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Question-120059

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Question Number 120059 by A8;15: last updated on 29/Oct/20
Answered by Lordose last updated on 29/Oct/20
Firstly, compute the indefinite integral I = ∫xsec(2x)dx = (1/4)∫asec(a)da {a=2x} I = (1/4)(alntan((a/2)+(π/4)) − ∫lntan((a/2)+(π/4))da) {IBP} set (a/2) + (π/4) = y ⇒ da=2dy I = (1/4)(alntan(y) − 2∫lntan(y)dy) Ω = ∫lntan(y)dy {tan(y) = ((1−e^(−2iy) )/(1+e^(−2iy) ))} Ω = ∫ln(1−e^(−2iy) )dy − ∫ln(1+e^(−2iy) )dy set w=e^(−2iy ) ⇒dw=((2w)/i)dy Ω = (i/2)∫ ((ln(1−w))/w)dw − (i/2)∫ ((ln(1+w))/w)dw Ω = −(i/2)Li_2 (w) − (i/2)(−Li_2 (−w)) + C Ω = (i/2)(Li_2 (−e^(−2iy) ) − Li_2 (e^(−2iy) )) I=(1/4)(alntan(y) − i(Li_2 (−e^(−2iy) )−Li_2 (e^(−2iy) ))) I = (1/2)xlntan(x+(π/4)) − (i/4)(Li_2 (−e^(−i(2x+(π/2))) ) − Li_2 (e^(−i(2x+(π/2))) )) + C ∫_0 ^( 1) xsec(2x)dx =( (1/2)lntan(((4+π)/4)) − (i/4)Li_2 (−e^(−i(((4+π)/2))) ) + (i/4)Li_2 (e^(−i(((4+π)/2))) )) + (i/4)(Li_2 (−e^(−((iπ)/2)) )− Li_2 (e^(−((iπ)/2)) )) ∫_0 ^( 1) xsec(2x)dx = ∞ The integral diverges.
$$ \\ $$$$\mathrm{Firstly},\:\mathrm{compute}\:\mathrm{the}\:\mathrm{indefinite}\:\mathrm{integral} \\ $$$$\mathrm{I}\:=\:\int\mathrm{xsec}\left(\mathrm{2x}\right)\mathrm{dx}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{asec}\left(\mathrm{a}\right)\mathrm{da}\:\left\{\mathrm{a}=\mathrm{2x}\right\} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{alntan}\left(\frac{\mathrm{a}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\:−\:\int\mathrm{lntan}\left(\frac{\mathrm{a}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mathrm{da}\right)\:\:\:\left\{\mathrm{IBP}\right\} \\ $$$$\mathrm{set}\:\frac{\mathrm{a}}{\mathrm{2}}\:+\:\frac{\pi}{\mathrm{4}}\:=\:\mathrm{y}\:\Rightarrow\:\mathrm{da}=\mathrm{2dy} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{alntan}\left(\mathrm{y}\right)\:−\:\mathrm{2}\int\mathrm{lntan}\left(\mathrm{y}\right)\mathrm{dy}\right) \\ $$$$\Omega\:=\:\int\mathrm{lntan}\left(\mathrm{y}\right)\mathrm{dy}\:\:\:\:\:\left\{\mathrm{tan}\left(\mathrm{y}\right)\:=\:\frac{\mathrm{1}−\mathrm{e}^{−\mathrm{2iy}} }{\mathrm{1}+\mathrm{e}^{−\mathrm{2iy}} }\right\} \\ $$$$\Omega\:=\:\int\mathrm{ln}\left(\mathrm{1}−\mathrm{e}^{−\mathrm{2iy}} \right)\mathrm{dy}\:−\:\int\mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{2iy}} \right)\mathrm{dy} \\ $$$$\mathrm{set}\:\mathrm{w}=\mathrm{e}^{−\mathrm{2iy}\:} \Rightarrow\mathrm{dw}=\frac{\mathrm{2w}}{\mathrm{i}}\mathrm{dy} \\ $$$$\Omega\:=\:\frac{\mathrm{i}}{\mathrm{2}}\int\:\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{w}\right)}{\mathrm{w}}\mathrm{dw}\:−\:\frac{\mathrm{i}}{\mathrm{2}}\int\:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{w}\right)}{\mathrm{w}}\mathrm{dw} \\ $$$$\Omega\:=\:−\frac{\mathrm{i}}{\mathrm{2}}\mathrm{Li}_{\mathrm{2}} \left(\mathrm{w}\right)\:−\:\frac{\mathrm{i}}{\mathrm{2}}\left(−\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{w}\right)\right)\:+\:\mathrm{C} \\ $$$$\Omega\:=\:\frac{\mathrm{i}}{\mathrm{2}}\left(\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{e}^{−\mathrm{2iy}} \right)\:−\:\mathrm{Li}_{\mathrm{2}} \left(\mathrm{e}^{−\mathrm{2iy}} \right)\right) \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{alntan}\left(\mathrm{y}\right)\:−\:\mathrm{i}\left(\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{e}^{−\mathrm{2iy}} \right)−\mathrm{Li}_{\mathrm{2}} \left(\mathrm{e}^{−\mathrm{2iy}} \right)\right)\right) \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{xlntan}\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)\:−\:\frac{\mathrm{i}}{\mathrm{4}}\left(\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{e}^{−\mathrm{i}\left(\mathrm{2x}+\frac{\pi}{\mathrm{2}}\right)} \right)\:−\:\mathrm{Li}_{\mathrm{2}} \left(\mathrm{e}^{−\mathrm{i}\left(\mathrm{2x}+\frac{\pi}{\mathrm{2}}\right)} \right)\right)\:+\:\mathrm{C} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{xsec}\left(\mathrm{2x}\right)\mathrm{dx}\:=\left(\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lntan}\left(\frac{\mathrm{4}+\pi}{\mathrm{4}}\right)\:−\:\frac{\mathrm{i}}{\mathrm{4}}\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{e}^{−\mathrm{i}\left(\frac{\mathrm{4}+\pi}{\mathrm{2}}\right)} \right)\:+\:\frac{\mathrm{i}}{\mathrm{4}}\mathrm{Li}_{\mathrm{2}} \left(\mathrm{e}^{−\mathrm{i}\left(\frac{\mathrm{4}+\pi}{\mathrm{2}}\right)} \right)\right)\:+\:\frac{\mathrm{i}}{\mathrm{4}}\left(\mathrm{Li}_{\mathrm{2}} \left(−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)−\:\mathrm{Li}_{\mathrm{2}} \left(\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \right)\right) \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{xsec}\left(\mathrm{2x}\right)\mathrm{dx}\:=\:\infty \\ $$$$\mathrm{The}\:\mathrm{integral}\:\mathrm{diverges}. \\ $$

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