k) ln(t)dt=Σ_(k=0) ^(∞) (−1)^k A_k A_k =∫_0 ^( 2−(√3)) x^(2k) ln(t)dt = [_(by parts ) (x^(2k+1) /(2k+1))ln(t)]_0 ^(2−(√3)) −(1/(2k+1))∫_0 ^( 2−(√3)) x^(2k) dt =((ln(2−(√3))(2−(√3))^(2k+1) )/(2k+1))− [(1/(2k+1)).(x^(2k+1) /(2k+1))]_0 ^(2−(√3)) =((ln(2−(√3))(2−(√3))^(2k+1) )/(2k+1))−(1/((2k+1)^2 ))×(2−(√3))^(2k+1) I=Σ_(k=0) ^(∞) (−1)^k A_k =Σ_(k=0) ^(∞) (−1)^k [((ln(2−(√3))(2−(√3))^(2k+1) )/(2k+1))−(1/((2k+1)^2 ))×(2−(√3))^(2k+1) ] =Σ_(k=0) ^(∞) (−1)^k [2−(√3))^(2k+1) ((((2k+1)ln(2−(√3))−1)/((2k+1)^2 ))) Since G=Σ_(k=1) ^(∞) (−1)^k (1/((2k+1)^2 )),we need to prove that: [2−(√3))^(2k+1) ((((2k+1)ln(2−(√3))−1)/((2k+1)^2 )))=(2/3)×(1/((2k+1)^2 )) ⇔[(2k+1)ln(2−(√3))−1]×(2−(√3))^(2k+1) =((−2)/3) ?.....
$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{12}}} \mathrm{ln}\left(\mathrm{tanx}\right)\mathrm{dx}.\mathrm{Put}\:\mathrm{tanx}\:=\mathrm{t} \\ $$$$\Rightarrow\mathrm{dt}=\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{dx}\Rightarrow\mathrm{I}=\int_{\mathrm{0}} ^{\:\mathrm{2}−\sqrt{\mathrm{3}}} \frac{\mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{2}−\sqrt{\mathrm{3}}} \mathrm{ln}\left(\mathrm{t}\right)\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\Sigma}}\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{x}^{\mathrm{2k}} \right)\mathrm{dt}= \\ $$$$=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\Sigma}}\left(−\mathrm{1}\right)^{\mathrm{k}} \int_{\mathrm{0}} ^{\:\mathrm{2}−\sqrt{\mathrm{3}}} \mathrm{x}^{\mathrm{2k}} \mathrm{ln}\left(\mathrm{t}\right)\mathrm{dt}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\Sigma}}\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{A}_{\mathrm{k}} \\ $$$$\mathrm{A}_{\mathrm{k}} =\int_{\mathrm{0}} ^{\:\mathrm{2}−\sqrt{\mathrm{3}}} \mathrm{x}^{\mathrm{2k}} \mathrm{ln}\left(\mathrm{t}\right)\mathrm{dt}\:\:\:\underset{\mathrm{by}\:\mathrm{parts}\:\:\:\:\:} {=\:\:\:\:\:\left[}\frac{\mathrm{x}^{\mathrm{2k}+\mathrm{1}} }{\mathrm{2k}+\mathrm{1}}\mathrm{ln}\left(\mathrm{t}\right)\right]_{\mathrm{0}} ^{\mathrm{2}−\sqrt{\mathrm{3}}} −\frac{\mathrm{1}}{\mathrm{2k}+\mathrm{1}}\int_{\mathrm{0}} ^{\:\mathrm{2}−\sqrt{\mathrm{3}}} \mathrm{x}^{\mathrm{2k}} \mathrm{dt} \\ $$$$=\frac{\mathrm{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2k}+\mathrm{1}} }{\mathrm{2k}+\mathrm{1}}−\:\:\left[\frac{\mathrm{1}}{\mathrm{2k}+\mathrm{1}}.\frac{\mathrm{x}^{\mathrm{2k}+\mathrm{1}} }{\mathrm{2k}+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$=\frac{\mathrm{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2k}+\mathrm{1}} }{\mathrm{2k}+\mathrm{1}}−\frac{\mathrm{1}}{\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} }×\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2k}+\mathrm{1}} \\ $$$$\mathrm{I}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\Sigma}}\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{A}_{\mathrm{k}} =\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\Sigma}}\left(−\mathrm{1}\right)^{\mathrm{k}} \left[\frac{\mathrm{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2k}+\mathrm{1}} }{\mathrm{2k}+\mathrm{1}}−\frac{\mathrm{1}}{\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} }×\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2k}+\mathrm{1}} \right] \\ $$$$=\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\Sigma}}\left(−\mathrm{1}\right)^{\mathrm{k}} \left[\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2k}+\mathrm{1}} \left(\frac{\left(\mathrm{2k}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)−\mathrm{1}}{\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$\mathrm{Since}\:\mathrm{G}=\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\Sigma}}\left(−\mathrm{1}\right)^{\mathrm{k}} \frac{\mathrm{1}}{\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} },\mathrm{we}\:\mathrm{need}\: \\ $$$$\mathrm{to}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\left[\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2k}+\mathrm{1}} \left(\frac{\left(\mathrm{2k}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)−\mathrm{1}}{\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} }\right)=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{1}}{\left(\mathrm{2k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Leftrightarrow\left[\left(\mathrm{2}\boldsymbol{\mathrm{k}}+\mathrm{1}\right)\boldsymbol{\mathrm{ln}}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)−\mathrm{1}\right]×\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}\boldsymbol{\mathrm{k}}+\mathrm{1}} \\ $$$$=\frac{−\mathrm{2}}{\mathrm{3}}\:?….. \\ $$
Commented by mathdave last updated on 20/Aug/20
u try iwill send my solution as well
$${u}\:{try}\:{iwill}\:{send}\:{my}\:{solution}\:{as}\:{well} \\ $$
Answered by mathdave last updated on 20/Aug/20
solution to show that ∫_0 ^(π/(12)) ln(tanx)dx=−(2/3)G recall to lemma (2) theorem which state ∫_0 ^(π/4) ln(tanx)dx=2∫_0 ^(π/(12)) ln(tan3x)dy by let y=3x and dx=(1/3)dy (2/3)∫_0 ^(π/4) ln(tany)dy=∫_0 ^(π/4) ln(tanx)dx let t=tany and dy=(dt/(1+t^2 )) ∵∫_0 ^(π/4) ln(tanx)dx=(2/3)∫_0 ^1 ((lnt)/(1+t^2 ))dt but the series of (1/(1+t^2 ))=(−1)^n Σ_(n=0) ^∞ t^(2n) =(2/3)∫_0 ^1 (−1)^n Σ_(n=0) ^∞ t^(2n) lnt using feynmann trick (d/da)∣_(a=1) I=(2/3)•(−1)^n Σ_(n=0) ^∞ ∫_0 ^1 t^(2n) .t^(a−1) dt=(2/3)•(−1)^n Σ_(n=0) ^∞ (1/((2n+a))) (d/da)∣_(a=1) I=−(2/3)•Σ_(n=0) ^∞ (((−1)^n )/((2n+a)^2 ))=−(2/3)•Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 )) but Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 ))=G (call catalan constant) ∵∫_0 ^(π/(12)) ln(tanx)dx=−(2/3)G Q.E.D by mathdave (20/08/2020)
$${solution} \\ $$$${to}\:{show}\:{that}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{12}}} \mathrm{ln}\left(\mathrm{tan}{x}\right){dx}=−\frac{\mathrm{2}}{\mathrm{3}}{G} \\ $$$${recall}\:{to}\:{lemma}\:\left(\mathrm{2}\right)\:{theorem}\:{which}\:{state}\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tan}{x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{12}}} \mathrm{ln}\left(\mathrm{tan3}{x}\right){dy} \\ $$$${by}\:{let}\:{y}=\mathrm{3}{x}\:\:\:{and}\:{dx}=\frac{\mathrm{1}}{\mathrm{3}}{dy} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tan}{y}\right){dy}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tan}{x}\right){dx} \\ $$$${let}\:\:{t}=\mathrm{tan}{y}\:\:{and}\:\:{dy}=\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\because\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tan}{x}\right){dx}=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$${but}\:{the}\:{series}\:{of}\:\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }=\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{t}^{\mathrm{2}{n}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{t}^{\mathrm{2}{n}} \mathrm{ln}{t} \\ $$$${using}\:{feynmann}\:{trick} \\ $$$$\:\frac{{d}}{{da}}\mid_{{a}=\mathrm{1}} {I}=\frac{\mathrm{2}}{\mathrm{3}}\bullet\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}{n}} .{t}^{{a}−\mathrm{1}} {dt}=\frac{\mathrm{2}}{\mathrm{3}}\bullet\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+{a}\right)} \\ $$$$\frac{{d}}{{da}}\mid_{{a}=\mathrm{1}} {I}=−\frac{\mathrm{2}}{\mathrm{3}}\bullet\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+{a}\right)^{\mathrm{2}} }=−\frac{\mathrm{2}}{\mathrm{3}}\bullet\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${but}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }={G}\:\:\:\left({call}\:{catalan}\:{constant}\right) \\ $$$$\because\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{12}}} \mathrm{ln}\left(\mathrm{tan}{x}\right){dx}=−\frac{\mathrm{2}}{\mathrm{3}}{G}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Q}.{E}.{D} \\ $$$${by}\:{mathdave}\:\left(\mathrm{20}/\mathrm{08}/\mathrm{2020}\right) \\ $$
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
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}.\mathrm{I}\:\mathrm{think}\:\mathrm{that}\:\mathrm{just}\:\mathrm{integrating}\: \\ $$$$\mathrm{by}\:\mathrm{parts}\:\mathrm{obtain}\:\mathrm{also}\:\mathrm{above}\:\mathrm{result}\: \\ $$$$\mathrm{but}\:\mathrm{don}'\mathrm{t}\:\mathrm{need}\:\mathrm{to}\:\mathrm{use}\:\mathrm{feynmann}'\mathrm{s}\:\mathrm{trick}\:, \\ $$$$\mathrm{it}\:\mathrm{seems}\:\mathrm{becomes}\:\mathrm{trouble}\:\mathrm{more} \\ $$$$\mathrm{However},\mathrm{Sir}'\mathrm{s}\:\mathrm{idea}\:\mathrm{can}\:\mathrm{using}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\mathrm{other}\:\mathrm{problems} \\ $$

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