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Question Number 163469 by Zaynal last updated on 07/Jan/22

Commented by smallEinstein last updated on 07/Jan/22

Answered by amin96 last updated on 07/Jan/22

$$\mathrm{\Omega }=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}{\int }_0^1{x}^{n-1}dx=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=-\zeta \left(2\right)$$

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