xsinx-1-cosx-dx-

Question Number 206200 by Shrodinger last updated on 09/Apr/24

$$\int\frac{{xsinx}}{\mathrm{1}−{cosx}}{dx} \\ $$

Answered by Frix last updated on 09/Apr/24

∫((xsin x)/(1−cos x))dx=i∫((x(e^(ix) +1))/(e^(ix) −1))dx= =i∫xdx+2i∫(x/(e^(ix) −1))dx i∫xdx=i(x^2 /2) 2i∫(x/(e^(ix) −1))dx =^(t=e^(ix) −1) −2i∫((ln (t+1))/(t(t+1)))dt= =2i∫((ln (t+1))/(t+1))dt−2i∫((ln (t+1))/t)dt= =i ln^2 (t+1) +2i Li_2 (t) = =−ix^2 +2i Li_2 (e^(ix) −1) ⇒ ∫((xsin x)/(1−cos x))dx=(2Li_2 (e^(ix) −1) −(x^2 /2))i+C

$$\int\frac{{x}\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}{dx}=\mathrm{i}\int\frac{{x}\left(\mathrm{e}^{\mathrm{i}{x}} +\mathrm{1}\right)}{\mathrm{e}^{\mathrm{i}{x}} −\mathrm{1}}{dx}= \\ $$$$=\mathrm{i}\int{xdx}+\mathrm{2i}\int\frac{{x}}{\mathrm{e}^{\mathrm{i}{x}} −\mathrm{1}}{dx} \\ $$$$\mathrm{i}\int{xdx}=\mathrm{i}\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{2i}\int\frac{{x}}{\mathrm{e}^{\mathrm{i}{x}} −\mathrm{1}}{dx}\:\overset{{t}=\mathrm{e}^{\mathrm{i}{x}} −\mathrm{1}} {=}\:−\mathrm{2i}\int\frac{\mathrm{ln}\:\left({t}+\mathrm{1}\right)}{{t}\left({t}+\mathrm{1}\right)}{dt}= \\ $$$$=\mathrm{2i}\int\frac{\mathrm{ln}\:\left({t}+\mathrm{1}\right)}{{t}+\mathrm{1}}{dt}−\mathrm{2i}\int\frac{\mathrm{ln}\:\left({t}+\mathrm{1}\right)}{{t}}{dt}= \\ $$$$=\mathrm{i}\:\mathrm{ln}^{\mathrm{2}} \:\left({t}+\mathrm{1}\right)\:+\mathrm{2i}\:\mathrm{Li}_{\mathrm{2}} \:\left({t}\right)\:= \\ $$$$=−\mathrm{i}{x}^{\mathrm{2}} +\mathrm{2i}\:\mathrm{Li}_{\mathrm{2}} \:\left(\mathrm{e}^{\mathrm{i}{x}} −\mathrm{1}\right) \\ $$$$\Rightarrow \\ $$$$\int\frac{{x}\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}{dx}=\left(\mathrm{2Li}_{\mathrm{2}} \:\left(\mathrm{e}^{\mathrm{i}{x}} −\mathrm{1}\right)\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\mathrm{i}+{C} \\ $$

Commented by TonyCWX08 last updated on 10/Apr/24

$${Looks}\:{like}\:{you}\:{also}\:{know}\:{the}\:{Dilogarithm}\:{Function}! \\ $$

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