1. ## Improper Integrals

Hey I am having trouble with improper integrals. I have like 5-6 problems that I am just super stuck on. I have read the chapter like 2-3 times and still don't understand it.

The problems:

1) int -2 to infinity 2 dx / (x^2-1)

this I think can be broken up into two integrals

2/(x+1) + 2/(x-1) and then these are both du/u problems.

so we have

lim b-> infinity int -2 to b 2/(x+1) + 2/(x-1)

now I am stuck.

2)

int -infinity to -2 2 dx/(x^2-1)

can anyone give me a shove in the right direction please?

3)

int -infinity to infinity 2x dx/(x^2+1)^2

another du/u Im assuming

but how do you deal with a -infinity to infinity improper integral

I would greatly appreciate any help to get these started

2. Originally Posted by ur5pointos2slo
Hey I am having trouble with improper integrals. I have like 5-6 problems that I am just super stuck on. I have read the chapter like 2-3 times and still don't understand it.

The problems:

1) int -2 to infinity 2 dx / (x^2-1)

this I think can be broken up into two integrals

2/(x+1) + 2/(x-1) and then these are both du/u problems.

so we have

lim b-> infinity int -2 to b 2/(x+1) + 2/(x-1)

now I am stuck.

2)

int -infinity to -2 2 dx/(x^2-1)

can anyone give me a shove in the right direction please?

3)

int -infinity to infinity 2x dx/(x^2+1)^2

another du/u Im assuming

but how do you deal with a -infinity to infinity improper integral

I would greatly appreciate any help to get these started
Lets take a look at the first integrand

We will simplify by partial fractions

$\displaystyle \frac{2}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}$

$\displaystyle 2 =A(x+1)+B(x-1)$

if x =1 $\displaystyle 2=A(1+1) \implies A=1$

if x = -1 $\displaystyle 2 =B(-1-1) \implies B=-1$

So now we have the integral

$\displaystyle \int_{-2}^{\infty}\frac{1}{x-1}-\frac{1}{x+1}dx$

This integral is double improper.

so we need to break it up into a few integrals

$\displaystyle \int_{-2}^{\infty}\frac{1}{x-1}dx-\int_{-2}^{\infty}\frac{1}{x+1}dx$

Since the bound of the integral has a zero of the denominator we need to break each one up into two integrals

I will do the first one

$\displaystyle \lim_{a \to 1^-}\int_{-2}^{1}\frac{1}{x-1}dx+\lim_{c \to 1^+}\lim_{b \to \infty}\int_{c}^{b}\frac{1}{1-x}dx$

Each of these is a U sub I hope this helps

3. Originally Posted by ur5pointos2slo
Hey I am having trouble with improper integrals. I have like 5-6 problems that I am just super stuck on. I have read the chapter like 2-3 times and still don't understand it.

The problems:

1) int -2 to infinity 2 dx / (x^2-1)

this I think can be broken up into two integrals

2/(x+1) + 2/(x-1) and then these are both du/u problems.

so we have

lim b-> infinity int -2 to b 2/(x+1) + 2/(x-1)

now I am stuck.

2)

int -infinity to -2 2 dx/(x^2-1)

can anyone give me a shove in the right direction please?

3)

int -infinity to infinity 2x dx/(x^2+1)^2

another du/u Im assuming

but how do you deal with a -infinity to infinity improper integral

1.) $\displaystyle 2\int_{-2}^{\infty} \frac{dx}{x^2-1}$ is undefined, as $\displaystyle \frac{1}{x^2-1}$ is not defined at $\displaystyle x=1$
2.) The same goes for $\displaystyle 2\int_{-\infty}^{-2}\frac{dx}{x^2-1}$ because $\displaystyle \frac{1}{x^2-1}$ is not defined at $\displaystyle x=-1$
3.) Let $\displaystyle u=x^2+1$. Thus $\displaystyle du=2x\,dx$. Your integral is now $\displaystyle \int\frac{du}{u^2} = -\frac{1}{u}$.
Plugging back in $\displaystyle x^2+1$ for $\displaystyle u$, we have $\displaystyle \lim_{b\to\infty}\left[-\frac{1}{x^2+1}\right|_{-b}^b = 0$