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Math Help - Improper Integrals

  1. #1
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    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

    another shove please?

    I would greatly appreciate any help to get these started
    Last edited by ur5pointos2slo; April 26th 2009 at 07:56 PM.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by ur5pointos2slo View Post
    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

    another shove please?

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

    We will simplify by partial fractions

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

    2 =A(x+1)+B(x-1)

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

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

    So now we have the integral

    \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

    \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

    \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
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  3. #3
    Super Member redsoxfan325's Avatar
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    Swampscott, MA
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    Quote Originally Posted by ur5pointos2slo View Post
    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

    another shove please?

    I would greatly appreciate any help to get these started
    1.) 2\int_{-2}^{\infty} \frac{dx}{x^2-1} is undefined, as \frac{1}{x^2-1} is not defined at x=1

    2.) The same goes for 2\int_{-\infty}^{-2}\frac{dx}{x^2-1} because \frac{1}{x^2-1} is not defined at x=-1

    3.) Let u=x^2+1. Thus du=2x\,dx. Your integral is now \int\frac{du}{u^2} = -\frac{1}{u}.

    Plugging back in x^2+1 for u, we have \lim_{b\to\infty}\left[-\frac{1}{x^2+1}\right|_{-b}^b = 0
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