Transformation set

The transformation theorem (also transformation formula ) describes in analysis the behavior of integrals under coordinate transformations . It is thus the generalization of integration through substitution on functions of higher dimensions. The transformation theorem is used as an aid in calculating integrals if the integral can be calculated more easily after transferring it to another coordinate system.

Formulation of the sentence

Let it be an open set and a diffeomorphism . Then the function on if and integrable if the function to be integrated. In this case: ${\displaystyle \Omega \subseteq \mathbb {R} ^{d}}$ ${\displaystyle \Phi \colon \Omega \to \Phi (\Omega )\subseteq \mathbb {R} ^{d}}$ ${\displaystyle f}$ ${\displaystyle \Phi (\Omega )}$ ${\displaystyle x\mapsto f(\Phi (x))\cdot \left|\det(D\Phi (x))\right|}$ ${\displaystyle \Omega }$

${\displaystyle \int _{\Phi (\Omega )}f(y)\,\mathrm {d} y=\int _{\Omega }f(\Phi (x))\cdot \left|\det(D\Phi (x))\right|\mathrm {d} x\;.}$

Here is the Jacobian matrix and the functional determinant of . ${\displaystyle D\Phi (x)}$ ${\displaystyle \det(D\Phi (x))}$ ${\displaystyle \Phi }$

Special cases

• If you choose 1 for the constant function , the left side of the formula simply represents the volume or -dimensional Lebesgue measure of the image set : ${\displaystyle f}$ ${\displaystyle d}$ ${\displaystyle \Phi (\Omega )}$

  ${\displaystyle \operatorname {vol} (\Phi (\Omega ))=\int _{\Omega }\left|\det(D\Phi (x))\right|\,\mathrm {d} x\;.}$

• Also, if the mapping is linear or affine , where is a matrix and , so is . Thus applies ${\displaystyle \Phi }$ ${\displaystyle \Phi (x)=Ax+b}$ ${\displaystyle A}$ ${\displaystyle d\times d}$ ${\displaystyle b\in \mathbb {R} ^{d}}$ ${\displaystyle D\Phi (x)=A}$

  ${\displaystyle \operatorname {vol} (\Phi (\Omega ))=\left|\det(A)\right|\cdot \operatorname {vol} (\Omega )\;.}$

More general form

With the conditions of the transformation theorem, the conditions that the mapping is a diffeomorphism can be weakened: ${\displaystyle \Phi }$

• It does not have to be continuously differentiable. It is sufficient if it is locally Lipschitz continuous . In this case it is differentiable almost everywhere and the functional determinant is locally restricted and locally integratable .
• It doesn't need to be injective . It is sufficient if this property applies almost everywhere.
• The functional determinant may also have the value zero. Because of Sard's theorem , the image of the set of points at which this happens is a null set under the map . ${\displaystyle \Phi }$

example

To show that the integral is over the Gaussian bell

${\displaystyle {\frac {1}{\sigma {\sqrt {2\pi }}}}\mathrm {e} ^{-{\frac {1}{2}}{\big (}{\frac {x-\mu }{\sigma }}{\big )}^{2}}}$

equals 1, it suffices to make the statement

${\displaystyle \left(\int _{-\infty }^{\infty }\mathrm {e} ^{-x^{2}}\,\mathrm {d} x\right)^{2}=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\mathrm {e} ^{-x^{2}-y^{2}}\,\mathrm {d} x\,\mathrm {d} y=\pi }$

to prove. Since the function is rotationally symmetrical , the calculation of the integral in polar coordinates instead of Cartesian coordinates is obvious: ${\displaystyle f(x,y)=\mathrm {e} ^{-x^{2}-y^{2}}=\mathrm {e} ^{-r^{2}}}$

It be and ${\displaystyle \Omega =\mathbb {R} _{>0}\times (0,2\pi )}$

${\displaystyle \Phi \colon \Omega \to \mathbb {R} ^{2},\quad (r,\varphi )\mapsto (r\cos \varphi ,r\sin \varphi ).}$

Then is the functional determinant

${\displaystyle \det D\Phi (r,\varphi )={\begin{vmatrix}\cos \varphi &-r\sin \varphi \\\sin \varphi &r\cos \varphi \end{vmatrix}}=r(\cos ^{2}\varphi +\sin ^{2}\varphi )=r.}$

The complement of is a zero set , so with results ${\displaystyle \Phi (\Omega )\subset \mathbb {R} ^{2}}$ ${\displaystyle f(x,y)=\mathrm {e} ^{-x^{2}-y^{2}}}$

${\displaystyle \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\mathrm {e} ^{-x^{2}-y^{2}}\,\mathrm {d} x\,\mathrm {d} y}$

${\displaystyle {}=\int _{\Phi (\Omega )}\mathrm {e} ^{-x^{2}-y^{2}}\,\mathrm {d} x\,\mathrm {d} y}$

${\displaystyle {}=\int _{\Omega }\mathrm {e} ^{-(r\cos \varphi )^{2}-(r\sin \varphi )^{2}}\cdot \det D\Phi (r,\varphi )\,\mathrm {d} r\,\mathrm {d} \varphi }$

${\displaystyle {}=\int _{\Omega }\mathrm {e} ^{-r^{2}}\cdot r\,\mathrm {d} r\,\mathrm {d} \varphi }$

${\displaystyle {}=\int _{0}^{2\pi }\int _{0}^{\infty }r\mathrm {e} ^{-r^{2}}\,\mathrm {d} r\,\mathrm {d} \varphi }$

${\displaystyle {}=\int _{0}^{2\pi }{\frac {1}{2}}\cdot \mathrm {d} \varphi =\pi .\,}$

The evaluation of the inner integral in the penultimate line can be justified, for example, by a substitution . ${\displaystyle t=r^{2}}$

literature

• Otto Forster : Analysis. Volume 3: Measure and integration theory, integral theorems in R ⁿ and applications , 8th improved edition. Springer Spectrum, Wiesbaden, 2017, ISBN 978-3-658-16745-5 .
• Konrad Königsberger : Analysis 2 , Springer, Berlin 2004, p. 211