Finite-interval oscillatory integral and phase convention

Let $a<b$, $\omega \in \mathbb{R}\setminus \{0\}$, $f:[a,b]\to \mathbb{C}$ be continuous, and $g:[a,b]\to \mathbb{R}$ be continuous. Define $I_{\omega }(f,g;[a,b]):={\int }_a^{b}f(x)e^{i\omega g(x)}dx\mathrm{.}$ The complex Riemann integral here means the real Riemann integral of the real part plus $i$ times the real Riemann integral of the imaginary part. Euler's formula $\exp (x+iy)=e^x(\cos y+i\sin y)$, $\mathrm{\mid }\exp (x+iy)\mathrm{\mid }=e^x$, and $e^{i\pi }+1=0$ and continuity of sine and cosine, which follows from The derivatives of sine and cosine are cosine and minus sine and differentiability implying continuity A function differentiable at $c$ is continuous at $c$, make $e^{i\omega g}$ continuous. The product is continuous, so both component integrals exist by A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion. A function is in $C^r[a,b]$ when it and its derivatives through order $r$ are continuous on $[a,b]$, with one-sided endpoint derivatives. A phase is nonstationary on $[a,b]$ when $g\in C^1[a,b]$ and $\mathrm{\mid }{g}^{\mathrm{\prime }}(x)\mathrm{\mid }\ge c$ everywhere for some fixed $c>0$. For such a phase and differentiable $h$, put $T_g^{0}h=h$ and $T_g^{j+1}h=(T_g^{j}h\mathrm{/}{g}^{\mathrm{\prime }}{)}^{\mathrm{\prime }}$ whenever that derivative exists.

This is an exact-real definition. A later numerical implementation must separately state its moment errors and any floating-point solve error.

The page uses real-valued $g$, complex-valued $f$, and real nonzero frequency $\omega$. A simple stationary point is treated in a separate local branch; it is not declared nonstationary by deleting the point where $g^{\mathrm{\prime }}=0$.