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

Is my understanding correct? and
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.

Please don't use \frac in exponents or limits of integrals. It looks bad and confusing, and it rarely appears in professional mathematics typesetting. — GNUSupporter 8964民主女神地下教會, Jan 26 '21 at 14:24
You integral diverges because for $x$ high enough $1-e^{1/x} \sim 1$ — EDX, Jan 26 '21 at 14:27
You should be able to show that for $x > 1$ you have $\frac{0.63212}x \le (1-e^{-{1}/{x}}) \le \frac1x$ — Henry, Jan 26 '21 at 14:41

score 3 · Answer 1 · 2021-01-26T14:35:24.680

3

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

edited Jan 26 '21 at 14:35

answered Jan 26 '21 at 14:33

Thanks for corrective – EDX Jan 26 '21 at 14:35

score 2 · Answer 2 · edited Nov 07 '23 at 08:24

2

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

edited Nov 07 '23 at 08:24

Rócherz

3,976

answered Jan 26 '21 at 14:30

EDX

1,797

$\displaystyle\lim_{n\to\infty}1-e^{-1/x} = 1-e^{0} = 0$? – Joshua Wang Jan 26 '21 at 14:32
Right ! I corrected it ! – EDX Jan 26 '21 at 14:36
How it tends to $1/x$? – Priya Jan 26 '21 at 15:04
Taylor Young is enough ! – EDX Jan 26 '21 at 16:57
Sorry what is Taylor Young ? – Priya Jan 27 '21 at 00:57
It is the asymptotic development proposed by Varun Vejalla – EDX Jan 27 '21 at 10:00

score 2 · Accepted Answer · answered Jan 26 '21 at 14:39

2

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

answered Jan 26 '21 at 14:39

Henry Lee

12,215

After the word However I did not understand why you take that integral $\int_{1}^\infty \frac{1}{x^n}dx$ – Priya Jan 26 '21 at 15:10
@NuwanPriyankara this is a standard integral for which values are known and we can use it as a comparison – Henry Lee Jan 26 '21 at 16:02
Yes, I know that so $\int_0^\infty \frac{1}{x}$ diverges but how do we compare it to this, here we need to get the idea of $e^{-1/x}$ on $[1,\infty)$ now. please explain little bit – Priya Jan 26 '21 at 16:30
Did you expand $1−e^{−1/x}=\frac{1}{x}−\frac{1}{2x^2}+ ...$ and did you use the Laurent series? – Priya Jan 27 '21 at 02:32

Varun Vejalla · Answer 4 · 2021-01-26T16:43:31.763

1

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.

edited Jan 26 '21 at 16:43

answered Jan 26 '21 at 16:32

Varun Vejalla

8,873

Here you use Maclaurin series, as I mentioned in the original problem can we use Maclaurin series expansion? – Priya Jan 26 '21 at 16:39
You can't use the Maclaurin series for $e^{-1/x}$ as is, but you can use the Laurent series expansion (which I did in my question). – Varun Vejalla Jan 26 '21 at 16:44
Ok got it, like this situation when we are using Laurent series do we need to consider or specify any condition to apply (because the Laurent series is a series that represent a complex function f(z) as a series and a Laurent series includes terms with negative as well as non-negative powers of z.) ? – Priya Jan 26 '21 at 16:56

score 0 · Answer 5 · answered Dec 13 '21 at 03:31

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.

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

5 Answers5