Question Number 69261 by mathmax by abdo last updated on 22/Sep/19
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{xtanx}\:{dx} \\ $$
Answered by mind is power last updated on 23/Sep/19
![=∫_0 ^1 x((e^(ix) −e^(−ix) )/(i(e^(ix) +e^(−ix) )))dx =∫_0 ^1 (x/i)(1−((2e^(−ix) )/(e^(ix) +e^(−ix) )))dx =∫_0 ^1 (x/i)−(2/i)∫x(e^(−2ix) /(1+e^(−2ix) ))dx =(1/(2i))−(2/i)∫_0 ^1 ((xe^(−2ix) )/(1+e^(−2ix) ))dx...a u′=(e^(−2ix) /(1+e^(−2ix) )) v=x u=((−1)/(2i))ln(1+e^(−2ix) ) v′=1 ∫_0 ^1 ((xe^(−2ix) )/(1+e^(−2ix) ))dx=[((−x)/(2i))ln(1+e^(−2ix) )]_0 ^1 +(1/(2i))∫_0 ^1 ln(1+e^(−2ix) )dx =−((ln(1+e^(−2i) ))/(2i))+(1/(2i))∫_0 ^1 ln(((1+e^(2ix) )/e^(2ix) ))dx =−((ln(1+e^(−2i) ))/(2i))+(1/(2i))∫_0 ^1 {ln(1+e^(2ix) )−ln(e^(2ix) )}dx =−((ln(1+e^(−2i) ))/(2i))+(1/(2i))∫_0 ^1 {Σ_(n≥1) (((−1)^(n+1) e^(2inx) )/n)−2ix}dx =−((ln(1+e^(−2i) ))/(2i))+(1/(2i))Σ_(n≥1) ∫_0 ^1 (((−1)^(n+1) e^(2inx) )/n)dx−∫_0 ^1 xdx =−((ln(1+e^(−2i) ))/(2i))+(1/(2i))Σ_(n≥1) [(((−1)^(n+1) e^(2inx) )/(2in^2 ))]_0 ^1 −(1/2) =−((ln(1+e^(−2i) ))/(2i))−(1/2)+(1/(2i))Σ_(n≥1) (((−1)^(n+1) e^(2in) )/(2in^2 ))+(1/(2i))Σ_(n≥1) (((−1)^n )/(2in^2 )) =−((ln(1+e^(−2i) ))/(2i))−(1/2)−(1/(2i))Σ_(n≥1) (((−e^(2i) )^n )/(2in^2 ))+(1/(2i))Σ_(n≥1) (((−1)^n )/(2in^2 ))=..b Σ_(n≥1) (((−1)^n )/n^2 )=Σ_(k≥0) ((−1)/((2k+1)^2 ))+Σ_(k≥1) (1/(4k^2 )) Σ_(k≥1) (1/k^2 )=Σ_(k≥1) (1/(4k^2 ))+Σ_(k≥0) (1/((2k+1)^2 ))⇒Σ_(k≥1) (1/((2k+1)^2 ))=(π^2 /8) ⇒Σ_(n≥1) (((−1)^n )/n^2 )=(π^2 /(24))−(π^2 /8)=−(π^2 /(12)) ⇒b=−((ln(1+e^(−2i) ))/(2i))−(1/2)−(1/(2i))Σ_(n≥1) (((−e^(2i) )^n )/(2in^2 ))−(π^2 /(48)) Li_2 (x)=Σ_(n≥1) (((x)^n )/n^2 )⇒Σ_(n≥1) (((−e^(2i) )^n )/n^2 )=Li_2 (−e^(2i) ) ⇒b=−((ln(1+e^(−2i) ))/(2i))−(1/2)+((Li_2 (−e^(2i) ))/4)+(π^2 /(48)) ⇒(1/(2i))−(2/i)∫_0 ^1 ((xe^(−2ix) )/(1+e^(−2ix) ))dx=(1/(2i))−(2/i)(−((ln(1+e^(2i) ))/(2i))−(1/2)+((Li_2 (−e^(2i) ))/4)+(π^2 /(48))) =(1/(2i))−ln(1+e^(−2i) )+(1/i)−((Li_2 (−e^(2i) ))/(2i))−(π^2 /(i24)) =(3/(2i))−ln(e^(2i) +1)+ln(e^(−2i) )−((Li_2 (−e^(2i) ))/(2i))+((iπ^2 )/(24)) =(i/2)−log(1+e^(2i) )+i((Li_2 (−e^(2i) ))/2)+i(π^2 /(24)) =(i/(24))(π^2 +12(Li_2 (−e^(2i) )+1+ilog(1+e^(2i) )))](https://www.tinkutara.com/question/Q69409.png)
$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}\frac{{e}^{{ix}} −{e}^{−{ix}} }{{i}\left({e}^{{ix}} +{e}^{−{ix}} \right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{{i}}\left(\mathrm{1}−\frac{\mathrm{2}{e}^{−{ix}} }{{e}^{{ix}} +{e}^{−{ix}} }\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{{i}}−\frac{\mathrm{2}}{{i}}\int{x}\frac{{e}^{−\mathrm{2}{ix}} }{\mathrm{1}+{e}^{−\mathrm{2}{ix}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}−\frac{\mathrm{2}}{{i}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xe}^{−\mathrm{2}{ix}} }{\mathrm{1}+{e}^{−\mathrm{2}{ix}} }{dx}…{a} \\ $$$${u}'=\frac{{e}^{−\mathrm{2}{ix}} }{\mathrm{1}+{e}^{−\mathrm{2}{ix}} }\:\:\:\:{v}={x} \\ $$$${u}=\frac{−\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{ix}} \right)\:\:{v}'=\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xe}^{−\mathrm{2}{ix}} }{\mathrm{1}+{e}^{−\mathrm{2}{ix}} }{dx}=\left[\frac{−{x}}{\mathrm{2}{i}}{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{ix}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{ix}} \right){dx} \\ $$$$=−\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}} \right)}{\mathrm{2}{i}}+\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\frac{\mathrm{1}+{e}^{\mathrm{2}{ix}} }{{e}^{\mathrm{2}{ix}} }\right){dx} \\ $$$$=−\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}} \right)}{\mathrm{2}{i}}+\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\mathrm{1}} \left\{{ln}\left(\mathrm{1}+{e}^{\mathrm{2}{ix}} \right)−{ln}\left({e}^{\mathrm{2}{ix}} \right)\right\}{dx} \\ $$$$=−\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}} \right)}{\mathrm{2}{i}}+\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\sum_{{n}\geqslant\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {e}^{\mathrm{2}{inx}} }{{n}}−\mathrm{2}{ix}\right\}{dx} \\ $$$$=−\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}} \right)}{\mathrm{2}{i}}+\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{n}\geqslant\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {e}^{\mathrm{2}{inx}} }{{n}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} {xdx} \\ $$$$=−\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}} \right)}{\mathrm{2}{i}}+\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{n}\geqslant\mathrm{1}} \left[\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {e}^{\mathrm{2}{inx}} }{\mathrm{2}{in}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=−\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}} \right)}{\mathrm{2}{i}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{n}\geqslant\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {e}^{\mathrm{2}{in}} }{\mathrm{2}{in}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{n}\geqslant\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{in}^{\mathrm{2}} } \\ $$$$=−\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}} \right)}{\mathrm{2}{i}}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{n}\geqslant\mathrm{1}} \frac{\left(−{e}^{\mathrm{2}{i}} \right)^{{n}} }{\mathrm{2}{in}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{n}\geqslant\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{in}^{\mathrm{2}} }=..{b} \\ $$$$\sum_{{n}\geqslant\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }=\sum_{{k}\geqslant\mathrm{0}} \frac{−\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }+\sum_{{k}\geqslant\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}{k}^{\mathrm{2}} } \\ $$$$\sum_{{k}\geqslant\mathrm{1}} \frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\sum_{{k}\geqslant\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}{k}^{\mathrm{2}} }+\sum_{{k}\geqslant\mathrm{0}} \frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }\Rightarrow\sum_{{k}\geqslant\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\Rightarrow\sum_{{n}\geqslant\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\Rightarrow{b}=−\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}} \right)}{\mathrm{2}{i}}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{n}\geqslant\mathrm{1}} \frac{\left(−{e}^{\mathrm{2}{i}} \right)^{{n}} }{\mathrm{2}{in}^{\mathrm{2}} }−\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$${Li}_{\mathrm{2}} \left({x}\right)=\sum_{{n}\geqslant\mathrm{1}} \frac{\left({x}\right)^{{n}} }{{n}^{\mathrm{2}} }\Rightarrow\sum_{{n}\geqslant\mathrm{1}} \frac{\left(−{e}^{\mathrm{2}{i}} \right)^{{n}} }{{n}^{\mathrm{2}} }={Li}_{\mathrm{2}} \left(−{e}^{\mathrm{2}{i}} \right) \\ $$$$\Rightarrow{b}=−\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}} \right)}{\mathrm{2}{i}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{{Li}_{\mathrm{2}} \left(−{e}^{\mathrm{2}{i}} \right)}{\mathrm{4}}+\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}{i}}−\frac{\mathrm{2}}{{i}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xe}^{−\mathrm{2}{ix}} }{\mathrm{1}+{e}^{−\mathrm{2}{ix}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}{i}}−\frac{\mathrm{2}}{{i}}\left(−\frac{{ln}\left(\mathrm{1}+{e}^{\mathrm{2}{i}} \right)}{\mathrm{2}{i}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{{Li}_{\mathrm{2}} \left(−{e}^{\mathrm{2}{i}} \right)}{\mathrm{4}}+\frac{\pi^{\mathrm{2}} }{\mathrm{48}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}−{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{i}} \right)+\frac{\mathrm{1}}{{i}}−\frac{{Li}_{\mathrm{2}} \left(−{e}^{\mathrm{2}{i}} \right)}{\mathrm{2}{i}}−\frac{\pi^{\mathrm{2}} }{{i}\mathrm{24}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}{i}}−{ln}\left({e}^{\mathrm{2}{i}} +\mathrm{1}\right)+{ln}\left({e}^{−\mathrm{2}{i}} \right)−\frac{{Li}_{\mathrm{2}} \left(−{e}^{\mathrm{2}{i}} \right)}{\mathrm{2}{i}}+\frac{{i}\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$=\frac{{i}}{\mathrm{2}}−{log}\left(\mathrm{1}+{e}^{\mathrm{2}{i}} \right)+{i}\frac{{Li}_{\mathrm{2}} \left(−{e}^{\mathrm{2}{i}} \right)}{\mathrm{2}}+{i}\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$$$=\frac{{i}}{\mathrm{24}}\left(\pi^{\mathrm{2}} +\mathrm{12}\left({Li}_{\mathrm{2}} \left(−{e}^{\mathrm{2}{i}} \right)+\mathrm{1}+{ilog}\left(\mathrm{1}+{e}^{\mathrm{2}{i}} \right)\right)\right) \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 24/Sep/19
$${thank}\:{you}\:{sir}. \\ $$
Commented by mind is power last updated on 24/Sep/19
$${y}'{re}\:{welcom} \\ $$