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What is the way to change the limits of q-integration of a double integral.

For exemple, what is the answer after change the order of integration of $$\int_0^1 \int_0^{x} f(x,y)\ d_qy\ d_qx$$

https://en.wikipedia.org/wiki/Jackson_integral

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  • $\begingroup$ No sorry. I missed the link $\endgroup$
    – Gio67
    Commented Jun 4, 2017 at 19:59

1 Answer 1

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The definition of a $q$ Integral is \begin{eqnarray*} \int_{0}^{x} f(x) d_{q}x =(1-q) x \sum_{k=0}^{\infty} q^{k}f(x^{k}x). \end{eqnarray*} Lets try calculating your double intergral where the function is a single monomial $f(x,y)=x^{\alpha}y^{\beta} $ & change the $q$ variable to $p$ for the $y$ intergral (We will change it back at the end) \begin{eqnarray*} \int_{0}^{1} d_{q}x \int_{0}^{x} d_{p}y x^{\alpha}y^{\beta} &= & \left[ \int_{0}^{1} d_{q}x (1-p) y \sum_{k=0}^{\infty} p^{k} x^{\alpha} (p^{k}y)^{\beta} \right]_0^{x} \\ \end{eqnarray*} \begin{eqnarray*} &= & (1-p) \sum_{k=0}^{\infty} p^{k(1+\beta)} \int_{0}^{1} d_{q} x x^{1+\alpha+\beta} \\ \end{eqnarray*} \begin{eqnarray*} &= & (1-p) \sum_{k=0}^{\infty} p^{k(1+\beta)} (1-q) \left[x \sum_{j=0}^{\infty} q^{j}(p^{j}x) ^{1+\alpha+\beta} \right]_{0}^{1} \\ \end{eqnarray*} \begin{eqnarray*} &= & (1-p) \sum_{k=0}^{\infty} p^{k(1+\beta)} (1-q) \sum_{j=0}^{\infty} q^{j(2+\alpha+\beta)} \\ \end{eqnarray*} \begin{eqnarray*} &= & \frac{ (1-p)}{ (1- p^{1+\beta})} \frac{ (1-q) }{(1- q^{2+\alpha+\beta})} \\ \end{eqnarray*} Now lets guess that the interchange of the integrals does the usual thing ... let us calculate \begin{eqnarray*} \int_{0}^{1} d_{p}y \int_{y}^{1} d_{q}x x^{\alpha}y^{\beta} &= & \left[\int_{0}^{1} d_{p}y (1-q) x \sum_{k=0}^{\infty} q^{j} (q^{j} x)^{\alpha} y^{\beta} \right]_y^{1} \\ \end{eqnarray*} \begin{eqnarray*} &= & \int_{0}^{1} d_{p} (1-q) \sum_{k=0}^{\infty} q^{j(1+\alpha)} \left( y^{1+\beta} -y^{1+\alpha+\beta} \right) \\ \end{eqnarray*} \begin{eqnarray*} &= & (1-q) \sum_{j=0}^{\infty} q^{j(1+\alpha)} \left[ (1-p) \sum_{k=0}^{\infty} p^{k} \left( (p^ky)^{1+\beta} -(p^{k}y)^{1+\alpha+\beta} \right) \right]_0^1 \\ \end{eqnarray*} \begin{eqnarray*} &= & (1-q) \sum_{j=0}^{\infty} q^{j(1+\alpha)} (1-p)y \sum_{k=0}^{\infty} \left( p^{k(1+\beta)} -p^{k(2+\alpha+\beta)} \right) \\ \end{eqnarray*} \begin{eqnarray*} = & \frac{ (1-q)}{ (1- q^{1+\alpha})} \left( \frac{ (1-p) }{(1- p^{1+\beta})} - \frac{ (1-p) }{(1- p^{2+\alpha+\beta})} \right) \end{eqnarray*} \begin{eqnarray*} = & \frac{ (1-q)}{ (1- q^{1+\alpha})} \frac{ p^{1+\beta} (1-p) (1-p^{1+\alpha}) }{(1- p^{1+\beta})(1- p^{2+\alpha+\beta})} \end{eqnarray*} So the answer is no ... trying invert the integrals ( in the usual way) introduces an extra factor of $p^{1+\beta}$ for each monomial.

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