How to calculate this n dimensional integral?

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The integral such as
enter image description here



It's easy to evaluate the first few items



Integrate[x1 (1/(-1+n))^n (1/v1)^(n/(-1+n)) x1^((2-n)/(-1+n)), x1, 0, v1]
Integrate[x1 Integrate[(1/(-1+n))^n (1/v1)^(n/(-1+n)) x1^((2-n)/(-1+n)) x2^((2-n)/(-1+n)), x2, 0, x1], x1, 0, v1]
Integrate[x1 Integrate[Integrate[(1/(-1+n))^n (1/v1)^(n/(-1+n)) x1^((2-n)/(-1+n)) x2^((2-n)/(-1+n)) x3^((2-n)/(-1+n)), x3, 0, x1], x2, 0, x1], x1, 0, v1]


but how do I calculate this integral where the number of integrals is aribtrary?






calculus-and-analysis







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  • Isn't this quantity just equal to$$ I = left( frac1n-1 right)^n nu_1^-n/(n-1) int_0^nu_1 x_1 x_1^(2-n)/(n-1) left[ int_0^x_1 y^(2-n)/(n-1) dy right]^n-1 , dx_1 ? $$
    – Michael Seifert
    1 hour ago















up vote
2
down vote

favorite












The integral such as
enter image description here



It's easy to evaluate the first few items



Integrate[x1 (1/(-1+n))^n (1/v1)^(n/(-1+n)) x1^((2-n)/(-1+n)), x1, 0, v1]
Integrate[x1 Integrate[(1/(-1+n))^n (1/v1)^(n/(-1+n)) x1^((2-n)/(-1+n)) x2^((2-n)/(-1+n)), x2, 0, x1], x1, 0, v1]
Integrate[x1 Integrate[Integrate[(1/(-1+n))^n (1/v1)^(n/(-1+n)) x1^((2-n)/(-1+n)) x2^((2-n)/(-1+n)) x3^((2-n)/(-1+n)), x3, 0, x1], x2, 0, x1], x1, 0, v1]


but how do I calculate this integral where the number of integrals is aribtrary?






calculus-and-analysis







share|improve this question





















  • Isn't this quantity just equal to$$ I = left( frac1n-1 right)^n nu_1^-n/(n-1) int_0^nu_1 x_1 x_1^(2-n)/(n-1) left[ int_0^x_1 y^(2-n)/(n-1) dy right]^n-1 , dx_1 ? $$
    – Michael Seifert
    1 hour ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











The integral such as
enter image description here



It's easy to evaluate the first few items



Integrate[x1 (1/(-1+n))^n (1/v1)^(n/(-1+n)) x1^((2-n)/(-1+n)), x1, 0, v1]
Integrate[x1 Integrate[(1/(-1+n))^n (1/v1)^(n/(-1+n)) x1^((2-n)/(-1+n)) x2^((2-n)/(-1+n)), x2, 0, x1], x1, 0, v1]
Integrate[x1 Integrate[Integrate[(1/(-1+n))^n (1/v1)^(n/(-1+n)) x1^((2-n)/(-1+n)) x2^((2-n)/(-1+n)) x3^((2-n)/(-1+n)), x3, 0, x1], x2, 0, x1], x1, 0, v1]


but how do I calculate this integral where the number of integrals is aribtrary?






calculus-and-analysis







share|improve this question













The integral such as
enter image description here



It's easy to evaluate the first few items



Integrate[x1 (1/(-1+n))^n (1/v1)^(n/(-1+n)) x1^((2-n)/(-1+n)), x1, 0, v1]
Integrate[x1 Integrate[(1/(-1+n))^n (1/v1)^(n/(-1+n)) x1^((2-n)/(-1+n)) x2^((2-n)/(-1+n)), x2, 0, x1], x1, 0, v1]
Integrate[x1 Integrate[Integrate[(1/(-1+n))^n (1/v1)^(n/(-1+n)) x1^((2-n)/(-1+n)) x2^((2-n)/(-1+n)) x3^((2-n)/(-1+n)), x3, 0, x1], x2, 0, x1], x1, 0, v1]


but how do I calculate this integral where the number of integrals is aribtrary?






calculus-and-analysis




calculus-and-analysis






share|improve this question













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asked 1 hour ago









mathe

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  • Isn't this quantity just equal to$$ I = left( frac1n-1 right)^n nu_1^-n/(n-1) int_0^nu_1 x_1 x_1^(2-n)/(n-1) left[ int_0^x_1 y^(2-n)/(n-1) dy right]^n-1 , dx_1 ? $$
    – Michael Seifert
    1 hour ago

















  • Isn't this quantity just equal to$$ I = left( frac1n-1 right)^n nu_1^-n/(n-1) int_0^nu_1 x_1 x_1^(2-n)/(n-1) left[ int_0^x_1 y^(2-n)/(n-1) dy right]^n-1 , dx_1 ? $$
    – Michael Seifert
    1 hour ago
















Isn't this quantity just equal to$$ I = left( frac1n-1 right)^n nu_1^-n/(n-1) int_0^nu_1 x_1 x_1^(2-n)/(n-1) left[ int_0^x_1 y^(2-n)/(n-1) dy right]^n-1 , dx_1 ? $$
– Michael Seifert
1 hour ago





Isn't this quantity just equal to$$ I = left( frac1n-1 right)^n nu_1^-n/(n-1) int_0^nu_1 x_1 x_1^(2-n)/(n-1) left[ int_0^x_1 y^(2-n)/(n-1) dy right]^n-1 , dx_1 ? $$
– Michael Seifert
1 hour ago











1 Answer
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Your integral is equal to
$$
I = left( frac1n-1 right)^n nu_1^-n/(n-1) int_0^nu_1 left x_1 x_1^(2-n)/(n-1) left[ int_0^x_1 y^(2-n)/(n-1) dy right]^n-1 , right dx_1,
$$
which is easy enough to solve for in Mathematica. Note that we must have $n > 1$ for convergence, and things simplify nicely if we assume that $nu_1 > 0$ as well.



subint[x1_, n_] = Integrate[y^((2 - n)/(n - 1)), y, 0, x1, Assumptions -> n > 1]
int[[Nu]1_, n_] = Simplify[(1/(n - 1))^n [Nu]1^(-n/(n - 1))
 Integrate[ x1^((2 - n)/(n - 1) + 1) subint[x1, n]^(n - 1), x1, 0, [Nu]1], Assumptions -> n > 1, [Nu]1 > 0]

(* [Nu]1/(-1 + 2 n) *)





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    Your integral is equal to
    $$
    I = left( frac1n-1 right)^n nu_1^-n/(n-1) int_0^nu_1 left x_1 x_1^(2-n)/(n-1) left[ int_0^x_1 y^(2-n)/(n-1) dy right]^n-1 , right dx_1,
    $$
    which is easy enough to solve for in Mathematica. Note that we must have $n > 1$ for convergence, and things simplify nicely if we assume that $nu_1 > 0$ as well.



    subint[x1_, n_] = Integrate[y^((2 - n)/(n - 1)), y, 0, x1, Assumptions -> n > 1]
    int[[Nu]1_, n_] = Simplify[(1/(n - 1))^n [Nu]1^(-n/(n - 1))
     Integrate[ x1^((2 - n)/(n - 1) + 1) subint[x1, n]^(n - 1), x1, 0, [Nu]1], Assumptions -> n > 1, [Nu]1 > 0]

    (* [Nu]1/(-1 + 2 n) *)





    share|improve this answer
























      up vote
      5
      down vote













      Your integral is equal to
      $$
      I = left( frac1n-1 right)^n nu_1^-n/(n-1) int_0^nu_1 left x_1 x_1^(2-n)/(n-1) left[ int_0^x_1 y^(2-n)/(n-1) dy right]^n-1 , right dx_1,
      $$
      which is easy enough to solve for in Mathematica. Note that we must have $n > 1$ for convergence, and things simplify nicely if we assume that $nu_1 > 0$ as well.



      subint[x1_, n_] = Integrate[y^((2 - n)/(n - 1)), y, 0, x1, Assumptions -> n > 1]
      int[[Nu]1_, n_] = Simplify[(1/(n - 1))^n [Nu]1^(-n/(n - 1))
       Integrate[ x1^((2 - n)/(n - 1) + 1) subint[x1, n]^(n - 1), x1, 0, [Nu]1], Assumptions -> n > 1, [Nu]1 > 0]

      (* [Nu]1/(-1 + 2 n) *)





      share|improve this answer






















        up vote
        5
        down vote










        up vote
        5
        down vote









        Your integral is equal to
        $$
        I = left( frac1n-1 right)^n nu_1^-n/(n-1) int_0^nu_1 left x_1 x_1^(2-n)/(n-1) left[ int_0^x_1 y^(2-n)/(n-1) dy right]^n-1 , right dx_1,
        $$
        which is easy enough to solve for in Mathematica. Note that we must have $n > 1$ for convergence, and things simplify nicely if we assume that $nu_1 > 0$ as well.



        subint[x1_, n_] = Integrate[y^((2 - n)/(n - 1)), y, 0, x1, Assumptions -> n > 1]
        int[[Nu]1_, n_] = Simplify[(1/(n - 1))^n [Nu]1^(-n/(n - 1))
         Integrate[ x1^((2 - n)/(n - 1) + 1) subint[x1, n]^(n - 1), x1, 0, [Nu]1], Assumptions -> n > 1, [Nu]1 > 0]

        (* [Nu]1/(-1 + 2 n) *)





        share|improve this answer












        Your integral is equal to
        $$
        I = left( frac1n-1 right)^n nu_1^-n/(n-1) int_0^nu_1 left x_1 x_1^(2-n)/(n-1) left[ int_0^x_1 y^(2-n)/(n-1) dy right]^n-1 , right dx_1,
        $$
        which is easy enough to solve for in Mathematica. Note that we must have $n > 1$ for convergence, and things simplify nicely if we assume that $nu_1 > 0$ as well.



        subint[x1_, n_] = Integrate[y^((2 - n)/(n - 1)), y, 0, x1, Assumptions -> n > 1]
        int[[Nu]1_, n_] = Simplify[(1/(n - 1))^n [Nu]1^(-n/(n - 1))
         Integrate[ x1^((2 - n)/(n - 1) + 1) subint[x1, n]^(n - 1), x1, 0, [Nu]1], Assumptions -> n > 1, [Nu]1 > 0]

        (* [Nu]1/(-1 + 2 n) *)






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        Michael Seifert

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