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Averaged measure in integrations

Take $\gamma(x)\equiv 1$, $f(x)\equiv x$; for given $n$, $g$, $A$, the first expression is some function $F(x)$, which must equal the second expression for some $B$; using the Cauchy formula, we thus ...
Carlo Beenakker's user avatar
2 votes

Cauchy reduction formula with measure (a variation)

I can reduce the four-fold integral to two integrations: \begin{align} & \int_0^{x_5} ~\int_0^{x_4} \alpha (x_3)~\int_0^{x_3} ~\int_0^{x_2} \alpha (x_1)~F(x_1)\,dx_1 dx_2 dx_3 dx_4\\ %&=\left(...
Carlo Beenakker's user avatar
6 votes

Closed form for $ \int_{0}^{1} \dotsi \int_{0}^{1} \frac{x_1^q + \dotsb + x_n^q}{x_1^p + \dotsb + x_n^p} \, \mathrm{d}x_1 \dotsm \mathrm{d}x_n $

Using the substitution $u_i=x_i^p$ and letting $r:=q/p$ and $a:=1-1/p\in[0,1)$, we see that the integral in question is \begin{align} I&=np^{-n}\int_0^1\cdots\int_0^1\frac{du_1\cdots du_n}{u_1^...
Iosif Pinelis's user avatar
4 votes

Closed form for $ \int_{0}^{1} \dotsi \int_{0}^{1} \frac{x_1^q + \dotsb + x_n^q}{x_1^p + \dotsb + x_n^p} \, \mathrm{d}x_1 \dotsm \mathrm{d}x_n $

Analogously to the answer for $p=1,q=2$ at https://mathoverflow.net/a/289068/11260 , one has for $p=1$ and general $q$ the integral expression $$\int_{0}^{1} \cdots \int_{0}^{1} \frac{x_1^q + \cdots + ...
Carlo Beenakker's user avatar
4 votes
Accepted

On a A089039 and pair of sequences with simple recursion

Let $d'(n) := d(n)/(n-1)! = b(n-1)+c(n)$ and $a'(n):=a(n)/(n-1)!$. From the given recurrences for $b(n)$ and $c(n)$, it follows that $d'(n)$ satisfies the following order-4 recurrence: $$d'(n) + (2n + ...
Max Alekseyev's user avatar

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