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Let's consider the following form of the Mellin integral transform: $$m_{pq} =\iint\limits_{D_R} \! x^p y^q f(x,y) \, dx\; dy, \, D_R={\{(x,y)\,|\,x^2 + y^2 \le R^2\}}$$ If we scale the domain of the function $f$ by a factor $\gamma$: $$f_\gamma(\gamma x, \gamma y) = f(x, y)$$ $$m_{pq}^{(\gamma)} =\iint\limits_{D_{\gamma R}} \! x^p y^q f_\gamma(x, y) \, dx \; dy$$ we obtain the following relationship: $$m_{pq}^{(\gamma)} = \iint\limits_{D_R} \! (\gamma x)^p (\gamma y)^q f_\gamma(\gamma x,\gamma y) \, d(\gamma x) d(\gamma y) = \gamma^{p+q+2} \iint\limits_{D_R} \! x^p y^q f(x,y) \, dx \; dy = \gamma^{p+q+2}\,m_{pq}$$ But how to express the relationship between $m_{pq}$ and $m_{pq}^{(\alpha)}$ obtained by a rotation transform of the domain of the function $f$: $$\begin{pmatrix} x^{(\alpha)} \\ y^{(\alpha)} \end{pmatrix} = \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$ $$f_{\alpha}(x^{(\alpha)},y^{(\alpha)}) = f(x,y)$$

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Surely rotating the domain will force the integral to take into account $f$'s values in regions it wasn't payed attention to before. Since new information is gained (and lost!) through rotation I don't see a nice relationship between $m$ and $m^{(\alpha)}$ in the works. On the other hand, imposing some kind of symmetry on $f$ could change this. – anon Mar 4 '12 at 14:48
up vote 0 down vote accepted

$$m_{pq}^{(\alpha)} = \iint\limits_{D_R} \! x^{(\alpha)p} y^{(\alpha)p} f_\alpha(x^{(\alpha)}, y^{(\alpha)}) \, dx^{(\alpha)} dy^{(\alpha)} =$$ $$ = \iint\limits_{D_R} \! (x\cos\alpha-y\sin\alpha)^p (x\sin\alpha+y\cos\alpha)^q f(x, y) \Bigl\lvert \frac{\partial(x^{(\alpha)}, y^{(\alpha)})}{\partial(x,y)} \Bigr\rvert \, dx dy = $$

$$ = \iint\limits_{D_R} \! (x\cos\alpha-y\sin\alpha)^p (x\sin\alpha+y\cos\alpha)^q f(x, y) \, dx dy =$$

$$ = \iint\limits_{D_R} \! \Bigl( \sum\limits_{i=0}^{p} (-1)^{p-i} \begin{pmatrix} p \\ i \end{pmatrix} x^i\cos^i\alpha \, y^{p-i}\sin^{p-i}\alpha \Bigr) \Bigl(\sum\limits_{j=0}^{q}\begin{pmatrix} q \\ j \end{pmatrix} x^j\sin^j\alpha \, y^{q-j}\cos^{q-j}\alpha\Bigr) \, f(x, y) \, dx dy$$

Assume that $q = 0$ then: $$ m_{p,0}^{(\alpha)} = \iint\limits_{D_R} \! \sum\limits_{i=0}^{p} (-1)^{p-i} \begin{pmatrix} p \\ i \end{pmatrix} x^i\cos^i\alpha \, y^{p-i}\sin^{p-i}\alpha \, f(x, y) \, dx dy = $$ $$ = \sum\limits_{i=0}^{p} (-1)^{p-i} \begin{pmatrix} p \\ i \end{pmatrix} \cos^i\alpha \, \sin^{p-i}\alpha \iint\limits_{D_R} \! x^i y^{p-i} f(x,y) \, dx\; dy = $$ $$ = \sum\limits_{i=0}^{p} (-1)^{p-i} \begin{pmatrix} p \\ i \end{pmatrix} \cos^i\alpha \, \sin^{p-i}\alpha \; m_{i,p-i} $$ If we continue the reasoning and consider some particular cases, we can obtain one of well-known invariants of the Mellin integral transform: $$m_{20}^{(\alpha)} + m_{02}^{(\alpha)} = m_{20} + m_{02}$$

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