Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 164129 by mnjuly1970 last updated on 14/Jan/22

prove 𝛗= Re (∫_0 ^( 1) Li_( 2) ( (1/x) ) )dx = ΞΆ (2) βˆ’βˆ’βˆ’m.nβˆ’βˆ’βˆ’

$$ \\ $$$$\:\:\:{prove} \\ $$$$ \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\:\mathscr{R}{e}\:\left(\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{Li}_{\:\mathrm{2}} \:\left(\:\frac{\mathrm{1}}{{x}}\:\right)\:\right){dx}\:=\:\zeta\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:βˆ’βˆ’βˆ’{m}.{n}βˆ’βˆ’βˆ’ \\ $$

Answered by mindispower last updated on 14/Jan/22

∫Li_2 ((1/x)) =xLi_2 ((1/x))βˆ’βˆ«_0 ^1 ln(1βˆ’(1/x))dx βˆ€x∈]0,1[ ln(1βˆ’(1/x))=ln((1/x)βˆ’1)+iΟ€ βˆ…=[xli_2 ((1/x))]_0 ^1 βˆ’βˆ«_0 ^1 ln((1/x)βˆ’1)dx =li_2 (1)βˆ’βˆ«_1 ^∞ ((ln(xβˆ’1))/x^2 )dx=ΞΆ(2)βˆ’A A=∫_0 ^∞ ((ln(z))/((1+z)^2 ))dz=∫_0 ^∞ βˆ’((ln(z))/((1+z)^2 ))dz=βˆ’Aβ‡’A=0 βˆ…=Li_2 (1)=ΞΆ(2)

$$\int{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$={xLi}_{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right)βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}βˆ’\frac{\mathrm{1}}{{x}}\right){dx} \\ $$$$\left.\forall{x}\in\right]\mathrm{0},\mathrm{1}\left[\:{ln}\left(\mathrm{1}βˆ’\frac{\mathrm{1}}{{x}}\right)={ln}\left(\frac{\mathrm{1}}{{x}}βˆ’\mathrm{1}\right)+{i}\pi\right. \\ $$$$\emptyset=\left[{xli}_{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} βˆ’\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\frac{\mathrm{1}}{{x}}βˆ’\mathrm{1}\right){dx} \\ $$$$={li}_{\mathrm{2}} \left(\mathrm{1}\right)βˆ’\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({x}βˆ’\mathrm{1}\right)}{{x}^{\mathrm{2}} }{dx}=\zeta\left(\mathrm{2}\right)βˆ’{A} \\ $$$${A}=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({z}\right)}{\left(\mathrm{1}+{z}\right)^{\mathrm{2}} }{dz}=\int_{\mathrm{0}} ^{\infty} βˆ’\frac{{ln}\left({z}\right)}{\left(\mathrm{1}+{z}\right)^{\mathrm{2}} }{dz}=βˆ’{A}\Rightarrow{A}=\mathrm{0} \\ $$$$\emptyset={Li}_{\mathrm{2}} \left(\mathrm{1}\right)=\zeta\left(\mathrm{2}\right) \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 15/Jan/22

thanks alot sir power...very nice solution..grateful

$$\:\:{thanks}\:{alot}\:{sir}\:{power}...{very}\:{nice} \\ $$$${solution}..{grateful} \\ $$

Commented by mindispower last updated on 15/Jan/22

withe Pleasur thanks have a nice day sir

$${withe}\:{Pleasur}\:{thanks} \\ $$$${have}\:{a}\:{nice}\:{day}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com