Question Number 209393 by Shrodinger last updated on 08/Jul/24
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left({tanx}\right)}{\mathrm{1}+{tanx}}{dx} \\ $$
Commented by Frix last updated on 09/Jul/24
$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$
Commented by Shrodinger last updated on 09/Jul/24
$${Yes}\:{sir}. \\ $$
Commented by Frix last updated on 09/Jul/24
![∫_0 ^(π/2) ((ln tan x)/(1+tan x))dx =^(t=tan x) ∫_0 ^∞ ((ln t)/((t+1)(t^2 +1)))dt= =∫_0 ^1 ((ln t)/((t+1)(t^2 +1)))dt+∫_1 ^∞ ((ln t)/((t+1)(t^2 +1)))dt= [∫_1 ^∞ ((ln t)/((t+1)(t^2 +1)))dt =^(u=(1/t)) −∫_0 ^1 ((uln u)/((u+1)(u^2 +1)))du] =∫_0 ^1 (((1−t)ln t)/((t+1)(t^2 +1)))dt =_(parts) ^(by) u′=((1−t)/((t+1)(t^2 +1))) → u=ln (t+1) −((ln (t^2 +1))/2) v=ln t → v′=(1/t) =[(ln (t+1) −((ln (t^2 +1))/2))ln t]_0 ^1 − {=0} −∫_0 ^1 ((ln (t+1))/t)dt+(1/2)∫_0 ^1 ((ln (t^2 +1))/t)dt= =[Li_2 (−t)−((Li_2 (−t^2 ))/4)]_0 ^1 =((3Li_2 (−1))/4)=−(π^2 /(16))](https://www.tinkutara.com/question/Q209402.png)
$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{ln}\:\mathrm{tan}\:{x}}{\mathrm{1}+\mathrm{tan}\:{x}}{dx}\:\overset{{t}=\mathrm{tan}\:{x}} {=}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{ln}\:{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}= \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}+\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{\mathrm{ln}\:{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}= \\ $$$$\:\:\:\:\:\left[\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{\mathrm{ln}\:{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}\:\overset{{u}=\frac{\mathrm{1}}{{t}}} {=}\:−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{u}\mathrm{ln}\:{u}}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{du}\right] \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\left(\mathrm{1}−{t}\right)\mathrm{ln}\:{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}\:\underset{\mathrm{parts}} {\overset{\mathrm{by}} {=}} \\ $$$${u}'=\frac{\mathrm{1}−{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}\:\rightarrow\:{u}=\mathrm{ln}\:\left({t}+\mathrm{1}\right)\:−\frac{\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}} \\ $$$${v}=\mathrm{ln}\:{t}\:\rightarrow\:{v}'=\frac{\mathrm{1}}{{t}} \\ $$$$=\left[\left(\mathrm{ln}\:\left({t}+\mathrm{1}\right)\:−\frac{\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}}\right)\mathrm{ln}\:{t}\right]_{\mathrm{0}} ^{\mathrm{1}} −\:\:\:\:\:\left\{=\mathrm{0}\right\} \\ $$$$\:\:\:\:\:−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\left({t}+\mathrm{1}\right)}{{t}}{dt}+\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{{t}}{dt}= \\ $$$$=\left[\mathrm{Li}_{\mathrm{2}} \:\left(−{t}\right)−\frac{\mathrm{Li}_{\mathrm{2}} \:\left(−{t}^{\mathrm{2}} \right)}{\mathrm{4}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{3Li}_{\mathrm{2}} \:\left(−\mathrm{1}\right)}{\mathrm{4}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$
Answered by mathmax last updated on 09/Jul/24
![I=_(tanx=t) ∫_0 ^∞ ((ln(t))/(1+t))(dt/(1+t^2 )) =∫_0 ^1 ((lnt)/((t+1)(t^2 +1)))dt +∫_1 ^∞ ((lnt)/((t+1)(t^2 +1)))(→t=(1/u)) =∫_0 ^1 ((lnt)/((t+1)(t^2 +1)))dt−∫_0 ^1 ((−lnu)/(((1/u)+1)((1/u^2 )+1)))(((−du)/u^2 )) =∫_0 ^1 ((lnt)/((t+1)(t^2 +1)))dt−∫_0 ^1 ((tlnt)/((t+1)(t^2 +1)))dt =∫_0 ^1 (((1−t)lnt)/((t+1)(t^2 +1)))dt f(t)=((1−t)/((t+1)(t^2 +1)))=(a/(t+1))+((bt+c)/(t^2 +1)) a=1 lim_∞ tf(t)=0=a+b ⇒b=−1 f(0)=1=a+c ⇒c=0 ⇒ f(t)=(1/(t+1))−(t/(t^2 +1)) ⇒ I=∫_0 ^1 ((1/(t+1))−(t/(t^2 +1)))lnt dt =∫_0 ^1 ((lnt)/(1+t))dt−∫_0 ^1 ((tlnt)/(1+t^2 ))dt we have ∫_0 ^1 ((lnt)/(1+t))dt =∫_0 ^1 lntΣ_(n=0) ^∞ (−1)^n t^n dt =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 t^n lnt dt=Σ(−1)^n u_n u_n =[(t^(n+1) /(n+1))lnt]_0 ^1 −∫_0 ^1 (t^(n+1) /(n+1))(dt/t) =−(1/((n+1)^2 )) ⇒∫_0 ^1 ((lnt)/(1+t))dt=−Σ_(n=0) ^∞ (((−1)^n )/((n+1)^2 )) =Σ_(n=1) ^∞ (((−1)^n )/n^2 )=η(2)=(2^(1−2) −1)ξ(2) =−(1/2).(π^2 /6)=−(π^2 /(12)) ∫_0 ^1 ((tlnt)/(1+t^2 ))dt=∫_0 ^1 tlntΣ_(n=0) ^∞ (−1)^n t^(2n) dt =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 t^(2n+1) lnt dt=Σ(−1)^n v_n v_n =[(1/(2n+2))t^(2n+2) lnt]_0 ^1 −∫_0 ^1 (t^(2n+2) /(2n+2))(dt/t) =−(1/((2n+2)^2 ))=−(1/(4(n+1)^2 )) ⇒ ∫_0 ^1 ((tlnt)/(1+t^2 ))dt=−(1/4)Σ_(n=0) ^∞ (((−1)^n )/((n+1)^2 )) =(1/4)Σ_(n=1) ^∞ (((−1)^n )/n^2 )=(1/4)(−(π^2 /(12))) ⇒ I=−(π^2 /(12))+(1/4).(π^2 /(12))=((1/4)−1)(π^2 /(12)) =−(3/4)×(π^2 /(12))=−(π^2 /(16)) ★I=−(π^2 /(16))★](https://www.tinkutara.com/question/Q209421.png)
$${I}=_{{tanx}={t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({t}\right)}{\mathrm{1}+{t}}\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}\:+\int_{\mathrm{1}} ^{\infty} \frac{{lnt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}\left(\rightarrow{t}=\frac{\mathrm{1}}{{u}}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{lnu}}{\left(\frac{\mathrm{1}}{{u}}+\mathrm{1}\right)\left(\frac{\mathrm{1}}{{u}^{\mathrm{2}} }+\mathrm{1}\right)}\left(\frac{−{du}}{{u}^{\mathrm{2}} }\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tlnt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{t}\right){lnt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt} \\ $$$${f}\left({t}\right)=\frac{\mathrm{1}−{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{a}}{{t}+\mathrm{1}}+\frac{{bt}+{c}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$${a}=\mathrm{1}\:\:\:{lim}_{\infty} {tf}\left({t}\right)=\mathrm{0}={a}+{b}\:\Rightarrow{b}=−\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1}={a}+{c}\:\Rightarrow{c}=\mathrm{0}\:\Rightarrow \\ $$$${f}\left({t}\right)=\frac{\mathrm{1}}{{t}+\mathrm{1}}−\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{{t}+\mathrm{1}}−\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\right){lnt}\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\mathrm{1}+{t}}{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tlnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\mathrm{1}+{t}}{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {lnt}\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {t}^{{n}} {dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}} {lnt}\:{dt}=\Sigma\left(−\mathrm{1}\right)^{{n}} {u}_{{n}} \\ $$$${u}_{{n}} =\left[\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{lnt}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\frac{{dt}}{{t}} \\ $$$$=−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\mathrm{1}+{t}}{dt}=−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }=\eta\left(\mathrm{2}\right)=\left(\mathrm{2}^{\mathrm{1}−\mathrm{2}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tlnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} {tlnt}\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {t}^{\mathrm{2}{n}} {dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}+\mathrm{1}} {lnt}\:{dt}=\Sigma\left(−\mathrm{1}\right)^{{n}} {v}_{{n}} \\ $$$${v}_{{n}} =\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}{t}^{\mathrm{2}{n}+\mathrm{2}} {lnt}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}{n}+\mathrm{2}} }{\mathrm{2}{n}+\mathrm{2}}\frac{{dt}}{{t}} \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{2}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tlnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=−\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right)\:\Rightarrow \\ $$$${I}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{4}}.\frac{\pi^{\mathrm{2}} }{\mathrm{12}}=\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}\right)\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{4}}×\frac{\pi^{\mathrm{2}} }{\mathrm{12}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\bigstar{I}=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\bigstar \\ $$$$ \\ $$