Question

improper integral question

determine whether the improper integral converges.

\[\int^3_1 \frac{x^2+2x+3}{x(x-1)^2}\text{d}x\]

Answer 1

ok so I can see that it is improper at \(x=1\).

Answer 2

and as \(x\rightarrow 1\), the integrand behaves like \[\frac{6}{(x-1)^2}\]

Answer 3

so if I can show that \[\int^3_1 \frac{x^2+2x+3}{x(x-1)^2}\text{d}x\quad\text{ behaves like }\quad\int^3_1 \frac{6}{(x-1)^2}\text{d}x\] I can say that it diverges right?

Answer 4

now how would I show this? my solution does this step: \[\lim_{x\rightarrow1}\frac{\frac{x^2+2x+3}{x(x-1)^2}}{\frac{6}{(x-1)^2}}=\lim_{x\rightarrow 1}\frac{x^2+2x+3}{6(x-1)^2}=1\in(0,\infty)\]what is this step accomplishing?

Answer 5

You can approach this more analytically rather than using some "approximate" integral. You could you partial fractions to break up the integral and integrate those individually which should be much easier.

Answer 6

Are you familiar with the limit comparison test for series? That last step looks a lot like the procedure for the test. http://en.wikipedia.org/wiki/Limit_comparison_test

Answer 7

I think it's a typo in that last line. but I can see that it equals 1.

Answer 8

I am familiar with the "idea" of the test. I just have yet to find something that really properly shows me how to use it. where in that wikipedia article does it say to take that limit?

Answer 9

Do you have to use this method to determine its convergence?

Answer 10

I suppose it doesn't, but if you can show that (near \(x=1\)) the two functions are similar enough, then you can make a conclusion about the first integral's divergence based on the fact that the second (comparison) integral diverges. For this particular problem, you need only consider the neighborhood of \(x=1\) because the end behavior of both integrands is apparent (i.e., \(\lim_{x\to\infty}(\text{integrand})=0\)).

Does that make sense? I'm not sure how else to put it.

Answer 11

Much less if I'm even putting it correctly...

Answer 12

Use partial fraction to get \[\frac{6}{(-1+x)^2}-\frac{2}{(-1+x)}+\frac{3}{x}\] and just the integrating 2/(-1+x) is clear that it won't converge

Answer 13

Since you'd get \[2 ln(x-1)\big|_1^3\] which doesn't exist when evaluating it for x=1 (ln(0) is undefined

Answer 14

hmm, well I would like to at least learn what is going on in this method.

Answer 15

and yes that does make sense siths and giggles. what I am struggling with is how to show that the functions behave the same way. how is taking the limit of the integrand over the comparison integrand showing that the integrals behave the same way?

Answer 16

Ok well I found this: http://math.duke.edu/~cbray/Stanford/2002-2003/Math%2042/limitcomp.pdf
k so it turns out this method is similar to the limit comparison test for infinite series

Answer 17

The idea is: you want to compare the integrand with one in which YOU KNOW converges or diverges. Then when taking the ration of that with your original integrand, that limit must be finite. If that's the case, the original integral:
1) converges if the one you KNEW actually does converge
2) diverges if the one you KNOW actually diverges

Answer 18

the integral 6(x-1)^2 is easy to verify that it diverges (with bounds from x to 3), so they that the ratio of that with the original integrand. Since he limit of this ratio was finite (more specifically, the limit belongs in the open interval (0, infinity), then we conclude the integral diverges

Answer 19

You usually make the comparison with the "dominant" part of the integral, but for which you know it converges/or diverges

Answer 20

Use comparison test.
Show that for some interval around 1, the integrand is bounded above by another function which more obviously converges.

Answer 21

yes, I get that. i don't understand how to do that.

Answer 22

so shouldn't it actually be \[\lim_{x \to 1}\frac{x^2+2x+3}{6x}\]

Answer 23

What function do you want to compare it to? Got any ideas?

Answer 24

uh probably a power function?

Answer 25

(x-1)^3?

Answer 26

How about a function like \[
\frac{(x-1)^2}{x(x-1)^2}
\]

Answer 27

You can determine some \(c\) \[
\frac{c(x-1)^2}{x(x-1)^2}
\]So that it is larger near \(1\)

Answer 28

ok, just a second

Answer 29

Hmmmm
Maybe you can compare it to a straight up constant.

Answer 30

Okay here is a decent function something like \[
\frac{c}{x(x-1)^2}
\]Do you think you could integrate that?

Answer 31

I don't see why I couldn't. but tbh I have no idea where these functions are coming from...

Answer 32

Basically we want a function that is easier to integrate. Does that make sense?

Answer 33

sure but that integral is still improper on [1,3]

Answer 34

The fact that it is improper is not a problem.

Answer 35

But that's the point. You want something that you know will converge or diverge, improper or not

Answer 36

If you can integrate a function, then you can tell if it converges/diverges.

Answer 37

ok but this integral to me seems like I would want to use partial fractions. it's not much simpler than the one I had before...

Answer 38

If you can integrate the one you had before. then just integrate it.

Answer 39

I would really like to learn the method I was asking about earlier though! I seem to be on a wild goose chase here haha! thanks for your help though.

Answer 40

The idea behind the method is to find a function that you know will bound it above/below and is easier to integrate. That's all you need to know.

Answer 41

thanks, that's Folland tells me at all as well!

but on my mid term that I failed I wrote "need to find a function that bounds above"

Answer 42

not even joking. my problem is finding that function easily.

Answer 43

I got one mark out of 10 at least because I guessed correctly haha.

Answer 44

maybe my book from calc 1,2 and 3 will shed some light onto this. sucks to be stuck on a problem that I HAVE A SOLUTION for this long haha

Answer 45

Well, first of all, if we want to bound about we need a function that has \((x-1)^2\) in the denominator. That is how you make sure it goes to infinity as \(x\) approaches 1.

Answer 46

I agree.

Answer 47

So \[
\frac{c}{(x-1)^2}
\]Is a good candidate. We'd let \[
c = \max \left[\frac{x^2+2x+3}{x}\right]
\]

Answer 48

We'd want to max it on some interval.

Answer 49

so 18/3=6

Answer 50

that part seems intuitive to me.

Answer 51

How do you get those numbers?

Answer 52

isn't 3 clearly the max on the interval we are talking about?

Answer 53

Take the derivative of it: \[
\left[\frac{x^2+2x+3}{x}\right]' = \frac{(2x+2)x-(x^2+2+4)}{x^2}
\]

Answer 54

If we wanna be rigorous we gotta find the critical numbers.

Our interval is \([1,3]\)

Answer 55

So we know \(1\) and \(3\) are critical number candidates, but there might be another.

Answer 56

\[

\left[\frac{x^2+2x+3}{x}\right]'
= \frac{2x^2+2x-(x^2+2x+4)}{x^2}
= \frac{x^2+4}{x^2 }

\]Okay, fortunately there are no roots here. So yeah, \(1\) and \(3\) are the only candidates.

Answer 57

Whoops it was \[
\frac{x^2-4}{x^2}=\frac{(x-2)(x+2)}{x^2}
\]So \(2\) is a also a critical number.

Answer 58

uh no?

Answer 59

Why not?

Answer 60

http://www.wolframalpha.com/input/?i=+d%2Fdx++%28x%5E2%2B2x%2B3%29%2Fx

Okay fine, the other candidate is \(\sqrt{3}\)

Answer 61

right. I know that the maximum is at 3 though, or 1 I guess

Answer 62

either way it is 6.

Answer 63

Okay then let's let \(c = 100\)

Answer 64

why not 6?

Answer 65

Because 100 is even higher.

Answer 66

alright so i'm comparing it to \[\frac{100}{(x-1)^2}\]

Answer 67

wait. why wouldn't I compare it to something smaller?

Answer 68

You compare it to something smaller to show it diverges. You compare it to something larger to show it converges.

Answer 69

exactly, so I need something smaller
\[\frac{1}{(x-1)^2}\]

Answer 70

Okay then do that.

Answer 71

nope. still fricken lost.

I just wish there was a step by step thing.

l understand I need to compare it. I know how to do partial fractions, I know hwo to minimize and maximize functions on intervals. I just don't understand a step-by step, brainless, test-taking way to find the correct function. Either way I am giving up for now. my exam is in 7 days and I just wasted 3 hours getting absolutely nowhere. thank you for your help though man. I honestly do appreciate it.

Answer 72

NEVER try to learn from "Folland- Advanced Calculus". Worst book ever written.

Answer 73

The thing is that you kind of "eyeball" your chosen function When doing tests of convergence for series (or integrals like here), there isn't always a "step-by-step" way in the sense that there is a method that will give you the EXACT process, using the exact same comparisons and whatnot.

Answer 74

In fact

Answer 75

This could not be

Answer 76

be

Answer 77

This could not be more true when solving second-order differential equations in which you choose "trial functions" that you hope will work when finding the answer,