Question Number 164129 by mnjuly1970 last updated on 14/Jan/22
$$ \\ $$$$\:\:\:{prove} \\ $$$$ \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\:\mathscr{R}{e}\:\left(\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{Li}_{\:\mathrm{2}} \:\left(\:\frac{\mathrm{1}}{{x}}\:\right)\:\right){dx}\:=\:\zeta\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:βββ{m}.{n}βββ \\ $$
Answered by mindispower last updated on 14/Jan/22
![β«Li_2 ((1/x)) =xLi_2 ((1/x))ββ«_0 ^1 ln(1β(1/x))dx βxβ]0,1[ ln(1β(1/x))=ln((1/x)β1)+iΟ β
=[xli_2 ((1/x))]_0 ^1 ββ«_0 ^1 ln((1/x)β1)dx =li_2 (1)ββ«_1 ^β ((ln(xβ1))/x^2 )dx=ΞΆ(2)βA A=β«_0 ^β ((ln(z))/((1+z)^2 ))dz=β«_0 ^β β((ln(z))/((1+z)^2 ))dz=βAβA=0 β
=Li_2 (1)=ΞΆ(2)](https://www.tinkutara.com/question/Q164146.png)
$$\int{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$={xLi}_{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right)β\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}β\frac{\mathrm{1}}{{x}}\right){dx} \\ $$$$\left.\forall{x}\in\right]\mathrm{0},\mathrm{1}\left[\:{ln}\left(\mathrm{1}β\frac{\mathrm{1}}{{x}}\right)={ln}\left(\frac{\mathrm{1}}{{x}}β\mathrm{1}\right)+{i}\pi\right. \\ $$$$\emptyset=\left[{xli}_{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} β\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\frac{\mathrm{1}}{{x}}β\mathrm{1}\right){dx} \\ $$$$={li}_{\mathrm{2}} \left(\mathrm{1}\right)β\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({x}β\mathrm{1}\right)}{{x}^{\mathrm{2}} }{dx}=\zeta\left(\mathrm{2}\right)β{A} \\ $$$${A}=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({z}\right)}{\left(\mathrm{1}+{z}\right)^{\mathrm{2}} }{dz}=\int_{\mathrm{0}} ^{\infty} β\frac{{ln}\left({z}\right)}{\left(\mathrm{1}+{z}\right)^{\mathrm{2}} }{dz}=β{A}\Rightarrow{A}=\mathrm{0} \\ $$$$\emptyset={Li}_{\mathrm{2}} \left(\mathrm{1}\right)=\zeta\left(\mathrm{2}\right) \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 15/Jan/22
$$\:\:{thanks}\:{alot}\:{sir}\:{power}…{very}\:{nice} \\ $$$${solution}..{grateful} \\ $$
Commented by mindispower last updated on 15/Jan/22
$${withe}\:{Pleasur}\:{thanks} \\ $$$${have}\:{a}\:{nice}\:{day}\:{sir} \\ $$