The Gaussian integral on the line with a certified sinh-sinh truncation tail

On the real axis, $x^{\mathrm{\prime }}(t)=(\pi \mathrm{/}2)\cosh t\cosh u(t)>0$. As $t\to \pm \mathrm{\infty }$, $x(t)\to \pm \mathrm{\infty }$. We show the tail envelope for $t\to +\mathrm{\infty }$; evenness gives the left tail. For $t\ge 1$, $\sinh t\ge c{e}^t$ with $c=(1-e^{-2})\mathrm{/}2>0$, so $u(t)\ge c_1{e}^t$ for $c_1=\pi c\mathrm{/}2$. For $u\ge 1$, $\sinh u\ge c{e}^u$, hence $x(t{)}^2\ge c^2{e}^{2u(t)}\ge c^2{e}^{2c_1{e}^t}$. Also $u(t)\le (\pi \mathrm{/}4)e^t$ and $\cosh t\le e^t$, $\cosh u\le e^u$, so for a fixed constant $C^{\mathrm{\prime }}>0$ and all sufficiently large $t$, $(\pi \mathrm{/}2)\cosh t\cosh u(t)\le e^{C^{\mathrm{\prime }}{e}^t}$. Put $v=e^t$. By [F8], $v\to \mathrm{\infty }$, and [F9] implies $c^2{e}^{2c_{1}v}\mathrm{/}v\to \mathrm{\infty }$. Thus $x(t{)}^2$ eventually exceeds both $2C^{\mathrm{\prime }}{e}^t$ and $2e^t$. It follows that $\mathrm{\Phi }(t)\le e^{C^{\mathrm{\prime }}{e}^t-x(t{)}^2}\le e^{-x(t{)}^2\mathrm{/}2}\le e^{-e^t}$ for all sufficiently large $t$.

For any proposed substrip halfwidth $a>0$, fix $0<y<\min (a,\pi \mathrm{/}2)$; then [F13] gives $\sin y,\cos y>0$. Extend $u(z)=(\pi \mathrm{/}2)\sinh z$ holomorphically; this extension is entire by [F10]. For $z=x+iy$, write $u=U+iV$ with $U=(\pi \mathrm{/}2)\sinh x\cos y$ and $V=(\pi \mathrm{/}2)\cosh x\sin y$. The continuous function $V(x)$ has positive derivative for $x>0$ and tends to infinity. For every sufficiently large integer $j$, its value $(j+1\mathrm{/}2)\pi$ lies strictly between $V(0)$ and some $V(R)$; [F11] gives a solution $x_j\in (0,R)$, and strict increase makes it unique. Defining $x_j$ as that unique solution avoids any choice of a sequence. Then $x_j\to \mathrm{\infty }$. At those points, [F7] gives $\mathrm{Re}({\sinh }^{2}u)=-{\cosh }^{2}U$ and $\mathrm{\mid }\cosh u\mathrm{\mid }=\sinh U$. Since $U(x_j)\to \mathrm{\infty }$, also $\mathrm{\mid }\cosh (x_j+iy){\mathrm{\mid }}^2={\sinh }^2{x}_j+{\cos }^{2}y\ge {\sinh }^2{x}_j$ by [F7] and ${\cosh }^2{x}_j-{\sinh }^2{x}_j=1$ from [F8]. Thus it follows that $\mathrm{\mid }\mathrm{\Phi }(x_j+iy)\mathrm{\mid }=\frac{\pi }{2}\mathrm{\mid }\cosh (x_j+iy)\mathrm{\mid }\sinh U(x_j)e^{{\cosh }^{2}U(x_j)}⟶\mathrm{\infty }\mathrm{.}$ Thus no bound $C{e}^{-\eta \mathrm{\mid }x\mathrm{\mid }}$ can hold on that horizontal line; this disproves the uniform strip hypothesis in [F6] for every proposed positive halfwidth.

The envelope in step 1.1 makes $\mathrm{\Phi }$ absolutely integrable on both tails. On every compact interval, $\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }$ is continuous and Riemann integrable by [F12], so $\mathrm{\Phi }$ is absolutely integrable on $\mathbb{R}$ as an improper Riemann integral. Apply [F2] and then [F4] to obtain $\int \mathrm{\Phi }=\sqrt{\pi }$. Applying [F3] with $A=B=\kappa =1$ gives the asserted omitted-sample bound when $N_{\pm }h\ge T$.

By [F5], the bound of the full-error theorem requires the strip premise as well as the tail premise. The real integral and truncation tail are justified in step 2.1, while step 1.2 disproves that strip premise. Thus no total quadrature error bound is asserted.