From 0 to ∞ \infty

Calculus Level 3

∫ 0 ∞ e − ( 1 + 1 x ) x d x = ? \int_0^\infty e- \left(1+\frac 1x\right)^x dx=?

e e e 2 \frac e2 ∞ \infty e 2 e^2 2 e 2e

X X by X X , 17, Taiwan

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1 solution

X X
May 6, 2018

lim ⁡ x → ∞ e − ( 1 + 1 x ) x 1 x = e 2 \displaystyle\lim_{x\to\infty}\frac{e- \left(1+\frac 1x\right)^x}{\frac1x}=\frac e2 ,so ∫ 0 1 e − ( 1 + 1 x ) x d x + ∫ 1 ∞ e − ( 1 + 1 x ) x d x ≈ e 2 ∫ 1 ∞ 1 x d x = ∞ \displaystyle\int_0^1 e- \left(1+\frac 1x\right)^x dx+\int_1^\infty e- \left(1+\frac 1x\right)^x dx\approx \frac e2\int_1^\infty \frac1x dx=\infty

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