Question-84379

Question Number 84379 by Power last updated on 12/Mar/20

Answered by mind is power last updated on 12/Mar/20

=∫((x2i)/(e^(ix) −e^(−ix) ))dx =∫((x2ie^(ix) )/((e^(2ix) −1)))dx=f(x) y=e^(ix) ⇒dy=ie^(ix) dx x=((ln(y))/i) =∫((2ln(y)dy)/(i(y−1)(y+1))) =(/i).∫ln(y){(1/(y−1))−(1/(y+1))}dy =−i∫((ln(y))/(y−1)).dy+i∫((ln(y))/(y+1))dy −u=y−1,v=y+1⇒ =−i∫((ln(1−u))/u)du+i∫((ln(v−1))/v)dv Li_2 (z)=−∫_0 ^z ((ln(1−x))/x)dx⇒ =iLi_2 (u)+i∫((ln(−1)+ln(1−v))/v)dv =iLi_2 (u)+i∫((iπ)/v)−i.−∫((ln(1−v))/v)dv =iLi_2 (u)−πln(v)−iLi_2 (−v)+c f(x)=iLi_2 (1−e^(ix) )−πln(v)−iLi_2 (1+e^(ix) )+c =−πln(v)+i{Li_2 (1−e^(ix) )−Li_2 (1+e^(ix) )}+c

$$=\int\frac{{x}\mathrm{2}{i}}{{e}^{{ix}} −{e}^{−{ix}} }{dx} \\ $$$$=\int\frac{{x}\mathrm{2}{ie}^{{ix}} }{\left({e}^{\mathrm{2}{ix}} −\mathrm{1}\right)}{dx}={f}\left({x}\right) \\ $$$${y}={e}^{{ix}} \Rightarrow{dy}={ie}^{{ix}} {dx} \\ $$$${x}=\frac{{ln}\left({y}\right)}{{i}} \\ $$$$=\int\frac{\mathrm{2}{ln}\left({y}\right){dy}}{{i}\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right)} \\ $$$$=\frac{}{{i}}.\int{ln}\left({y}\right)\left\{\frac{\mathrm{1}}{{y}−\mathrm{1}}−\frac{\mathrm{1}}{{y}+\mathrm{1}}\right\}{dy} \\ $$$$=−{i}\int\frac{{ln}\left({y}\right)}{{y}−\mathrm{1}}.{dy}+{i}\int\frac{{ln}\left({y}\right)}{{y}+\mathrm{1}}{dy} \\ $$$$−{u}={y}−\mathrm{1},{v}={y}+\mathrm{1}\Rightarrow \\ $$$$=−{i}\int\frac{{ln}\left(\mathrm{1}−{u}\right)}{{u}}{du}+{i}\int\frac{{ln}\left({v}−\mathrm{1}\right)}{{v}}{dv} \\ $$$${Li}_{\mathrm{2}} \left({z}\right)=−\int_{\mathrm{0}} ^{{z}} \frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx}\Rightarrow \\ $$$$={iLi}_{\mathrm{2}} \left({u}\right)+{i}\int\frac{{ln}\left(−\mathrm{1}\right)+{ln}\left(\mathrm{1}−{v}\right)}{{v}}{dv} \\ $$$$={iLi}_{\mathrm{2}} \left({u}\right)+{i}\int\frac{{i}\pi}{{v}}−{i}.−\int\frac{{ln}\left(\mathrm{1}−{v}\right)}{{v}}{dv} \\ $$$$={iLi}_{\mathrm{2}} \left({u}\right)−\pi{ln}\left({v}\right)−{iLi}_{\mathrm{2}} \left(−{v}\right)+{c} \\ $$$${f}\left({x}\right)={iLi}_{\mathrm{2}} \left(\mathrm{1}−{e}^{{ix}} \right)−\pi{ln}\left({v}\right)−{iLi}_{\mathrm{2}} \left(\mathrm{1}+{e}^{{ix}} \right)+{c} \\ $$$$=−\pi{ln}\left({v}\right)+{i}\left\{{Li}_{\mathrm{2}} \left(\mathrm{1}−{e}^{{ix}} \right)−{Li}_{\mathrm{2}} \left(\mathrm{1}+{e}^{{ix}} \right)\right\}+{c} \\ $$$$ \\ $$

Commented by Power last updated on 12/Mar/20

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by Kamel Kamel last updated on 12/Mar/20

Put: t=tan((x/2)) ∫(x/(sin(x)))dx=∫((Arctan(t))/t)dt=Ti_2 (tan((x/2)))+c /c∈R Ti_2 (x) is Inverse Arctangent integral.

$${Put}:\:{t}={tan}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$\int\frac{{x}}{{sin}\left({x}\right)}{dx}=\int\frac{{Arctan}\left({t}\right)}{{t}}{dt}={Ti}_{\mathrm{2}} \left({tan}\left(\frac{{x}}{\mathrm{2}}\right)\right)+{c}\:/{c}\in\mathbb{R} \\ $$$${Ti}_{\mathrm{2}} \left({x}\right)\:{is}\:{Inverse}\:{Arctangent}\:{integral}. \\ $$

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