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Integrating a two variable equation

  1. Aug 2, 2010 #1
    1. The problem statement, all variables and given/known data

    This is the problem I was given:

    [tex]I_{n} = \int^{0}_{\infty}(1 + x^{2})^{-n} dx[/tex]

    I was told to "deduce that"

    [tex]I_{n} = 2n(I_{n} - I_{n + 1})[/tex]

    so I can "Hence or otherwise show that"

    [tex]\int^{0}_{\infty}(1 + x^{2})^{-4} dx = \frac{5\pi}{32}[/tex]

    2. Relevant equations

    I don't even know what I am being asked to do. I have relegated myself to failing this problem. Originally, I figured I would just try to find the original function of the integrand and see if that left me somewhere that made more sense, but I can't find anything in my notes that explains how to solve this with two variables. I am not looking for an answer, but rather maybe a hint as to where I should be looking for an integration technique.

    3. The attempt at a solution

    The first thing that I gather is that I have to separate the parts, so I could do something like: if

    [tex] z = 1 + x^{2} [/tex]

    then,

    [tex] dz = 2x dx [/tex]

    and,

    [tex]I_{n} = \int^{0}_{\infty}z(x)^{-n} dz [/tex]


    That obviously looks like something that came out of the chain rule, so I first go backwards on the power using [tex] \int a^{x} = \frac{a^{x+1}}{x+1} [/tex]

    which combined with the chain rule fives me,
    [tex]\int z(x)^{-n} dz = \frac{z(x)^{1-n}}{1-n}[/tex]

    So far, so good. I know the integral of [tex](1 + x^{2})[/tex] is [tex](x + \frac{x^{3}}{3})[/tex]

    and here is where it all falls apart. I have no idea how to put those two parts together, and I don't know what to review / re-read to figure it out. Can anyone just at least tell me what kind of problem this is so I know what I am supposed to be searching for? As an econ student my calc background is very basic, we never had to deal with this sort of thing.
     
  2. jcsd
  3. Aug 2, 2010 #2

    Mark44

    Staff: Mentor

    All of your limits of integration are upside-down. The integrals should look like this:
    [tex]I_{n} = \int_0^{\infty}(1 + x^{2})^{-n} dx[/tex]

    This (above) isn't right. It looks like you just replaced dx with dz, but dz = 2x dx. That 2x has to come from somewhere.

    Instead of an ordinary substitution like you tried, a trig substitution might be what is called for.
     
  4. Aug 2, 2010 #3

    rock.freak667

    User Avatar
    Homework Helper

    Try using integrating by parts with u=(1+x2)-n.


    After which, you will need to use something along the lines of x= x+9-9 :wink:
     
  5. Aug 2, 2010 #4
    Yikes. That was a latex typo, my apologies. Thank you for pointing out that dx error, i was under the impression that the 2xwould come from the derivative of [tex]x^{2}[/tex]. I am learning this all from a set of notes and some videos, all by myself (no teacher or classmates) so it feels very overwhelming.

    I thought about the trig substitution too. I remember that:
    [tex] f(x) = tan^{-1}(x) [/tex]
    and
    [tex]f'(x) = \frac{1}{1+x^{2}} [/tex]


    but I was unable tie that part together to n part. I get that [tex]\frac{1}{(1+x^{2})^{n}}[/tex] could just be expanded to [tex](\frac{1}{(1+x^{2})})_{n} (\frac{1}{(1+x^{2})})_{n-1} \cdots (\frac{1}{(1+x^{2})})_{0}[/tex]

    but that still leaves me pretty confused.

    At the core of it, I recognize that my calculus is poor. In fact, I have absolutely no clue what to do with the "n" term, I've never integrated something with two variables, so I don't know the special rules or formulas it takes to combine them. Looking through the notes I have on integrals, there's nothing with more than one variable in it, so I don't know where to look for more information. is that what they call partial differential equations? If I'm beyond help on this one, how about a suggestion for a textbook?
     
  6. Aug 2, 2010 #5

    rock.freak667

    User Avatar
    Homework Helper

    Think of 'n' as a constant rather than a variable. If you try integration by parts, it works out a bit easier than the trig. substitution IMO.
     
  7. Aug 2, 2010 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Oftentimes, when you're asked to deduce a recurrence relation involving integrals, integration by parts, as rock.freak suggested, is a good place to start.
     
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