1. ## Direct Comparison Theorem

Use the Comparison Theorem to determine whether the integral is convergent or divergent.

So I'm thinking that I can compare that integrand to

$\displaystyle \int_1^\infty \frac{e^{-x}}{x}dx$

If that integral is divergent, then my original integral is also divergent. But, I get stuck on evaluating my substitution

$\displaystyle \frac{2 + e^{-x}}{x} \geq \frac{e^{-x}}{x} dx}$

$\displaystyle \lim_{t \to \infty} \int_1^t \frac{e^{-x}}{x} dx$

$\displaystyle let u = e^{-x} du = -e^{-x}dx$
$\displaystyle let dv = \frac{dx}{x} v = ln(x)$

$\displaystyle \lim_{t \to \infty} \int_1^t \frac{e^{-x}}{x} dx = \lim_{t \to \infty} e^{-x}\ln{x} + \int_1^t e^{-x}}\ln{x} dx$

And I'm not getting anywhere... Can anybody suggest something to get me on the right path? Thanks.

2. ## Re: Direct Comparison Theorem

note that $\displaystyle \frac{2+e^{-x}}{x} > \frac{1}{x}$ for $\displaystyle x > 1$

and, $\displaystyle \int_1^\infty \frac{1}{x} \, dx$ diverges ...

Thanks!