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0-ln-sinh-x-dx-




Question Number 117811 by Lordose last updated on 13/Oct/20
∫_( 0) ^( 𝛑) ln∣sinh(x)∣dx
$$\int_{\:\mathrm{0}} ^{\:\boldsymbol{\pi}} \boldsymbol{\mathrm{ln}}\mid\boldsymbol{\mathrm{sinh}}\left(\boldsymbol{\mathrm{x}}\right)\mid\boldsymbol{\mathrm{dx}} \\ $$
Answered by mindispower last updated on 13/Oct/20
=∫_0 ^Ο€ ln(sh(x))dx  sh(x)=((e^x βˆ’e^(βˆ’x) )/2)  =∫_0 ^Ο€ (ln(e^x  (1βˆ’e^(βˆ’2x) ))βˆ’ln(2))dx  =∫_0 ^Ο€ (x+ln(1βˆ’e^(βˆ’2x) )βˆ’ln(2))dx  =[(1/2)x^2 βˆ’xln(2)]_0 ^Ο€   _(=A) +∫_0 ^Ο€ (ln(1βˆ’e^(βˆ’2x) )dx)_(=B)   ln(1βˆ’e^(βˆ’2x) )=βˆ’Ξ£_(nβ‰₯0) (e^(βˆ’2(n+1x) /(n+1))  ∫_0 ^Ο€ ln(1βˆ’e^(βˆ’2x) )dx=βˆ’Ξ£_(nβ‰₯0) (1/(n+1))∫_0 ^Ο€ e^(βˆ’2(n+)x) dx  Ξ£_(nβ‰₯0) ((e^(βˆ’2(n+1)Ο€) βˆ’1)/(2(n+1)^2 ))=Ξ£_(nβ‰₯1) (((e^(βˆ’2Ο€) )^n )/n^2 )βˆ’Ξ£_(nβ‰₯1) (1/(2n^2 ))  Li_s (x)=Ξ£_(kβ‰₯1) (x^k /k^s )  we get=(1/2)(Li_2 (e^(βˆ’2Ο€) )βˆ’Li_2 (1))=(1/2)Li_2 (e^(βˆ’2Ο€) )βˆ’(1/2)ΞΆ(2)  =(1/2)Li_2 (e^(βˆ’2Ο€) )βˆ’(Ο€^2 /(12))  A=(1/2)Ο€^2 βˆ’Ο€ln(2)  A+B=((5Ο€^2 )/(12))βˆ’Ο€ln(2)+Li_2 (e^(βˆ’2Ο€) )  ∫_0 ^Ο€ ln(sh(x))dx=((5Ο€^2 )/(12))βˆ’Ο€ln(2)+Li_2 (e^(βˆ’2Ο€) )
$$=\int_{\mathrm{0}} ^{\pi} {ln}\left({sh}\left({x}\right)\right){dx} \\ $$$${sh}\left({x}\right)=\frac{{e}^{{x}} βˆ’{e}^{βˆ’{x}} }{\mathrm{2}} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \left({ln}\left({e}^{{x}} \:\left(\mathrm{1}βˆ’{e}^{βˆ’\mathrm{2}{x}} \right)\right)βˆ’{ln}\left(\mathrm{2}\right)\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \left({x}+{ln}\left(\mathrm{1}βˆ’{e}^{βˆ’\mathrm{2}{x}} \right)βˆ’{ln}\left(\mathrm{2}\right)\right){dx} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} βˆ’{xln}\left(\mathrm{2}\right)\right]_{\mathrm{0}} ^{\pi} \:\underset{={A}} {\:}+\int_{\mathrm{0}} ^{\pi} \left({ln}\left(\mathrm{1}βˆ’{e}^{βˆ’\mathrm{2}{x}} \right){dx}\underset{={B}} {\right)} \\ $$$${ln}\left(\mathrm{1}βˆ’{e}^{βˆ’\mathrm{2}{x}} \right)=βˆ’\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{e}^{βˆ’\mathrm{2}\left({n}+\mathrm{1}{x}\right.} }{{n}+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}βˆ’{e}^{βˆ’\mathrm{2}{x}} \right){dx}=βˆ’\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\pi} {e}^{βˆ’\mathrm{2}\left({n}+\right){x}} {dx} \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{e}^{βˆ’\mathrm{2}\left({n}+\mathrm{1}\right)\pi} βˆ’\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({e}^{βˆ’\mathrm{2}\pi} \right)^{{n}} }{{n}^{\mathrm{2}} }βˆ’\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} } \\ $$$${Li}_{{s}} \left({x}\right)=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{x}^{{k}} }{{k}^{{s}} } \\ $$$${we}\:{get}=\frac{\mathrm{1}}{\mathrm{2}}\left({Li}_{\mathrm{2}} \left({e}^{βˆ’\mathrm{2}\pi} \right)βˆ’{Li}_{\mathrm{2}} \left(\mathrm{1}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}{Li}_{\mathrm{2}} \left({e}^{βˆ’\mathrm{2}\pi} \right)βˆ’\frac{\mathrm{1}}{\mathrm{2}}\zeta\left(\mathrm{2}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{Li}_{\mathrm{2}} \left({e}^{βˆ’\mathrm{2}\pi} \right)βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\pi^{\mathrm{2}} βˆ’\pi{ln}\left(\mathrm{2}\right) \\ $$$${A}+{B}=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{12}}βˆ’\pi{ln}\left(\mathrm{2}\right)+{Li}_{\mathrm{2}} \left({e}^{βˆ’\mathrm{2}\pi} \right) \\ $$$$\int_{\mathrm{0}} ^{\pi} {ln}\left({sh}\left({x}\right)\right){dx}=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{12}}βˆ’\pi{ln}\left(\mathrm{2}\right)+{Li}_{\mathrm{2}} \left({e}^{βˆ’\mathrm{2}\pi} \right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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