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Question Number 206364 by Skabetix last updated on 12/Apr/24

Answered by TonyCWX08 last updated on 13/Apr/24

$$Ionlyknow$$$$e^{\pi i}=-1$$$$\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}\frac{\sin \left(x\right)}{x}dx=\pi$$$${lim}_{n\to \mathrm{\infty }}{(1+\frac{1}{n})}^n=e$$$$Thelastoneisabithard...$$

Answered by Berbere last updated on 13/Apr/24

$${\int }_{-\pi }^{\pi }{e}^{\frac{1}{tan\left(x\right)}}dx⩾{\int }_0^{\frac{\pi }{2}}{e}^{\frac{1}{tan\left(x\right)}}dx={\int }_0^{\mathrm{\infty }}{e}^{\frac{1}{y}}\frac{dy}{1+y^2}={\int }_0^{\mathrm{\infty }}\frac{y^2}{1+y^2}{e}^{y}dy$$$$>{\int }_0^{\mathrm{\infty }}\frac{y^2}{1+y^2}dy={[y-{\tan }^{-1}\left(y\right)]}_0^{\mathrm{\infty }}=\mathrm{\infty }diverge$$

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