Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Whats the name of this function?

  1. Oct 12, 2008 #1
    this site:
    gives the integral of this function:


    but the answer contains some function I've never heard of before and cant find in the documentation.

    it looks like 2F1 (6.5, 1; 7.5; 8.02014*10^-11x)

    whats the name of this function?
  2. jcsd
  3. Oct 12, 2008 #2
    Well right at the bottom of the page that gives the anti-derivative, you have two links to the function named Hypergeometric2F1 (Hypergeometric Function).
  4. Oct 13, 2008 #3


    User Avatar
    Science Advisor

    Good site.

    I just tried [tex]\int \frac{x^{11/2}}{1-x}[/tex] and I got a result in terms of only standard functions (just log sqrt and polymonials).

    Your integral can easily be put in the above form with a change of variable so I'm not sure why you got a Hypergeometric?

    BTW. The result I got was :

    [tex]\frac{-2 \sqrt{x} \, (3465 + 1155 x + 693 x^2 + 495 x^3 + 385 x^4 + 315 x^5 )} {3465}\, - \, \log(-1 + \sqrt{x}) \,+\, \log(1 + \sqrt{x}) [/tex]

    Note that I entered your integral (or one trivially close to it) in the form that I thought would be least likely to confuse the program. I find that this is usually a good idea if you're hoping to get an answer in it's simplest form.

    BTW. Just substitute x = 1.25E+10 u to put your itegral into the above form.
    Last edited: Oct 13, 2008
  5. Oct 13, 2008 #4


    User Avatar
    Science Advisor

    BTW: I just checked and it was only the 11/2 versus the 5.5 that was needed to give the simpler result.

    Also it seems that there's plenty of factors that cancel in the above.
    [tex] -2 \sqrt{x} \, (1 \,+\, \frac{1}{3}\, x \,+\, \frac{1}{5}\, x^2 \,+\, \frac{1}{7}\, x^3 \, +\, \frac{1}{9}\, x^4 \, +\, \frac{1}{11} x^5 ) - \, \log(-1 + \sqrt{x}) \,+\, \log(1 + \sqrt{x}) [/tex]
    Last edited: Oct 13, 2008
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook