Oscillatory integrals and filon levin methods — Examples

Direct differentiation gives $p_{\epsilon }(0)=1$, $p_{\epsilon }(1)=0$, $p_{\epsilon }^{\mathrm{\prime }}(0)=-i\omega$, and $p_{\epsilon }^{\mathrm{\prime }}(1)=\epsilon$. Therefore $L_{\omega }{p}_{\epsilon }(0)=0$ and $L_{\omega }{p}_{\epsilon }(1)=\epsilon$. The basis functions are linearly independent because $p_{\epsilon }$ is nonconstant, and [F1] gives $A_{\epsilon }=\left[\begin{array}{cc}i\omega & i\omega \\ i\omega & i\omega +\epsilon \end{array}\right],\det A_{\epsilon }=i\omega \epsilon \mathrm{\ne }0.$

For $f=0$, the exact right-hand side and solution are zero. Multiplying $A_{\epsilon }$ by $\widehat{c}=(-1,1)$ gives $\rho =(0,\epsilon )$, while the associated polynomial is $\widehat{p}=-{\psi }_1+{\psi }_2=p_{\epsilon }$. Its endpoint values give $Q(\widehat{c})=0\cdot e^{i\omega }-1=-1$. Hence the exact integral is $0$ and the endpoint error has magnitude $1$ although $\mathrm{\parallel }\rho {\mathrm{\parallel }}_2=\epsilon$.

Solving $A_{\epsilon }z=(0,1)$ gives $z=(-1\mathrm{/}\epsilon ,1\mathrm{/}\epsilon )$, so $\mathrm{\parallel }{A}_{\epsilon }^{-1}{\mathrm{\parallel }}_2\ge \sqrt{2}\mathrm{/}\epsilon$. Also $\mathrm{\parallel }{A}_{\epsilon }{\mathrm{\parallel }}_2\ge \mathrm{\parallel }{A}_{\epsilon }(1,0{)}^T{\mathrm{\parallel }}_2=\sqrt{2}\omega$. Thus ${\kappa }_2(A_{\epsilon })\ge 2\omega \mathrm{/}\epsilon \to \mathrm{\infty }$ as $\epsilon \downarrow 0$. The factor in [F2] records this amplification; the residual norm alone cannot certify a small quadrature error.

Here $g^{\mathrm{\prime }}(x)=x$, $g^{\mathrm{\prime }\mathrm{\prime }}(0)=1$, and $g(\pm 1)=1\mathrm{/}2$. Applying [F1] at $x_0=0$ gives the stationary term $\sqrt{2\pi \mathrm{/}\omega }{e}^{i\pi \mathrm{/}4}$, while the endpoint bracket is $\frac{1}{i\omega }(\frac{e^{i\omega \mathrm{/}2}}{1}-\frac{e^{i\omega \mathrm{/}2}}{-1})=\frac{2e^{i\omega \mathrm{/}2}}{i\omega }\mathrm{.}$ The remainder bound is the one in [F1].

Since $g^{\mathrm{\prime }}(0)=0$, no $c>0$ satisfies [F2] on $[-1,1]$. By [F3], the endpoint term and remainder together are bounded by $2\mathrm{/}\omega +C{\omega }^{-3\mathrm{/}2}$, both smaller than the stationary term's magnitude $\sqrt{2\pi \mathrm{/}\omega }$ by a factor tending to zero. Thus the leading scale is ${\omega }^{-1\mathrm{/}2}$, not the nonstationary integer-power expansion.

Since $e^{i2\pi N}=(e^{i\pi }{)}^{2N}=(-1{)}^{2N}=1$ and $e^{i\pi N}=(-1{)}^N=-1$ for odd $N$, by [F5]--[F7], [F2] and [F3] give ${\mu }_0=0$. For $f=1$ the endpoint interpolant is constant, so [F1] returns the exact integral $0$, while the one-sample value is $M_{\omega }(1)=e^{i\pi N}=-1$.

For $f=x^2$, the degree-one endpoint interpolant is $p(x)=x$. Step 1.1 gives $e^{i\omega }=1$, so [F2] gives ${\mu }_1=1\mathrm{/}(i\omega )$ and ${\mu }_2=1\mathrm{/}(i\omega )-2\mathrm{/}(i\omega {)}^2$. Thus $I_{\omega }(f,g)-Q_1={\mu }_2-{\mu }_1=-2\mathrm{/}(i\omega {)}^2=2\mathrm{/}{\omega }^2$. The hypotheses of [F4] hold, so this exact error has the asserted second-order frequency decay.