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A limit of an integral question

  1. Jan 24, 2008 #1
    http://img225.imageshack.us/my.php?image=29117598dh7.png

    i know that the integral of 0 to 0 interval equals to 0.

    i get 0/0 form

    how do i make the derivative of this integral

    i know the it should cancel out the integral sign
    but there is another variable "t"

    what should i do in this
    case??

    another way that i thought of is
    to solve this integral and then solve the limit
     
    Last edited: Jan 24, 2008
  2. jcsd
  3. Jan 24, 2008 #2
    Yeah just take the derivative of the integral, and the result will be the exact function under the integral sign, but now instead of the variable t, the result should contain the variable x. this is a well known theorem, and it can be proven, but it requres a little work.

    So when u take the derivative of the integral on the denominator the result will be
    [tex]ln(1+x^{2})[/tex]
     
  4. Jan 24, 2008 #3
    By the way, you cannot actually solve the integral, because it does not have an elementary function, you just can express in terms of some gamma function, or just find an approximation for it. But there is no close form of it.
     
    Last edited: Jan 24, 2008
  5. Jan 24, 2008 #4
    Actually this is not true. Integrating by parts you can calulate the integral

    [tex]\int \ln(1+x^2)\,d\,x=x\,\ln(1+x^2)-2\int \frac{x^2}{1+x^2}\,d\,x=x\,\ln(1+x^2)-2\int \frac{1+x^2}{1+x^2}\,d\,x+2\,\int\frac{d\,x}{1+x^2}=x\,\ln(1+x^2)-2\,x+2\,\arctan x[/tex]

    But for the OP, you don't have to do this! :smile:
     
  6. Jan 24, 2008 #5
    Pardone me! I was thinking of completely something else! Yeah, you are so right, it actually is just an easy thing to do, but i had something else in my mind, i mean another function, when i said that. My bad lol, i appologize for not paying so much attention to things!
    Thnx for pointing this out by the way.
     
  7. Jan 24, 2008 #6
    Do not! That's what makes us humans! :smile:
     
  8. Jan 24, 2008 #7

    Gib Z

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    I believe thats a little something called the Fundamental Theorem of Calculus =]
     
  9. Jan 24, 2008 #8
    Well maybe it is, but i am not used with English notation or naming of theorems. But i know how we call it in my native language, and we did not call it The Fundamental Theorem of Calculus, i mean even if we translated this into my language.
     
  10. Jan 24, 2008 #9

    Gib Z

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    O sorry my bad! I didn't know english was not your first language. Your english is very good =]
     
  11. Jan 24, 2008 #10
    Well no need to appologize, and thnx for commenting on my English!
     
  12. Jan 25, 2008 #11
  13. Jan 25, 2008 #12

    Gib Z

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    That indeed is correct.
     
  14. Jan 25, 2008 #13
  15. Jan 26, 2008 #14
  16. Jan 26, 2008 #15

    Gib Z

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    Do you think they'll ever develop a drug that cures my stupidity? I was specifically checking that for errors and I didn't see that. Sigh
     
  17. Jan 26, 2008 #16
    damn chain rule
     
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