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For a class on Monte-Carlo simulation methods I want to do a comparison of Monte-Carlo integration and numerical integration on sparse grids. For this purpose I need a function $f$ defined over the $n$-dimensional hypercube $[-1,1]^n$ of which I know the value of the integral $\int_{[-1,1]^n} f$ in order to compare the calculation errors of the different methods. I thought about integrating the unit $n$-sphere as this is defined over the hypercube and the exact volume of the unit $n$-sphere is given by

$$V_n=\frac{\pi^{\frac{n}{2}}}{\Gamma (\frac{n}{2} +1)}$$

Now integrating the whole $n$-sphere will result in the integral being 0 as it is symmetric. That's why I thought about using only the upper unit $n$-sphere. Then the value of the integral would be given by $V_n/2$. However, I did not manage to come up with a function describing this upper half of the $n$-sphere. In the 2-dimensional case the upper sphere $U_2$ would be given by $U_2=\left\{ x_2= \sqrt{1-x_1^2},x_2\geq 0\right\}$ or in the more general case of the unit $n$-ball $U_n=\left\{ \sqrt{1-\sum_{i=1}^{n-1}{x}_{i}^{2}},x_n\geq 0 \right\} $.

Is there a way to express $U_n$ as a function that I could plug into my implemented integration algorithms to integrate it over the $n$-dimensional hypercube?

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    $\begingroup$ The assertion "[the unit $n$-sphere] is defined over the hypercube..." is not true if $n \geq 2$. (The "shadow" of an ordinary hemisphere is a disk, not a square.) Separately, it's not clear what you mean by "integrating the whole $n$-sphere will result in the integral being $0$ as it is symmetric". $\endgroup$ Commented Jan 19, 2017 at 21:22
  • $\begingroup$ So on which set is the unit $n$-sphere defined? $\endgroup$
    – YukiJ
    Commented Jan 19, 2017 at 21:26
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    $\begingroup$ I think you want: The unit sphere is the union of two graphs $x_{n+1} = \pm\sqrt{1 - \|x\|^{2}}$ over the unit ball, the set of $x = (x_{1}, \dots, x_{n})$ in $\mathbf{R}^{n}$ where $\|x\|^{2} \leq 1$. $\endgroup$ Commented Jan 19, 2017 at 21:39
  • $\begingroup$ @YukiJ: Offering a bounty but not awarding it is like making a promise but not keeping it: a very ugly kind of behaviour. $\endgroup$
    – Alex M.
    Commented Jan 28, 2017 at 9:18
  • $\begingroup$ I am sorry @AlexM. But I Received an e-mail that the bounty will be automatically given to the best answer? Why has this not happened?! $\endgroup$
    – YukiJ
    Commented Jan 28, 2017 at 9:20

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Why don't you try polynomials - the simplest of the non-trivial examples? Choose a degree $d \in \Bbb N$ and a set of coefficients $\{a_{i_1, \dots, i_n} \mid i_1 + \dots + i_n \le d\} \subset \Bbb C$ and consider the associated polynomial $p(x_1, \dots, x_n) = \sum _{i_1 + \dots + i_n \le d} a_{i_1, \dots, i_n} x_1 ^{i_1} \dots x_n ^{i_n}$. Then

$$\int \limits _{[-1,1]^n} p(x_1, \dots, x_n) \ \Bbb d x_1 \dots \Bbb d x_n = \int \limits _{-1} ^1 \dots \int \limits _{-1} ^1 \sum _{i_1 + \dots + i_n \le d} a_{i_1, \dots, i_n} x_1 ^{i_1} \dots x_n ^{i_n} \ \Bbb d x_1 \dots \Bbb d x_n = \\ \sum _{i_1 + \dots + i_n \le d} a_{i_1, \dots, i_n} \frac {1 + (-1)^{i_1}} {i_1 + 1} \dots \frac {1 + (-1)^{i_n}} {i_n + 1} ,$$

a formula simple enough to let you compare its value with the one that you obtain numerically.

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