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Question Number 160427 by qaz last updated on 29/Nov/21

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }{\int }_0^1\underset{\mathrm{i}=1}{\sum ^{\mathrm{n}}}\frac{\underset{\mathrm{j}=1}{\prod ^{\mathrm{n}}}(\mathrm{x}+\frac{\mathrm{j}}{{\mathrm{n}}^2})}{\mathrm{x}+\frac{\mathrm{i}}{{\mathrm{n}}^2}}\mathrm{dx}=?$$

Answered by Kamel last updated on 30/Nov/21

$$=\sqrt{e}$$

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