Let $f:\Bbb R^n\to \Bbb R^n$ be a continuous function such that $\int_{\Bbb R^n}|f(x)|dx\lt\infty$. Let $A$ be a real $n\times n$ invertible matrix and for $x,y\in\Bbb R^n$, let $\langle x,y\rangle$ denote the standard inner product in $\Bbb R^n$. Then

$$\int_{\Bbb R^n}f(Ax)e^{i\langle y,x\rangle}\,dx\overset{?}{=}\ \begin{array}{l} (1)\ \int_{\Bbb R^n} f(x) e^{i\langle(A^{-1})^T y,x\rangle} \frac{dx}{|\det A|} \\ (2)\ \int_{\Bbb R^n} f(x) e^{i\langle A^T y,x\rangle} \frac{dx}{|\det A|}\\ (3)\ \int_{\Bbb R^n}f(x)e^{i\langle(A^T)^{-1} y,x\rangle}\,dx \\ (4)\ \int_{\Bbb R^n}f(x)e^{i\langle A^{-1} y,x\rangle}\frac {dx}{|\det A|} \end{array}$$


Let $Ax=X\Rightarrow x=A^{-1}X$, then $dx=\frac {dX}{|\det A|}$. Using all these, we have:

$\int_{\Bbb R^n}f(Ax)e^{i\langle y,x\rangle}dx=\int_{\Bbb R^n}f(X)e^{i\langle y,A^{-1}X\rangle}\frac{dX}{|\det A|}=\int_{\Bbb R^n}f(X)e^{i\langle (A^{-1})^Ty,X\rangle}\frac{dX}{|\det A|}$.

Now if we again put $x=X\Rightarrow dx=dX$, then we get the last expression as,

$\int_{\Bbb R^n}f(x)e^{i\langle(A^{-1})^T y,x\rangle}\frac {dx}{|\det A|}$. That means 1) is true.

  • $\begingroup$ What are your thoughts? $\endgroup$ – NoName Jun 24 '15 at 4:58
  • $\begingroup$ I have confusion about $Ax=X$ then $dx=\frac {dX}{|detA|}$, this. How it comes? $\endgroup$ – Chiranjeev_Kumar Jun 24 '15 at 5:03
  • $\begingroup$ Please change your title to something which describes your question's mathematical content. $\endgroup$ – Dorebell Jun 24 '15 at 5:08
  • $\begingroup$ i have seen this, it is used in solving the problems so i used it can u tell me how it comes? $\endgroup$ – Chiranjeev_Kumar Jun 24 '15 at 5:09
  • $\begingroup$ @Dorebell Instead of commanding the OP to do it, maybe you could do it for them and then leave a comment gently explaining the change you made and why it's important. $\endgroup$ – layman Jun 24 '15 at 5:11

For $\mathbf{x},\mathbf{y} \in \mathbb{R}^n$, let $\left\langle \mathbf{x},\mathbf{y} \right\rangle$ be the Euclidean dot product on $\mathbb{R}^n$. Given a function $f \in \mathrm{L}^1(\mathbb{R}^n)$, you can define its Fourier Transform $\hat{f}$ as follows :

$$ \forall \boldsymbol\xi \in \mathbb{R}^n, \, \hat{f}(\boldsymbol\xi) = \int_{\mathbb{R}^n} f(\mathbf{x}) e^{-2i\pi \left\langle \mathbf{x},\boldsymbol\xi \right\rangle} \, d \mathbf{x}. $$

Now, consider a $n \times n$ invertible real matrix $\mathbf{A} \in \mathrm{GL}(n,\mathbb{R})$. The mapping $\mathbf{x} \in \mathbb{R}^n \mapsto \mathbf{A}\mathbf{x}$ is a $\mathcal{C}^1$ diffeomorphism of $\mathbb{R}^n$ and the Change of variable theorem gives, for all $\boldsymbol\xi \in \mathbb{R}^n$ :

$$ \hat{f}(\boldsymbol\xi) = \int_{\mathbb{R}^n} f(\mathbf{A}\mathbf{x}) e^{-2i\pi \left\langle \mathbf{A}\mathbf{x},\boldsymbol\xi \right\rangle} \vert \det \mathbf{A} \vert \, d \mathbf{x}. $$

Given that $\left\langle \mathbf{A}\mathbf{x},\boldsymbol\xi \right\rangle = \left\langle \mathbf{x},\mathbf{A}^{\top}\boldsymbol\xi \right\rangle$, we have ::

$$ \hat{f}(\boldsymbol\xi) = \int_{\mathbb{R}^n} f(\mathbf{A}\mathbf{x}) e^{-2i\pi \left\langle \mathbf{x},\mathbf{A}^{\top}\boldsymbol\xi \right\rangle} \vert \det \mathbf{A} \vert \, d \mathbf{x}. $$

As a consequence :

$$ \int_{\mathbb{R}^{n}} f(\mathbf{A}\mathbf{x}) e^{-2i\pi \left\langle \mathbf{x}, \boldsymbol\xi \right\rangle} \, d \mathbf{x} = \frac{1}{\vert \det \mathbf{A} \vert} \hat{f}\big( (\mathbf{A}^{\top})^{-1} \boldsymbol\xi \big). $$

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