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

$$$$$$prove$$$$$$$$\mathit{\varphi }=\mathcal{R}e\left({\int }_0^1{\mathrm{Li}}_2\left(\frac{1}{x}\right)\right)dx=\zeta \left(2\right)$$$$---m.n---$$

Answered by mindispower last updated on 14/Jan/22

$$\int {Li}_2\left(\frac{1}{x}\right)$$$$={xLi}_2\left(\frac{1}{x}\right)-{\int }_0^{1}ln(1-\frac{1}{x})dx$$$$\mathrm{\forall }x\in ]0,1[ln(1-\frac{1}{x})=ln(\frac{1}{x}-1)+i\pi$$$$\mathrm{\varnothing }={\left[{xli}_2\left(\frac{1}{x}\right)\right]}_0^1-{\int }_0^{1}ln(\frac{1}{x}-1)dx$$$$={li}_2\left(1\right)-{\int }_1^{\mathrm{\infty }}\frac{ln(x-1)}{x^2}dx=\zeta \left(2\right)-A$$$$A={\int }_0^{\mathrm{\infty }}\frac{ln\left(z\right)}{{(1+z)}^2}dz={\int }_0^{\mathrm{\infty }}-\frac{ln\left(z\right)}{{(1+z)}^2}dz=-A\Rightarrow A=0$$$$\mathrm{\varnothing }={Li}_2\left(1\right)=\zeta \left(2\right)$$$$$$

Commented by mnjuly1970 last updated on 15/Jan/22

$$thanksalotsirpower...verynice$$$$solution..grateful$$

Commented by mindispower last updated on 15/Jan/22

$$withePleasurthanks$$$$haveanicedaysir$$

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