Prove $\int_0^\infty \frac{e^{\frac{-1}{x}}-1}{x^{\frac{2}{3}}}dx$ converges

Question

Prove the following integral converges: $$\int_0^\infty \frac{e^{\frac{-1}{x}}-1}{x^{\frac{2}{3}}}dx$$

My attempt:

First, we prove the convergence of $\int_0^1 \frac{e^{\frac{-1}{x}}-1}{x^{\frac{2}{3}}}dx$

In this interval: $e^{\frac{-1}{x}}-1<0,$ therefore we can use the Limit Comparison Test with $-x^{\frac{2}{3}}$ which converges and we conclude the convergence of the integral.

I'm not sure how to deal with $\int_1^\infty \frac{e^{\frac{-1}{x}}-1}{x^{\frac{2}{3}}}dx$

I tried splitting the integral, but $\int_1^\infty \frac{1}{x^{\frac{2}{3}}}dx$ diverges, so I got stuck.

Any help appreciated.

Answer 1

Using,

$$e^{x} > x+1$$

For all $x \neq 0$, we have $e^{-1/x}>(-1/x)+1$. Hence for $x>0$,

$$\frac{e^{-1/x}-1}{x^{2/3}}>-\frac{1}{x^{5/3}}$$

At the same time $e^{-1/x}<1$ for $x>0$. So we have that for $x>0$,

$$0>\frac{e^{-1/x}-1}{x^{2/3}}=f(x)>-\frac{1}{x^{5/3}}$$

By convergence of $\int_{1}^{\infty} \frac{1}{x^{5/3}} dx$ the comparison test for integrals ensures the convergence of $\int_{1}^{\infty} f(x) dx$.

Answer 2

Note that $1-e^{-u} \sim u$, as $u \to 0$, by Taylor expansion.

Consequently $e^{-1/x}-1 \sim -x^{-1}$ as $x \to \infty$.

Therefore $x^{-2/3}(e^{-1/x}-1) \sim -x^{-5/3}$ as $x \to \infty$.

What can you conclude?

Answer 3

I suppose that life would be simpler changing variable $x=\frac 1y$ which makes $$I=\int_0^\infty\frac{e^{-1/x}-1}{x^{2/3}}\,dx=\int_0^\infty \frac{e^{-y}-1}{y^{4/3}}\,dy$$

Close to $y=0$, Taylor expansion gives $$\frac{e^{-y}-1}{y^{4/3}}=-\frac{1}{y^{1/3}}+\frac{1}{2}y^{2/3}+O\left(y^{4/3}\right)$$ so, no problem.