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Integrating a quadratic funtion raised to m/n

  1. Oct 30, 2009 #1
    I couldn't figure out what to do with this type of integration [tex]\int (a x^{2}+ bx +c) ^{\frac{m}{n}}dx[/tex] here m, n are integer numbers. Integration limit -∞ to +∞.
    Will Binomial expansion work?
    Please give me some clue.
     
  2. jcsd
  3. Oct 30, 2009 #2
    What makes you think that these integrals converge? The function to be integrated does not even converge to zero as x-> +/- infinity...!
     
  4. Oct 30, 2009 #3
    Actually I had to solve an integral [tex]\int(z^{2} -8z + 36)^\frac{-3}{2} dz [/tex] limit - infinity to + infinity in a physics problem. dz is the differential length of a conducting wire extending from -infinity to + infinity.
    Can you plz help me how can I solve above integral?
     
  5. Oct 30, 2009 #4
    Well, the easiest way is probably to look it up it an table of integrals:smile:

    Wikipedia states that
    826cfb9c30857b16d144b1b5b4c7cc04.png

    where s is the square root of x^2-a^2.....taking m=0, n=1 should help you.


    maybe that is not the bsst way to go.

    I rather suggest the following; use simple substitutions to reduce the integrand to
    [tex]
    (x^2+1)^{-3/2}
    [/tex]

    then use [itex]x=\sinh(z)[/itex] to obtain
    [tex]
    \int_{\mathbb{R}}{\frac{dz}{\cosh(z)^2}}
    [/tex]

    Once you have that, we can think about the next steps:smile:
     
    Last edited: Oct 30, 2009
  6. Oct 30, 2009 #5
    you gave me a good idea to look it up in integral table. I found one http://www.integral-table.com/

    I found quite similar integral in no. 40 in the table's list ( sorry it's quite difficult to write that integral along with its result here) . But How can I obtain that result?
     
  7. Oct 30, 2009 #6
    have you simplified your integral to something which looks similar to
    [tex]
    \int_{\mathbb{R}}{\frac{dz}{\cosh(z)^2}}
    [/tex]

    ?
     
  8. Oct 30, 2009 #7
    I'm puzzled at the substitution method u suggested. 1st would I let,
    z^2- 8 z + 36 = u
    2z -8 =du/dz
    but dz = du/ (2z-8)
    see there's a 'z' in the denominator on right hand side. So, letting z^2- 8 z + 36 = u won't work. What I can do now?
     
  9. Oct 30, 2009 #8
    Well....x^2-8x+36 = (x-4)^2+20

    thus your first substitution would be

    y = x-4.

    then

    w = y/sqrt(20)

    and finally

    sinh(z)= w.
     
  10. Oct 30, 2009 #9
    yeah. Thank u so much . Now I think I can manage the rest of the things.
     
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