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Question Number 136381 by mnjuly1970 last updated on 21/Mar/21

$$......sdvancedcslculus......$$$$ifx\in {\mathbb{R}}^{+}and::$$$$\mathit{\varphi }\left(x\right)={\int }_0^x\frac{e^t-1}{t}ln\left(\frac{x}{t}\right)dt$$$$thenprovethat::$$$$\mathrm{\Psi }={\int }_0^{\mathrm{\infty }}{e}^{-x}\mathit{\varphi }\left(x\right)dx=\zeta \left(2\right)$$

Answered by Ñï= last updated on 21/Mar/21

$$\mathrm{\Phi }\left(x\right)={\int }_0^x\frac{e^t-1}{t}ln\left(\frac{x}{t}\right)dt$$$$={\int }_0^1\frac{1-e^{xt}}{t}lntdt$$$$={\int }_0^1\frac{1-e^{x(1-t)}}{1-t}ln(1-t)dt$$$$\mathrm{\Psi }={\int }_0^{\mathrm{\infty }}{e}^{-x}{\int }_0^1\frac{1-e^{x(1-t)}}{1-t}ln(1-t)dtdx$$$$={\int }_0^{\mathrm{\infty }}{\int }_0^1\frac{e^{-x}-e^{-xt}}{1-t}ln(1-t)dtdx$$$$={\int }_0^1\frac{1-\frac{1}{t}}{1-t}ln(1-t)dt$$$$=-{\int }_0^1\frac{ln(1-t)}{t}dt$$$$={Li}_2\left(1\right)$$

Commented by mnjuly1970 last updated on 21/Mar/21

$$thankyousir...grateful..$$

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