Taylor series of $e^{-{1}/{x}}$
Does the integral converge? \begin{align*} \int_0^\infty (1-e^{-{1}/{x}})dx \end{align*}
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 after 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 $0$ is not 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.
Solution 1:
As EDX hoped to write: $1-e^{-1/x} \sim 1/x$, so the integral is divergent.
Solution 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