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Introduction
Best viewed on a desktop, if you must use a phone, maximize your browser in landscape mode and sorry some of the math won’t fully display on a mobile yet.In this insight article, we will build all the machinery necessary to evaluate unit hypercube integrals by a novel technique. We will first state a theorem on Dirichlet integrals, second develop a sequence of nested sets that point-wise converges to a unit hypercube, and thirdly make these two pieces compatible by means of a Dominated Convergence Theorem, and lastly demonstrate the technique of integration. Note: The same technique is outlined (in the same way) in the expanded insight article entitled A Path to Fractional Integral Representations of Some Special Functions.
The Integrals of Dirichlet
Dirichlet integrals as I learned them from an Advanced Calculus book are just that formula evaluating the integral to Gamma functions, they are not a type of integral like Riemann integral, more just a formula that would go on a table of integrals. Content is the 4+-dimensional version of volume (some writers use hypervolume instead of content).For the proof of this Dirichlet Integrals Theorem, I refer the reader to the text Special Functions by Askey, Andrews, and Roy. The result due to Dirichlet is given by
Theorem 1.1: Dirichlet Integrals
If ##t,{\alpha _p},{\beta _q},\Re \left[ {{\gamma _r}} \right] > 0\forall p,q,r## and ##V_t^n: = \left\{ {\left( {{z_1},{z_2}, \ldots ,{z_n}} \right) \in {\mathbb{R}^n}|{z_j} \geq 0\forall j,\sum\limits_{k = 1}^n {{{\left( {\frac{{{z_k}}}{{{\alpha _k}}}} \right)}^{{\beta _k}}} \leq t} } \right\}##, then$$\iint {\mathop \cdots \limits_{V_t^n} \int {\prod\limits_{\lambda = 1}^n {\left( {z_\lambda ^{{\gamma _\lambda } - 1}} \right)} d{z_n} \ldots d{z_2}d{z_1}} } = {t^{\sum\limits_{p = 1}^n {\frac{{{\gamma _p}}}{{{\beta _p}}}} }}{{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } \mathord{\left/{\vphantom {{\prod\limits_{q = 1}^n {\left[ {\frac{{\alpha _q^{{\gamma _q}}}}{{{\beta _q}}}\Gamma \left( {\frac{{{\gamma _q}}}{{{\beta _q}}}} \right)} \right]} } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}} \right. } {\Gamma \left( {1 + \sum\limits_{k = 1}^n {\frac{{{\gamma _k}}}{{{\beta _k}}}} } \right)}}$$
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