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Math Help - Definite Integration

  1. #1
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    Definite Integration

    Hi,

    Could somebody please help me in getting a closed form expression for integrating the following expression:

    (1+(x^m))^n

    limits of integration are from 0 to 1. Integration is to be done with respect to x and m and n are positive real numbers.

    Is it to do with the incomplete beta function, but the upper limit is coming out to be -1, which should lie in the interval 0 to 1.

    Thanks in anticipation.

    Dekar
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  2. #2
    Eater of Worlds
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    It appears to be hypergeometric. I ran it through Maple and it gave me.

    \int{(1+x^{m})^{n}}dx=xhypergeom\left([\frac{1}{m},-n],[1+\frac{1}{m}],-x^{m}\right).
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  3. #3
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    Thanks Galactus,

    Can you please help me in getting the expansion of this function?

    Thanks

    Dekar
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  4. #4
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    Quote Originally Posted by dekar View Post
    Thanks Galactus,

    Can you please help me in getting the expansion of this function?

    Thanks

    Dekar
    See here.

    -Dan
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  5. #5
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    Quote Originally Posted by dekar View Post
    Hi,

    Could somebody please help me in getting a closed form expression for integrating the following expression:

    (1+(x^m))^n

    limits of integration are from 0 to 1. Integration is to be done with respect to x and m and n are positive real numbers.
    I think the following works for n,m>0.

    \int_0^1 (1+x^m)^n dx
    Let t=x^m \implies t' = mx^{m-1}
    Thus,
    \frac{1}{m}\int_0^1 (1+t)^n t^{-1 +\frac{1}{m}}dt= \frac{1}{m}\mbox{B}\left( n+1, \frac{1}{m} \right)
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  6. #6
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    Hi,

    It won't work because it is (1+t), to have it to get reduced to the beta function we need (1-t),

    Thats why I guess, It may go to Incomplete Beta Function if we substitute say p=-t at this stage, but the upper limit then becomes -1.

    Dekar
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  7. #7
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    Quote Originally Posted by dekar View Post
    Hi,

    It won't work because it is (1+t), to have it to get reduced to the beta function we need (1-t),

    Thats why I guess, It may go to Incomplete Beta Function if we substitute say p=-t at this stage, but the upper limit then becomes -1.

    Dekar
    Yes, I made a mistake.
    You can see more here.
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  8. #8
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    So isn't there a better (a less messy) way to arrive at the closed form solution other than that hypergeometric?

    I need it a program friendly way to code this.

    Any help is highly appriciated.

    Thanks

    Dekar
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