Determine whether this improper integral converges or not

Question

The expression is: $$\int_1^\infty \frac{1}{x-e^{-x}}dx$$ I know that to prove if the the integral converges, it must have a limit and if it diverges it doesn't. I have gotten to this bit: $$\int_1^\infty \frac{1}{x-e^{-x}}dx=\lim_{t\to \infty}\int_1^t\frac{1}{x-e^{-x}}dx$$ But I have no idea how to integrate $\int_1^t\frac{1}{x-e^{-x}}dx$ to find the limit to see if it converges or not. Are there any tests I can use instead to see if it converges or not?

Answer 1

We have that

$$\frac{1}{x-e^{-x}}\sim \frac 1x$$

therefore the given integral diverges by limit comparison test with $\int_1^\infty \frac{1}{x}dx$.

As an alternative note that

$$\frac{1}{x-e^{-x}}\ge \frac 1x$$

then the given integral diverges also by comparison test.

Answer 2

This is not a "detailed proof" but I feel that it is sufficient. As mentioned, using a comparison test would give you your answer. For example consider the function: $g(x) = \frac{1}{x+1}$. It is easy to see that $$\frac{1}{x + 1}\leq\frac{1}{x-e^{-x}}$$ for all $x \geq 1$. We know that $$\int_1^{\infty}\frac{1}{x + 1}dx = \infty.$$ Because your function $\frac{1}{x-e^{-x}}$ is bounded below by a integral that diverges, it also diverges, which means:$$\int_1^{\infty}\frac{1}{x-e^{-x}}dx=\infty.$$