1. ## Converging integrals

Determine whether or not the following intervals converge,showing your reasoning

a)

b)

2. Hello,

The first one can be compared to a Riemann integral :

$\int x^{-\frac{1}{2}} dx = \int \frac{1}{x^{\frac{1}{2}}} dx$

Which diverges.

You can also remember that an antiderivative of $x^n$ is $\frac{x^{n+1}}{n+1}$.

Here, $n=\frac{-1}{2}$

So $\int_0^\infty x^{-\frac{1}{2}} dx=\frac{1}{1-\frac{1}{2}} [x^{1/2}]_0^\infty$

Which also diverges.

An antiderivative for $\frac{1}{x^2+1}$ is arctan function.

$\int_1^\infty \frac{1}{x^2+1} dx=[Arctan(x)]_1^\infty=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$

Edit : 5th post ^^

3. For the second one observe that $\frac{1}
{{1 + x^2 }} \le \frac{1}
{{x^2 }},$
hence the integral converges by direct comparison.

4. Originally Posted by Krizalid
For the second one observe that $\frac{1}
{{1 + x^2 }} \le \frac{1}
{{x^2 }},$
hence the integral converges by direct comparison.
You need to be careful here. You should actually say,
$0\leq \frac1{1+x^2}\leq \frac1{x^2}$.

You need the $\geq 0$ part, I am sure you can see why.

Or a stronger way to say this is,
$\left| \frac1{1+x^2}\right| \leq \frac1{x^2}$.

In general if $|f(x)|\leq g(x)$ and the integral of $g(x)$ converges then $f(x)$ integral will converge also.