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Let a function $f(x)$ be continuous and strictly positive on $[0, 1]$. How to compute the following integral : $$\idotsint_{[0, 1]^n} \frac{f(x_1) + \dots + f(x_k)}{f(x_1) + \dots + f(x_n)} \,dx_1\cdots dx_n, ~k = 1,2,\ldots, n$$

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closed as off-topic by levap, 6005, Dominik, Claude Leibovici, E. Joseph Dec 22 '16 at 13:35

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Start with this: $$ \begin{align} & \idotsint_{[0, 1]^n} \frac{f(x_1) + \dots + f(x_k)}{f(x_1) + \dots + f(x_n)} \,dx_1\cdots dx_n \\[10pt] = {} & \idotsint_{[0, 1]^n} \frac{f(x_1)}{f(x_1) + \dots + f(x_n)} \,dx_1\cdots dx_n + \cdots \cdots \\[6pt] & \qquad\qquad\qquad\qquad {} \cdots \cdots + \idotsint_{[0, 1]^n} \frac{f(x_k)}{f(x_1) + \dots + f(x_n)} \,dx_1\cdots dx_n. \end{align} $$ Then I'd ask whether there's a difference between the values of the $k$ terms in that last sum of integrals. If not, then the sum is just $k$ times the first term in the sum of $k$ terms. And if $k=n$, then you have $$ \idotsint_{[0, 1]^n} \frac{f(x_1) + \dots + f(x_n)}{f(x_1) + \dots + f(x_n)} \,dx_1\cdots dx_n = \idotsint_{[0, 1]^n} 1 \,dx_1\cdots dx_n, $$ and that's easy to evaluate. So each of the integrals in that sum of integrals would be $1/n$ times the integral that's easy to evaluate, so finally we'd get $k/n$ times the integral that's easy to evaluate.

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  • $\begingroup$ What if there is no difference between the values of the k terms in that sum of integrals? $\endgroup$ – Parket Dec 13 '16 at 9:35
  • $\begingroup$ @Parket : As I said, that means the sum is just $k$ times one of the integrals. $\endgroup$ – Michael Hardy Dec 13 '16 at 18:34

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