Taylor series of $e^{-1/x}$

Question

Does the integral converge? $$\int_0^\infty (1-e^{-1/x})\,\Bbb dx$$

Idea: First I tried to expand the $e^{-1/x}$ using the Maclaurin series and evaluated the integral, but it was not a good result and then I realized the result is not correct, because we cannot expand $e^{-1/x}$ by the Maclaurin series (because all derivatives of $e^{-1/x}$ at $x=0$ are zero but $e^{-1/x}$ is not a zero function so the approximate centered at $0$ doesn’t work).

We can find the Taylor series expansion of $e^{-1/x}$ centered at some point (like $x=1$). According to that, the integral is divergent. I just want to know

1. Is my understanding correct? and
2. Can we expand $e^{-1/x}$ by the Taylor series centered at some point (except zero) even the function is non-analytic?

And is this the right way to evaluate this kind of integral? It will be a great pleasure to me if someone can give a little explanation.

Answer 1

Since we have $x\in[0,\infty)$ we can see that: $$x\to0^+,\frac1x\to+\infty\therefore e^{-1/x}\to0$$ so for small values of $x$, $1-e^{-1/x}\approx 1$. Now if we look at large values of $x$: $$x\to+\infty,\frac1x\to0^+\therefore e^{-1/x}\to1$$ and so for large values of $x$, $1-e^{-1/x}\to0$. However, $$\int_1^\infty\frac{1}{x^n}dx$$ only converges for $n>1$ so for your integral to converge you need to show that your function tends to zero faster than $1/x$, which it does not so the integral will diverge

Answer 2

Your integral diverges because $$1-e^{-1/x} \sim 1/x$$ for $x$ high enough.

Answer 3

As EDX hoped to write: $1-e^{-1/x} \sim 1/x$, so the integral is divergent.

Answer 4

You have that $$e^{x}=1+x+\frac{x^2}{2}+o(x^3)$$

so $$1-e^{-1/x} = \frac{1}{x}-\frac{1}{2x^2}-O\left(\frac{1}{x^3}\right) < \frac{1}{x}-\frac{1}{2x^2}$$

And because $\int_1^{\infty}\left(\frac{1}{x} - \frac{1}{2x^2}\right) dx$ diverges, $\int_1^{\infty}\left(1-e^{-1/x}\right)dx$ will diverge as well.

Edit: I dropped the integral from $0$ to $1$ because the integrand is bounded by $1$, so it is convergent for that interval.

Answer 5

I wouldn't like to give an answer directly here. I have answered several questions related to the function $\textrm{e}^{x^\alpha}$ for $\alpha\in\mathbb{R}$ several times. However, I would like to list several references in which you can find the answer and more.

1. https://qifeng618.wordpress.com/2018/05/10/some-papers-related-to-the-function-e1-x-and-the-lah-numbers/
2. https://mathoverflow.net/questions/405015
3. https://math.stackexchange.com/a/4262516/945479
4. https://math.stackexchange.com/a/4262498/945479
5. https://math.stackexchange.com/a/4262657/945479
6. https://math.stackexchange.com/a/4263059/945479
7. https://math.stackexchange.com/a/4263072/945479
8. https://math.stackexchange.com/a/4263062/945479