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(e^(-1/x))/(x^2)<\frac1{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(e^(-1/x))/(x^2)\,\mathrm dx=\int_(-1/2)^0e^y\,\mathrm dy=e^0-e^(-1/2)=1-\frac1{\sqrt e}`