# Comparison Test Question?

1. Jun 14, 2010

1. The problem statement, all variables and given/known data

I'm just curious as to how to think about the following form of equation.

2. Relevant equations
$$\int_{3}^{\infty } \frac{1}{x + e^x} \,dx$$

3. The attempt at a solution

What you're trying to do is to test it;

$$\frac{1}{x \ + \ e^x} \ < \ \frac{1}{x}$$

$$\frac{1}{x}$$ diverges

$$\frac{1}{x \ + \ e^x} \ < \ \frac{1}{e^x}$$

$$\lim_{t \to \infty} \int_{3}^{t} e^{-x}\,dx \ = \ \lim_{t \to \infty} - e^{-x} | \ _3 ^t \ = \ \lim_{t \to \infty} - {\frac{1}{e^t} \ + \ \frac{1}{e^3}$$

so this converges to $$\frac{1}{e^3}$$.

I don't get how this means the original eq. will also converge?

Both $$\frac{1}{x}$$ and $$\frac{1}{e^x}$$ are bigger than the original eq. with one converging and the other diverging.

$$\frac{1}{x}$$ is bigger than $$\frac{1}{e^x}$$ but the test of $$\frac{1}{x^p}$$ is ringing in my ears as a kind of explanation, but I am confused.

Is there an easy way to link all of this together?

2. Jun 14, 2010

### Tedjn

Indeed, because for [strike]positive x[/strike] x > 1 when p > 1, 1/xp is always less than 1/x. In effect, its graph has less area underneath it for the integration. So there is no contradiction that 1/x leaves just too much area, i.e. it diverges, while 1/xp for p > 1 does not, i.e. it converges. Meanwhile, 1/ex just hugs the x-axis massively more than 1/xp for any constant p, hence it of course converges. Finally, 1/(x+ex) lies even closer to the x-axis and leaves even less area; it converges.

Knowing 1/x diverges does not mean we know 1/(x+ex) diverges or converges. The graph of the latter has less area, but by how much? Enough to stop the integral from diverging? However, knowing that 1/ex converges does tell us that 1/(x+ex) converges, since it has less area than a finite amount.

EDIT: It occurred to me that I also need x > 1, not just x positive. Sorry for the mistake.

Last edited: Jun 14, 2010
3. Jun 14, 2010