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Consider a real-valued function $g: \mathbb{R}^q\rightarrow \mathbb{R}$. I want to show that $\int g(x)dx=\frac{1}{h^q}\int g(\frac{x}{h})dx$. I guess I should use the change of variable rule but I don't see how to get $\frac{1}{h^q}$. Any hint would be really appreciated.

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    $\begingroup$ What do you mean by $dx$? Should be $dx_1dx_2\cdots dx_n$ or $dA$. $\endgroup$ – Gregory Grant Jan 24 '16 at 12:18
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Try rewriting that as $$ \int g(u)~du~=~\frac{1}{h^q}\int g(\frac{x}{h})~dx. $$ Now the substitution $u = \frac{x}{h}$ (which really means $u_i = \frac{x_i}{h}$ for $i = 1, \ldots, q$), plus the change of variable theorem, does the trick.

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  • $\begingroup$ Ok, thanks, but what is $\frac{d(u_1,...,u_q)}{d(x_1,...,x_q)}$? $\endgroup$ – STF Jan 24 '16 at 12:24
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    $\begingroup$ Since $u_i = x_i / k$, the matrix is diagonal, namely $\frac{1}{k} \mathbf{I}$. $\endgroup$ – John Hughes Jan 24 '16 at 13:37
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$dx=dx_1\cdots dx_q$. Change each of the $x_i$ as $x_i =t_i/h$. Therefore $$ dx=dx_1\cdots dx_q=\frac{dt_1}{h}\cdots\frac{dt_q}{h}=\frac{1}{h^q}dt_1\cdots dt_q\ . $$

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  • $\begingroup$ Ok, many thanks, when you write $dx=dx_1... dx_q$ are you using Fubini's Theorem? $\endgroup$ – STF Jan 24 '16 at 12:50
  • $\begingroup$ The real question is: what do you mean by $\int g(x)dx$ [this being a multidimensional integral, according to your question]? The only way I can interpret your question is that your $\int g(x)dx$ is a bad notation for $\int d\mathbf{x}g(\mathbf{x})$, where $\mathbf{x}$ is a vector with $q$ components. Then I just use the change of variable theorem en.wikipedia.org/wiki/… whose Jacobian matrix in your case is diagonal (therefore its determinant is just the product of the diag.. entries). Fubini's theorem has no place here. $\endgroup$ – Pierpaolo Vivo Jan 24 '16 at 12:59
  • $\begingroup$ Just to be more precise, the Jacobian matrix $J_{ij}=\frac{\partial x_i}{\partial t_j}=(1/h)\delta_{ij}$ so its determinant is $(1/h)^q$. $\endgroup$ – Pierpaolo Vivo Jan 24 '16 at 13:24
  • $\begingroup$ Just to make sure I understood: $\int_{\mathbb{R}^q} dx g(x):=\int_{\mathbb{R}^q} d(x_1,...,x_q)g(x_1,...,x_q)$. Is this different from $\int_{\mathbb{R}}... \int_{\mathbb{R}} g(x_1,...,x_q)dx_1 dx_2... dx_q$ unless $\int_{\mathbb{R}^q} |g(x_1,...,x_q)|d(x_1,...,x_q)<\infty$ (which is Fubini's Theorem)? $\endgroup$ – STF Jan 24 '16 at 14:09

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