Why disappear the integral over $(-\infty , \infty)$ I am reading 'Introduction to nonlinear dispersive equations' by Linares and Ponce.  In section 4.2 (local smoothing effects) I found this equalities difficult to understand. I need some puntual explanation. Making the variable change $2\pi \xi ^2 = r$, they write for $f_+ = \chi_{\mathbb{R}^+} f$
$$ c\int_{-\infty}^\infty \begin{vmatrix} \displaystyle\int_{-\infty}^\infty |\xi |^{1/2} e^{2\pi i x\xi} e^{-4\pi ^2 i t \xi ^2 }\widehat{f_+}(\xi)\ d\xi \end{vmatrix}^2 \ dt = \\ = c \int_{-\infty}^\infty \begin{vmatrix} \displaystyle\int_{0}^\infty r^{1/4} e^{  i x\sqrt{2\pi\xi}} e^{-2\pi i t r }\widehat{f}(\sqrt{r/(2\pi)}) \frac{1}{r^{1/2}}\ dr \end{vmatrix}^2 \ dt = \\\\ \overset{???}{=}c \int_{0}^\infty \begin{vmatrix} \displaystyle   e^{  i x\sqrt{2\pi\xi}}  \widehat{f}(\sqrt{r/(2\pi)}) \frac{1}{r^{1/4}}\end{vmatrix}^2\ dr =\\\\
c\int_{-\infty}^\infty |\widehat{f_+}(\xi)|^2\ d\xi = c\Vert \widehat{f_+} \Vert ^2 _ 2 =c\Vert f_+ \Vert ^2 _ 2 $$
What I cannot understand is the third step: why the integral over $(-\infty,\infty)$ dissapear? and it shouldn't be an inequality?
Any help is welcome! Thanks in advance.
 A: Your equality if of form $\int_{-\infty}^\infty \left|\int_0^\infty g(r)e^{-2\pi i t r}\ dr\right|^2\ dt = \int_0^\infty \left|g(r)\right|^2\ dr$. Let us use $|f|^2 = f^* f$ and get
$$\int_{-\infty}^\infty \left|\int_0^\infty g(r)e^{-2\pi i t r}\ dr\right|^2\ dt =\\
\int_{-\infty}^\infty \left(\int_0^\infty g(r_1)e^{-2\pi i t r_1}\ dr_1\right)\cdot \left(\int_0^\infty g^*(r_2)e^{2\pi i t r_2}\ dr_2\right)\ dt =\\
\int_0^\infty\ dr_1\int_0^\infty\ dr_2\ g(r_1)g^*(r_2)\int_{-\infty}^\infty e^{2\pi i t (r_2 - r_1)}\ dt$$
Replace $r_2 - r_1 = v$, then we get
$$\int_0^\infty\ dr_1 g(r_1)\int_{-\infty}^\infty\ dt \int_{-r_1}^\infty\ dv\ e^{2\pi i t v} g^*(v + r_1) =\\
\int_0^\infty g(r) g^*(r)\ dr =
\int_0^\infty |g(r)|^2\ dr
$$
This uses the fact that $\int_{-\infty}^\infty\ dt \int_{-q}^\infty\ dv\ e^{2\pi i v} h(v) = h(0)$. It's true at least if $h$ is differentiable in $0$ and $\frac{h(x)}{x}$ is integrable on $\mathbb{R} \setminus [-1; 1]$, as explained in, for example, this excellent post.
