derailed in solving the convergence of $\int_{x=2}^\infty \frac{1}{xe^x} dx$ with a Taylor expansion

Question

I'm trying to reason through whether $\int_{x=2}^\infty \frac{1}{xe^x} dx$ converges.

Intuitively, It would seem that since $\int_{x=2}^\infty \frac{1}{x} dx$ diverges, then multiplying the denominator by something to make it smaller would make it converge.

If I apply a Taylor expansion to the $\frac{1}{e^x}$, then I get

$\displaystyle \int_{x=2}^\infty \frac{1}{x} e^{-x} dx = \int_{x=2}^\infty \frac{1}{x}(1 + (-x) + \frac{(-x)^2}{2} + \cdots) dx$

My initial thought was to multiply everything out and just look at the leading order behavior, $ x^{n-1}$, but this would seem to mean the integral diverges.

What is wrong with this approach?

Answer 1

Just notice that,

$$ x \geq 2 \implies \frac{1}{x}\leq \frac{1}{2}\implies \frac{1}{xe^{x}}\leq \frac{1}{2e^{x}}= \frac 1 2 e^{-x}. $$

So you can compare with the function $ \frac 1 2 e^{-x} $.

Answer 2

I believe the easiest thing to do here is note that for all $x\geq 2$ one has $\frac{1}{xe^x}\leq e^{-x}$.

As for your approach, beyond what @user1337 rightfully said, note that we can't just decide that everything (asymptotically-)smaller than $\frac{1}{x}$ yields a converging integral, because (for example) $$\int_2^\infty\frac{dx}{x\log x}=\infty.$$ However, it is true that $$\int_2^\infty\frac{dx}{x(\log x)^{1+\epsilon}}<\infty,$$ so every function which grows at least as fast as $(\log x)^{1+\epsilon}$ (which includes, as we already know by the standard $p$-test, $x^\epsilon$) would do, so $e^x$ surely works.

Answer 3

Using the Taylor expansion was fine, the problem is in the "leading order" approach.

You chose to express $\frac{1}{e^x}$ as an infinite sum - you cannot truncate it for no reason.

Answer 4

The Taylor expansion works. Note from $e^x=1+x+\frac{x^2}{2!}+\cdots$, we have $e^x\gt x$ for positive $x$.

It follows that $$\frac{1}{xe^x}\lt \frac{1}{x^2}.$$ But we know that $\displaystyle\int_2^\infty \frac{1}{x^2}\,dx$ converges, and therefore by Comparison so does our integral.

As has been pointed out, there are better ways to handle the problem without using the Taylor series.