# Thread: Stuck on Another Direct Comparison Integral

1. ## Stuck on Another Direct Comparison Integral

I don't seem to be able to do these very well.

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

$\displaystyle \int_0^1 \frac{e^{-x}}{\sqrt{x}}dx$

I see two functions that might be of assistance:

$\displaystyle on [0,1], e^{-x} \leq 1$ so $\displaystyle \frac{e^{-x}}{\sqrt{x}} \leq \frac{1}{\sqrt{x}}$

But that won't help because $\displaystyle \int_0^1 \frac{1}{\sqrt{x}}dx$ is divergent, so it says nothing about a function that is less than it.

Then, I looked at the fact that on [0,1] $\displaystyle \sqrt{x} \leq 1$ so then $\displaystyle \frac{e^{-x}}{\sqrt{x}} \geq e^{-x}$. But $\displaystyle \int e^{-x}$ is a proper integral, so it says nothing about an integral that is bigger than it.

AAAAAGH. I can't seem to get any of these figure out. What am I doing wrong?

2. ## Re: Stuck on Another Direct Comparison Integral

Originally Posted by joatmon

I don't seem to be able to do these very well.

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

$\displaystyle \int_0^1 \frac{e^{-x}}{\sqrt{x}}dx$

I see two functions that might be of assistance:

$\displaystyle on [0,1], e^{-x} \leq 1$ so $\displaystyle \frac{e^{-x}}{\sqrt{x}} \leq \frac{1}{\sqrt{x}}$

But that won't help because $\displaystyle \int_0^1 \frac{1}{\sqrt{x}}dx$ is divergent, so it says nothing about a function that is less than it.

Then, I looked at the fact that on [0,1] $\displaystyle \sqrt{x} \leq 1$ so then $\displaystyle \frac{e^{-x}}{\sqrt{x}} \geq e^{-x}$. But $\displaystyle \int e^{-x}$ is a proper integral, so it says nothing about an integral that is bigger than it.

AAAAAGH. I can't seem to get any of these figure out. What am I doing wrong?
for $\displaystyle x \in [0,1]$ we have $\displaystyle 1 \ge e^{-x}\ge e^{-1}$

Therefore for $\displaystyle \varepsilon \in (0,1]$:

$\displaystyle \int_{\varepsilon}^1 \frac{1}{\sqrt{x}}dx\ge \int_{\varepsilon}^1 \frac{e^{-x}}{\sqrt{x}}dx \ge \int_{\varepsilon}^1 \frac{e^{-1}}{\sqrt{x}}dx$

Hence the integral converges/diverges as $\displaystyle \int_{0}^1 \frac{1}{\sqrt{x}}dx$ converges/diverges

CB