Question

Determine whether the integral is convergent or divergent. ∫[infinity] 2 e^−1/x / x^2 dx : O Convergent O divergent If it is convergent, evaluate it. (If the quantity diverges, enter DIVERGES.)

Answer 1

Let $f(x)=e^{-1/x}$. Then $f'(x)=\frac1{x^2}e^{-1/x}>0$ for all $x\ge2$, so $f$ is strictly increasing. As $x\to\infty$, $e^{-1/x}\to e^0=1$, so $f$ is bounded above by 1. This is to say,

$e^{-1/x}<1\implies\dfrac{e^{-1/x}}{x^2}<\dfrac1{x^2}$

and the integral of $\frac1{x^2}$ converges over the same domain, so this integral must also converge by comparison.

We have, by setting $y=-\frac1x$,

$\displaystyle\int_2^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx=\int_{-1/2}^0e^y\,\mathrm dy=e^0-e^{-1/2}=1-\frac1{\sqrt e}$