Question Number 124827 by mnjuly1970 last updated on 06/Dec/20
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:....\:{nice}\:\:\:{calculus}\:... \\ $$$$\:\:\:\:\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{log}\left(\mathrm{1}+{tan}\left({x}\right)\right)}{{tan}\left({x}\right)}{dx}=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}}\:\checkmark \\ $$$$ \\ $$
Answered by mindispower last updated on 06/Dec/20
![f(t)=∫_0 ^∞ ((ln(1+t.tan(x)))/(tan(x)))dx q=f(1),q f′(t)=∫_0 ^(π/2) (dx/(1+t.tan(x)))=∫_0 ^(π/2) ((cos(x))/(cos(x)+tsin(x)))dx cos(x)=α(cos(x)+tsin(x))+β(−sin(x)+tcos(x) ⇒α+βt=1 αt−β=0 α=(1/(1+t^2 )),β=(t/(1+t^2 )) f′(t)=(1/(1+t^2 ))∫_0 ^(π/2) dx+(t/(1+t^2 ))∫_0 ^(π/2) ((−sin(x)+tcos(x))/(cos(x)+tsin(x)))dx =(π/(2(1+t^2 )))+[(t/(1+t^2 ))ln(cos(x)+tsin(x))]_0 ^(π/2) =(π/(2(1+t^2 )))+((tln(t))/(1+t^2 )) f(0)=∫_0 ^(π/2) ((ln(1+otan(x)))/(tan(x)))dx=0 q=∫_0 ^1 (π/(2(1+t^2 )))+((tln(t))/(1+t^2 ))dt =(π/2)tan^(−1) (1)+Σ_(k≥0) ∫_0 ^1 (−1)^k t^(2k+1) ln(t)dt =(π/2).(π/4)+Σ_(k≥0) (((−1)^(k+1) )/((2k+2)^2 ))=(π/8)−(1/4)Σ_(k≥0) (((−1)^k )/((k+1)^2 )) =(π/8)+(1/4)(Σ_(k≥0) (1/((2k+2)^2 ))−(1/4)Σ_(k≥0) (1/((2k+1)^2 ))) =(π^2 /8)+(1/(16))ζ(2)−(3/(16))ζ(2) =(π^2 /8)−(1/8).(π^2 /6)=((5π^2 )/(48))=∫_0 ^(π/2) ((ln(1+tan(x)))/(tan(x)))dx](Q124855.png)
$${f}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{t}.{tan}\left({x}\right)\right)}{{tan}\left({x}\right)}{dx} \\ $$$${q}={f}\left(\mathrm{1}\right),{q} \\ $$$${f}'\left({t}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+{t}.{tan}\left({x}\right)}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({x}\right)}{{cos}\left({x}\right)+{tsin}\left({x}\right)}{dx} \\ $$$${cos}\left({x}\right)=\alpha\left({cos}\left({x}\right)+{tsin}\left({x}\right)\right)+\beta\left(−{sin}\left({x}\right)+{tcos}\left({x}\right)\right. \\ $$$$\Rightarrow\alpha+\beta{t}=\mathrm{1} \\ $$$$\alpha{t}−\beta=\mathrm{0} \\ $$$$\alpha=\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} },\beta=\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${f}'\left({t}\right)=\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}+\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{−{sin}\left({x}\right)+{tcos}\left({x}\right)}{{cos}\left({x}\right)+{tsin}\left({x}\right)}{dx} \\ $$$$=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}+\left[\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{ln}\left({cos}\left({x}\right)+{tsin}\left({x}\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}+\frac{{tln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${f}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left(\mathrm{1}+{otan}\left({x}\right)\right)}{{tan}\left({x}\right)}{dx}=\mathrm{0} \\ $$$${q}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\pi}{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}+\frac{{tln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\pi}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)+\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} {t}^{\mathrm{2}{k}+\mathrm{1}} {ln}\left({t}\right){dt} \\ $$$$=\frac{\pi}{\mathrm{2}}.\frac{\pi}{\mathrm{4}}+\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{\left(\mathrm{2}{k}+\mathrm{2}\right)^{\mathrm{2}} }=\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\pi}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{4}}\left(\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{2}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{16}}\zeta\left(\mathrm{2}\right)−\frac{\mathrm{3}}{\mathrm{16}}\zeta\left(\mathrm{2}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left(\mathrm{1}+{tan}\left({x}\right)\right)}{{tan}\left({x}\right)}{dx} \\ $$
Commented by mnjuly1970 last updated on 06/Dec/20
$${again}\:\:{thank}\:{you} \\ $$$${sir}\:{mindspower} \\ $$$$ \\ $$
Commented by mindispower last updated on 06/Dec/20
$${always}\:{pleasur}\:{sir}\: \\ $$
Answered by mathmax by abdo last updated on 06/Dec/20
![let f(a)=∫_0 ^(π/2) ((log(1+atanx))/(tanx))dx with a>0⇒f^′ (a)=∫_0 ^(π/2) ((tanx)/((1+atanx)tanx))dx =∫_0 ^(π/2) (dx/(1+atanx))=_(tanx=t) ∫_0 ^∞ (dt/((1+t^2 )(1+at))) =_(at=z) ∫_0 ^∞ (dz/(a(1+(z^2 /a^2 ))(z+1))) =a ∫_0 ^∞ (dz/((z^2 +a^2 )(z+1))) let decompose F(z)=(1/((z+1)(z^2 +a^2 ))) F(z)=(α/(z+1)) +((βz+n)/(z^2 +a^2 )) α =(1/(a^2 +1)) ,lim_(z→+∞) zF(z)=0=α+β ⇒β=−(1/(a^2 +1)) F(0)=(1/a^2 ) =α+(n/a^2 ) ⇒1=a^2 α+n ⇒n=1−(a^2 /(a^2 +1))=(1/(a^2 +1)) ⇒ F(z)=(1/((a^2 +1)(z+1))) +((−(1/(a^2 +1))z+(1/(a^2 +1)))/(z^2 +a^2 )) ⇒ ∫_0 ^∞ F(z)dz =(1/(a^2 +1))∫_0 ^∞ (dz/(z+1))−(1/(2(a^2 +1)))∫_0 ^∞ ((2z−2)/(z^2 +a^2 ))dz =(1/(a^2 +1))∫_0 ^∞ ((1/(z+1))−(1/2)×((2z)/(z^2 +1)))dz +(1/(a^2 +1))∫_0 ^∞ (dz/(z^2 +a^2 ))(→z=au) =(1/(a^2 +1))[ln∣((z+1)/( (√(z^2 +1))))∣]_0 ^∞ +(1/(a^2 +1))∫_0 ^∞ ((adu)/(a^2 (u^2 +1))) =(1/(a(a^2 +1)))×(π/2) ⇒f^′ (a)=(π/(2(1+a^2 ))) ⇒f(a)=(π/2) arctan(a)+C f(0)=C ⇒f(a)=(π/2) arctan(a) and ∫_0 ^(π/2) ((log(1+tanx))/(tanx))dx =(π/2)arctan(1) =(π/2).(π/4)=(π^2 /8)](Q124882.png)
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{log}\left(\mathrm{1}+\mathrm{atanx}\right)}{\mathrm{tanx}}\mathrm{dx}\:\mathrm{with}\:\mathrm{a}>\mathrm{0}\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{tanx}}{\left(\mathrm{1}+\mathrm{atanx}\right)\mathrm{tanx}}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{atanx}}=_{\mathrm{tanx}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{at}\right)}\:=_{\mathrm{at}=\mathrm{z}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dz}}{\mathrm{a}\left(\mathrm{1}+\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\right)\left(\mathrm{z}+\mathrm{1}\right)} \\ $$$$=\mathrm{a}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dz}}{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)\left(\mathrm{z}+\mathrm{1}\right)}\:\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}+\mathrm{1}\right)\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)} \\ $$$$\mathrm{F}\left(\mathrm{z}\right)=\frac{\alpha}{\mathrm{z}+\mathrm{1}}\:+\frac{\beta\mathrm{z}+\mathrm{n}}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} } \\ $$$$\alpha\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\:,\mathrm{lim}_{\mathrm{z}\rightarrow+\infty} \mathrm{zF}\left(\mathrm{z}\right)=\mathrm{0}=\alpha+\beta\:\Rightarrow\beta=−\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\: \\ $$$$\mathrm{F}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\:=\alpha+\frac{\mathrm{n}}{\mathrm{a}^{\mathrm{2}} }\:\Rightarrow\mathrm{1}=\mathrm{a}^{\mathrm{2}} \alpha+\mathrm{n}\:\Rightarrow\mathrm{n}=\mathrm{1}−\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{z}+\mathrm{1}\right)}\:+\frac{−\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}\mathrm{z}+\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{F}\left(\mathrm{z}\right)\mathrm{dz}\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dz}}{\mathrm{z}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}\right)}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2z}−\mathrm{2}}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} }\mathrm{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{1}}{\mathrm{z}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2z}}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\right)\mathrm{dz}\:\:+\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dz}}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} }\left(\rightarrow\mathrm{z}=\mathrm{au}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\left[\mathrm{ln}\mid\frac{\mathrm{z}+\mathrm{1}}{\:\sqrt{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}}\mid\right]_{\mathrm{0}} ^{\infty} \:+\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{adu}}{\mathrm{a}^{\mathrm{2}} \left(\mathrm{u}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}\right)}×\frac{\pi}{\mathrm{2}}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi}{\mathrm{2}}\:\mathrm{arctan}\left(\mathrm{a}\right)+\mathrm{C} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{C}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi}{\mathrm{2}}\:\mathrm{arctan}\left(\mathrm{a}\right)\:\mathrm{and}\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{log}\left(\mathrm{1}+\mathrm{tanx}\right)}{\mathrm{tanx}}\mathrm{dx}\:=\frac{\pi}{\mathrm{2}}\mathrm{arctan}\left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{2}}.\frac{\pi}{\mathrm{4}}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$
Commented by mnjuly1970 last updated on 06/Dec/20
$${thank}\:{you}\:{so}\:{much}\:{sir}\:{max}... \\ $$
Commented by mathmax by abdo last updated on 06/Dec/20
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir} \\ $$