Question-206364

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

Answered by TonyCWX08 last updated on 13/Apr/24

I o n l y k n o w

e^{π i} = - 1

\underset{- \infty}{\int^{\infty}} \frac{\sin (x)}{x} d x = π

{l i m}_{n \to \infty} {(1 + \frac{1}{n})}^{n} = e

T h e l a s t o n e i s a b i t h a r d \dots

Answered by Berbere last updated on 13/Apr/24

\int_{- π}^{π} e^{\frac{1}{t a n (x)}} d x ⩾ \int_{0}^{\frac{π}{2}} e^{\frac{1}{t a n (x)}} d x = \int_{0}^{\infty} e^{\frac{1}{y}} \frac{d y}{1 + y^{2}} = \int_{0}^{\infty} \frac{y^{2}}{1 + y^{2}} e^{y} d y

> \int_{0}^{\infty} \frac{y^{2}}{1 + y^{2}} d y = {[y - \tan^{- 1} (y)]}_{0}^{\infty} = \infty d i v e r g e

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