Question Number 113589 by mathdave last updated on 14/Sep/20
![prove the following integral ∫_0 ^(π/2) (x^2 /(sinx))dx=2πG−(7/2)ζ(3) ∫_0 ^1 ln[((x+(√(1−x^2 )))/(x−(√(1−x^2 ))))]^2 (x/(1−x^2 ))dx=((.π^2 )/2) ∫_0 ^∞ ((e^x lnx)/(1+e^(2x) ))dx=(π/2)ln[((Γ((3/4)))/(Γ((1/4))))(√(2π))]](https://www.tinkutara.com/question/Q113589.png)
$${prove}\:{the}\:{following}\:{integral} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}^{\mathrm{2}} }{\mathrm{sin}{x}}{dx}=\mathrm{2}\pi{G}−\frac{\mathrm{7}}{\mathrm{2}}\zeta\left(\mathrm{3}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left[\frac{{x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right]^{\mathrm{2}} \frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\frac{.\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{{x}} \mathrm{ln}{x}}{\mathrm{1}+{e}^{\mathrm{2}{x}} }{dx}=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left[\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\sqrt{\mathrm{2}\pi}\right] \\ $$
Answered by maths mind last updated on 15/Sep/20
![∫_0 ^(π/2) (x^2 /(sin(x)))dx=∫_0 ^(π/2) (x^2 /(2tg((x/2))))(1+tg^2 ((x/2)))dx=∫_0 ^(π/2) x^2 ((d(tg((x/2))))/(tg((x/2)))) IBP=[x^2 ln(tg((x/2)))]_0 ^(π/2) −∫_0 ^(π/2) 2xln(tg((x/2)))dx =−2∫_0 ^(π/2) xln(tg((x/2)))dx we use fourier serie of ln(tg(x)),x∈]0,(π/2)[ ln(tg(x))=−2Σ_(k≥0) ((cos(2(2k+1)x))/(2k+1)) we get =−2∫_0 ^(π/2) xΣ_(k≥0) .−2((cos((2k+1)x))/(2k+1))dx =4Σ_(k≥0) ∫_0 ^(π/2) x((cos((2k+1)x))/(2k+1)),∫_0 ^(π/2) x((cos(2k+1)x)/(2k+1))dx =[x((sin(2k+1)x)/((2k+1)^2 ))]_0 ^(π/2) −∫_0 ^(π/2) ((sin(2k+1)x)/((2k+1)^2 ))dx =(π/2).(((−1)^k )/((2k+1)^2 ))+[((cos(2k+1)x)/((2k+1)^3 ))]_0 ^(π/2) =(π/2).(((−1)^k )/((2k+1)^2 ))−(1/((2k+1)^3 )) 4Σ_(k≥0) ∫_0 ^(π/2) x((cos((2k+1)x))/(2k+1))=4Σ_(k≥0) ((π/2).(((−1)^k )/((2k+1)^2 ))−(1/((2k+1)^3 ))) =2πΣ(((−1)^k )/((2k+1)^2 ))−4Σ_(k≥0) (1/((2k+1)^3 )) G=Σ_(k≥0) (((−1)^k )/((2k+1)^2 )),ζ(3)=Σ(1/k^3 )=Σ_(k=0) ^∞ ((1/((2k+2)^3 ))+(1/((2k+1)^3 ))) =(1/8)Σ_(k≥0) (1/((k+1)^3 ))+Σ(1/((2k+1)^3 ))⇒Σ_(k≥0) (1/((2k+1)^3 ))=(7/8)ζ(3) we get So 2πG−4.(7/8)ζ(3)=2πG−(7/2)ζ(3)](https://www.tinkutara.com/question/Q113734.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}^{\mathrm{2}} }{{sin}\left({x}\right)}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}^{\mathrm{2}} }{\mathrm{2}{tg}\left(\frac{{x}}{\mathrm{2}}\right)}\left(\mathrm{1}+{tg}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right){dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}^{\mathrm{2}} \frac{{d}\left({tg}\left(\frac{{x}}{\mathrm{2}}\right)\right)}{{tg}\left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$${IBP}=\left[{x}^{\mathrm{2}} {ln}\left({tg}\left(\frac{{x}}{\mathrm{2}}\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{xln}\left({tg}\left(\frac{{x}}{\mathrm{2}}\right)\right){dx} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {xln}\left({tg}\left(\frac{{x}}{\mathrm{2}}\right)\right){dx} \\ $$$$\left.{we}\:{use}\:{fourier}\:{serie}\:{of}\:{ln}\left({tg}\left({x}\right)\right),{x}\in\right]\mathrm{0},\frac{\pi}{\mathrm{2}}\left[\right. \\ $$$${ln}\left({tg}\left({x}\right)\right)=−\mathrm{2}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{cos}\left(\mathrm{2}\left(\mathrm{2}{k}+\mathrm{1}\right){x}\right)}{\mathrm{2}{k}+\mathrm{1}} \\ $$$${we}\:{get} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\underset{{k}\geqslant\mathrm{0}} {\sum}.−\mathrm{2}\frac{{cos}\left(\left(\mathrm{2}{k}+\mathrm{1}\right){x}\right)}{\mathrm{2}{k}+\mathrm{1}}{dx} \\ $$$$=\mathrm{4}\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\frac{{cos}\left(\left(\mathrm{2}{k}+\mathrm{1}\right){x}\right)}{\mathrm{2}{k}+\mathrm{1}},\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\frac{{cos}\left(\mathrm{2}{k}+\mathrm{1}\right){x}}{\mathrm{2}{k}+\mathrm{1}}{dx} \\ $$$$=\left[{x}\frac{{sin}\left(\mathrm{2}{k}+\mathrm{1}\right){x}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\left(\mathrm{2}{k}+\mathrm{1}\right){x}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\pi}{\mathrm{2}}.\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }+\left[\frac{{cos}\left(\mathrm{2}{k}+\mathrm{1}\right){x}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{3}} }\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi}{\mathrm{2}}.\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\mathrm{4}\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\frac{{cos}\left(\left(\mathrm{2}{k}+\mathrm{1}\right){x}\right)}{\mathrm{2}{k}+\mathrm{1}}=\mathrm{4}\underset{{k}\geqslant\mathrm{0}} {\sum}\left(\frac{\pi}{\mathrm{2}}.\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{3}} }\right) \\ $$$$=\mathrm{2}\pi\Sigma\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{4}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${G}=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} },\zeta\left(\mathrm{3}\right)=\Sigma\frac{\mathrm{1}}{{k}^{\mathrm{3}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{2}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{3}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{3}} }+\Sigma\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{3}} }\Rightarrow\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right) \\ $$$${we}\:{get}\:{So} \\ $$$$\mathrm{2}\pi{G}−\mathrm{4}.\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)=\mathrm{2}\pi{G}−\frac{\mathrm{7}}{\mathrm{2}}\zeta\left(\mathrm{3}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$
Answered by maths mind last updated on 15/Sep/20
![∫_0 ^1 ln[((x+(√(1−x^2 )))/(x−(√(1−x^2 ))))]^2 (x/(1−x^2 ))dx,x=sin(t) =∫_0 ^(π/2) ln(((sin(t)+cos(t))/(sin(t)−cos(t))))^2 ((sin(t)cos(t))/(cos^2 (t)))dt =∫_0 ^(π/2) ln(((1+tg(t))/(−1+tg(t))))^2 tg(t)dt =∫_0 ^(π/4) 2ln(((tg(t)+1)/(1−tg(t))))tg(t)dt+∫_(π/4) ^(π/2) 2ln(((tg(t)+1)/(tg(t)−1)))tg(t)dt =2∫_0 ^(π/4) tg(t)ln(tg((π/4)+t))dt−2∫_(π/4) ^(π/2) tg(t)ln(((tg(t)−1)/(tg(t)+1)))dt =−2∫_0 ^(π/4) ln(tg(t))tg((π/4)−t)dt−2∫_(π/4) ^(π/2) tg(t)ln(tg(t−(π/4)))dt_(t−(π/4)=r) =−2∫_0 ^(π/4) ln(tg(t))((1+tg(t))/(1+tg(t)))dt−2∫_0 ^(π/4) tg(r+(π/4))ln(tg(r))dr =−2∫_0 ^(π/4) ln(tg(t))((1−tg(t))/(1+tg(t)))dt−2∫_0 ^(π/4) ln(tg(t))((1+tg(t))/(1−tg(t)))dt =−2∫_0 ^(π/4) ln(tg(t))[((1−tg(t))/(1+tg(t)))+((1+tg(t))/(1−tg(t)))]dt =−2∫_0 ^(π/4) [((2+2tg^2 (t))/(1−tg^2 (t)))]ln(tg(t))dt tg(t)=x⇒ =−4∫_0 ^1 ((ln(x))/(1−x^2 ))dx=−4∫_0 ^1 ln(x)Σ_(k≥0) x^(2k) dx =−4Σ_(k≥0) ∫_0 ^1 x^(2k) ln(x)=−4Σ_(k≥0) {[(x^(2k+1) /(2k+1))ln(x)]_0 ^1 −(1/(2k+1))∫_0 ^1 x^(2k) dx =4Σ_(k≥0) (1/((2k+1)^2 ))=4{ζ(2)−(1/4)ζ(2)}=3ζ(2)=3.(π^2 /6)=(π^2 /2)](https://www.tinkutara.com/question/Q113869.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left[\frac{{x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right]^{\mathrm{2}} \frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }{dx},{x}={sin}\left({t}\right) \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{{sin}\left({t}\right)+{cos}\left({t}\right)}{{sin}\left({t}\right)−{cos}\left({t}\right)}\right)^{\mathrm{2}} \frac{{sin}\left({t}\right){cos}\left({t}\right)}{{cos}^{\mathrm{2}} \left({t}\right)}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}+{tg}\left({t}\right)}{−\mathrm{1}+{tg}\left({t}\right)}\right)^{\mathrm{2}} {tg}\left({t}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{2}{ln}\left(\frac{{tg}\left({t}\right)+\mathrm{1}}{\mathrm{1}−{tg}\left({t}\right)}\right){tg}\left({t}\right){dt}+\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{ln}\left(\frac{{tg}\left({t}\right)+\mathrm{1}}{{tg}\left({t}\right)−\mathrm{1}}\right){tg}\left({t}\right){dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tg}\left({t}\right){ln}\left({tg}\left(\frac{\pi}{\mathrm{4}}+{t}\right)\right){dt}−\mathrm{2}\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} {tg}\left({t}\right){ln}\left(\frac{{tg}\left({t}\right)−\mathrm{1}}{{tg}\left({t}\right)+\mathrm{1}}\right){dt} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({tg}\left({t}\right)\right){tg}\left(\frac{\pi}{\mathrm{4}}−{t}\right){dt}−\mathrm{2}\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} {tg}\left({t}\right){ln}\left({tg}\left({t}−\frac{\pi}{\mathrm{4}}\right)\right){dt}_{{t}−\frac{\pi}{\mathrm{4}}={r}} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({tg}\left({t}\right)\right)\frac{\mathrm{1}+{tg}\left({t}\right)}{\mathrm{1}+{tg}\left({t}\right)}{dt}−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tg}\left({r}+\frac{\pi}{\mathrm{4}}\right){ln}\left({tg}\left({r}\right)\right){dr} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({tg}\left({t}\right)\right)\frac{\mathrm{1}−{tg}\left({t}\right)}{\mathrm{1}+{tg}\left({t}\right)}{dt}−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({tg}\left({t}\right)\right)\frac{\mathrm{1}+{tg}\left({t}\right)}{\mathrm{1}−{tg}\left({t}\right)}{dt} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({tg}\left({t}\right)\right)\left[\frac{\mathrm{1}−{tg}\left({t}\right)}{\mathrm{1}+{tg}\left({t}\right)}+\frac{\mathrm{1}+{tg}\left({t}\right)}{\mathrm{1}−{tg}\left({t}\right)}\right]{dt} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left[\frac{\mathrm{2}+\mathrm{2}{tg}^{\mathrm{2}} \left({t}\right)}{\mathrm{1}−{tg}^{\mathrm{2}} \left({t}\right)}\right]{ln}\left({tg}\left({t}\right)\right){dt}\:\:\:{tg}\left({t}\right)={x}\Rightarrow \\ $$$$=−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right)\underset{{k}\geqslant\mathrm{0}} {\sum}{x}^{\mathrm{2}{k}} {dx} \\ $$$$=−\mathrm{4}\underset{{k}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{k}} {ln}\left({x}\right)=−\mathrm{4}\underset{{k}\geqslant\mathrm{0}} {\sum}\left\{\left[\frac{{x}^{\mathrm{2}{k}+\mathrm{1}} }{\mathrm{2}{k}+\mathrm{1}}{ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{k}} {dx}\right. \\ $$$$=\mathrm{4}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{4}\left\{\zeta\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{4}}\zeta\left(\mathrm{2}\right)\right\}=\mathrm{3}\zeta\left(\mathrm{2}\right)=\mathrm{3}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great}\:\mathrm{sir} \\ $$