Singular and infinite interval quadrature

On $[a,b]\setminus \{c\}$ the numerator and denominator are continuous and the denominator is nonzero, so their quotient $g$ is continuous at every point other than $c$.

By the derivative limit in [F2], the quotient defining $g(x)$ tends to the assigned value $f^{\mathrm{\prime }}(c)=g(c)$ as $x\to c$. Thus $g$ is continuous at $c$ as well, and hence on the whole interval.

For $u_0=B{e}^{\kappa S}$, finite-interval substitution from [F3] gives ${\int }_S^R{e}^{-B{e}^{\kappa t}}dt=\frac{1}{\kappa }{\int }_{u_0}^{B{e}^{\kappa R}}\frac{e^{-u}}{u}du\mathrm{.}$ For $u\ge u_0>0$, $1\mathrm{/}u\le 1\mathrm{/}{u}_0$, so the right side is at most $e^{-u_0}\mathrm{/}(\kappa u_0)$ after taking $R\to \mathrm{\infty }$; the endpoint limit follows from [F4]. Since $S\ge 0$, $e^{\kappa S}\ge 1$, and each continuous one-sided tail is bounded by $e^{-B{e}^{\kappa S}}\mathrm{/}(\kappa B{e}^{\kappa S})\le e^{-B{e}^{\kappa S}}\mathrm{/}(\kappa B)$.

Apply step 1.1 with $S=N_{-}h$ to $t\mapsto \mathrm{\mid }\mathrm{\Phi }(-t)\mathrm{\mid }$ and with $S=N_{+}h$ to $t\mapsto \mathrm{\mid }\mathrm{\Phi }(t)\mathrm{\mid }$, using the envelope. Adding proves the asserted integral bound and absolute integrability of both tails.

The function $g(t)=A{e}^{-B{e}^{\kappa t}}$ is nonincreasing for $t\ge 0$, since its derivative is $-AB\kappa e^{\kappa t}{e}^{-B{e}^{\kappa t}}<0$ by [F1]–[F2]. For every integer $k>N_{+}$, $hg(kh)\le {\int }_{(k-1)h}^{kh}g(t)dt$; summing finite partial tails and taking their increasing limit bounds the right omitted samples by ${\int }_{N_{+}h}^{\mathrm{\infty }}g(t)dt$. The same comparison for $g(-t)$ bounds the left tail, and step 1.1 gives the stated sum bound.

The two bounds in steps 2.1 and 2.2 are finite and have the displayed right-hand side, proving both claims.

For $t>0$, induction using [F2]–[F3] gives $u^{(j)}(t)=t^{\gamma -j}{P}_j(\log t)$, where $P_0(z)=z^m$ and $P_{j+1}(z)=(\gamma -j)P_j(z)+P_j^{\mathrm{\prime }}(z)$. Each $P_j$ is a polynomial of degree at most $m$.

If $\delta >0$ and $P$ is a polynomial, put $y=-\log t$. For every real $M$, $0<t<\exp (-M)$ implies $\log t<-M$ by [F1] and [F7], so $y\to +\mathrm{\infty }$ as $t\downarrow 0$. Then $t^{\delta }\mathrm{\mid }P(\log t)\mathrm{\mid }=e^{-\delta y}\mathrm{\mid }P(-y)\mathrm{\mid }$: for some $C>0$ and integer $d$, $\mathrm{\mid }P(-y)\mathrm{\mid }\le C(1+y^d)$ for $y\ge 1$, while [F4] and [F6] give $(1+y^d)e^{-\delta y}\to 0$. Since $\gamma -j>0$ for every $j\le r$, step 1.1 therefore shows $u^{(j)}(t)\to 0$ at zero.

Set $u^{(j)}(0)=0$ for $j\le r$. Continuity at zero follows from step 2.1. For $j<r$, the difference quotient of this extension at zero is $t^{\gamma -j-1}{P}_j(\log t)$, which tends to zero by step 2.1 because $\gamma -j-1>0$; hence the derivative at zero equals the defined $u^{(j+1)}(0)$. Thus the derivatives through order $r$ exist and are continuous on $[0,1]$ by the definition [F5].

Put $q=\gamma +1>0$. For $0<\epsilon <1$, the function $x\mapsto x^q\mathrm{/}q$ is differentiable on $[\epsilon ,1]$ with derivative $x^{q-1}=x^{\gamma }$ by [F2]. Applying [F5] gives ${\int }_{\epsilon }^1{x}^{\gamma }dx=(1-{\epsilon }^q)\mathrm{/}q$.

As $\epsilon \downarrow 0$, $\log \epsilon \to -\mathrm{\infty }$: for any real $M$, $0<\epsilon <\exp (M)$ implies $\log \epsilon <M$ by [F1] and [F3]. Hence $q\log \epsilon \to -\mathrm{\infty }$ and ${\epsilon }^q=\exp (q\log \epsilon )\to 0$ by [F1] and [F4].

Taking $\epsilon \downarrow 0$ in step 1.1 using step 1.2 proves convergence of the improper integral and gives its value $1\mathrm{/}q=1\mathrm{/}(\gamma +1)$.

Define $K_h(z)=\frac{\pi i}{h}\frac{e^{2\pi iz\mathrm{/}h}+1}{e^{2\pi iz\mathrm{/}h}-1}$. By [F2] and [F3], its poles are exactly $z=kh$, $k\in \mathbb{Z}$: if the denominator vanishes, its modulus gives $\mathrm{Im}z=0$, and Euler's formula then gives $2\pi z\mathrm{/}h=2j\pi$. At each such point the denominator has derivative $2\pi i\mathrm{/}h$ and the numerator is $2$, so $K_h$ has residue $1$. By [F8], $F{K}_h$ is holomorphic away from these points and has a simple pole with residue $F(kh)$ when $F(kh)\mathrm{\ne }0$, or a removable singularity with residue zero when $F(kh)=0$; fill in each removable value. Thus it is meromorphic on the strip.

Fix $N\ge 0$ and $R=(N+\frac{1}{2})h$. Let ${\mathrm{\Gamma }}_R$ traverse the rectangle $[-R,R]+i[-a,a]$ counterclockwise, beginning at $R-ia$. It is a closed piecewise linear contour inside $S_d$. No $kh$ lies on its boundary: the horizontal edges have nonzero imaginary part and $R\mathrm{/}h=N+1\mathrm{/}2$ is not an integer. Every exterior point belongs to the unbounded component of the complement of this rectangle boundary, since it can be joined to infinity along an outward horizontal or vertical ray; hence its index is zero by [F6]. For an interior point $p$, each ray from $p$ meets the boundary exactly once, and along a positive edge $z(t)$ one has $\mathrm{Im}(z^{\mathrm{\prime }}(t)\mathrm{/}(z(t)-p))>0$, because $p$ lies strictly to the left of that edge. The continuous argument therefore increases through exactly one full turn, so [F6] gives index one. In particular ${\mathrm{\Gamma }}_R$ is null-homologous in $S_d$. Together with the meromorphicity and boundary check above, this is precisely the admissibility required in [F4]. Its enclosed possible poles are exactly $kh$, $-N\le k\le N$, each of index one; removable points contribute zero. Thus ${\int }_{{\mathrm{\Gamma }}_R}F{K}_{h}dz=2\pi i{\sum }_{k=-N}^{N}F(kh)$.

On the upper edge, $\mathrm{\mid }{e}^{2\pi i(x+ia)\mathrm{/}h}\mathrm{\mid }=\rho :=e^{-2\pi a\mathrm{/}h}<1$ and $\mid K_h(x+ia)+\frac{\pi i}{h}\mid =\frac{2\pi }{h}\mid \frac{e^{2\pi i(x+ia)\mathrm{/}h}}{e^{2\pi i(x+ia)\mathrm{/}h}-1}\mid \le \frac{2\pi \mathrm{/}h}{e^{2\pi a\mathrm{/}h}-1}\mathrm{.}$ On the lower edge, the analogous estimate is $\mid K_h(x-ia)-\pi i\mathrm{/}h\mid \le (2\pi \mathrm{/}h)\mathrm{/}(e^{2\pi a\mathrm{/}h}-1)$. On either vertical edge, $e^{2\pi iz\mathrm{/}h}$ is negative real and $\mathrm{\mid }{K}_h(z)\mathrm{\mid }\le \pi \mathrm{/}h$. The vertical integrals therefore tend to zero as $N\to \mathrm{\infty }$ by the uniform strip bound in the statement.

[F7] applies to $F$ on each filled rectangle $[-R,R]+i[0,a]$ and $[-R,R]+i[-a,0]$, since both compact rectangles lie inside the open strip $S_d$ where $F$ is holomorphic. Their positively oriented boundaries have no poles at all, including their real edges. Their zero contour integrals and the same vanishing vertical-side bound for $F$ show ${\int }_{\mathbb{R}}F(x\pm ia)dx=I$. Replacing $K_h$ on the two long edges of step 2.1 by its constants $\mp \pi i\mathrm{/}h$, the residue identity becomes $h{\sum }_{k=-N}^{N}F(kh)=\frac{1}{2}({\int }_{-R}^{R}F(x-ia)dx+{\int }_{-R}^{R}F(x+ia)dx)+E_N,$ where $\mathrm{\mid }{E}_N\mathrm{\mid }\le ({\int }_{-R}^R\mathrm{\mid }F(x-ia)\mathrm{\mid }dx+{\int }_{-R}^R\mathrm{\mid }F(x+ia)\mathrm{\mid }dx)\mathrm{/}(e^{2\pi a\mathrm{/}h}-1)+o(1)$ as $N\to \mathrm{\infty }$.

The real and horizontal sums and integrals converge absolutely by the exponential bound. Letting $N\to \mathrm{\infty }$ in step 4.1, the average tends to $I$ and the error term is at most $(I_{-}+I_{+})\mathrm{/}(e^{2\pi a\mathrm{/}h}-1)$. This proves the assertion.

By [F9], the continuous partial derivatives are bounded on the compact rectangle; they are Borel and their integrals exist by [F1, F6]–[F7]. Define $H(x)={\int }_{[c,d]}G(x,y)d\nu (y)$. For $0\le j<2n$, the difference quotient of ${\mathrm{\partial }}_x^{j}G$ in $x$ differs from ${\mathrm{\partial }}_x^{j+1}G(x,y)$ by a quantity tending uniformly in $y$ to zero: by [F4], the quotient is the average of ${\mathrm{\partial }}_x^{j+1}G(u,y)$ between $x$ and $x+h$, and [F5] makes that derivative uniformly continuous. Integrating the uniform error changes it by at most $M_{\nu }$ times its supremum by [F7], so induction gives $H\in C^{2n}$ and $H^{(2n)}(x)=\int {\mathrm{\partial }}_x^{2n}G(x,y)d\nu (y)$.

Expand $Q_{\mu }(H)-Q_{\mu }{Q}_{\nu }G={\sum }_i{w}_i(\int G(x_i,y)d\nu (y)-Q_{\nu }(G(x_i,\cdot )))$. The positive weights in [F2] sum to $M_{\mu }$, and [F3] bounds each parenthesized error by $K_{\nu }\mathrm{\parallel }{\mathrm{\partial }}_y^{2n}G{\mathrm{\parallel }}_{\mathrm{\infty }}$. Therefore this term has absolute value at most $M_{\mu }{K}_{\nu }\mathrm{\parallel }{\mathrm{\partial }}_y^{2n}G{\mathrm{\parallel }}_{\mathrm{\infty }}$.

Apply [F3] to $H$ under $\mu$. The derivative formula in step 1.1 gives $\mid \int Hd\mu -Q_{\mu }(H)\mid \le K_{\mu }{M}_{\nu }\mathrm{\parallel }{\mathrm{\partial }}_x^{2n}G{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$

The iterated integral minus the tensor rule is the sum of the errors in steps 2.1 and 1.2. The triangle inequality gives the stated bound.

For an increment $(h,k)$, $D(r+h,s+k)-D(r,s)=(h,sh+rk+hk)$. The linear part $(h,sh+rk)$ has matrix $\left(\begin{array}{cc}1& 0\\ s& r\end{array}\right)$, while the remainder has Euclidean norm $\mathrm{\mid }hk\mathrm{\mid }\le (h^2+k^2)\mathrm{/}2$; divided by $\mathrm{\parallel }(h,k){\mathrm{\parallel }}_2=\sqrt{h^2+k^2}$ it tends to zero. Thus this is the total derivative. Its entries are continuous, so the polynomial extension is $C^1$ on ${\mathbb{R}}^2$ by [F2]; its determinant is $r$ by [F3].

If $0\le r,s\le 1$, then $0\le rs\le r\le 1$, so $D(r,s)\in T$. Conversely, for $(x,y)\in T$ with $x>0$, $r=x$ and $s=y\mathrm{/}x$ lie in $[0,1]$ and map to $(x,y)$; when $x=0$, $y=0$ and every $(0,s)$ maps to the vertex. For $r>0$, the displayed inverse is unique. This proves the image, inverse, and collapsed-edge claims.

For each integer $n\ge 2$, define $g_n(x)=x^{\gamma }$ on $[1\mathrm{/}n,1]$ and $g_n(x)=0$ on $[0,1\mathrm{/}n)$. For every real $c$, its strict superlevel is all of $[0,1]$ if $c<0$, and $[1\mathrm{/}n,1]\cap \{x>0:x^{\gamma }>c\}$ if $c\ge 0$. The latter is Borel because the power is continuous on $(0,\mathrm{\infty })$ by [F4]; hence [F2] makes $g_n$ measurable, and $g_n\ge 0$. By [F9], $x^{\gamma }$ is bounded on $[1\mathrm{/}n,1]$, so $g_n$ is bounded and continuous on each side of its only possible jump at $1\mathrm{/}n$; isolating that point in an interval of arbitrarily small length and using uniform continuity on the two remaining compact pieces makes its Darboux gap arbitrarily small, so [F6] gives Riemann integrability.

The equality of Riemann and Lebesgue integrals in [F7], followed by [F4] and [F5] on $[1\mathrm{/}n,1]$, gives $\int g_{n}d\lambda ={\int }_{1\mathrm{/}n}^1{x}^{\gamma }dx=(1-n^{-(\gamma +1)})\mathrm{/}(\gamma +1)$. The sequence $g_n$ is nondecreasing and converges pointwise to $g_{\gamma }$.

The increasing-limit clause of [F2] makes $g_{\gamma }$ measurable. Apply monotone convergence [F8] to the sequence $(g_{k+1}{)}_{k\ge 1}$ from step 2.1. Its integral limit is $1\mathrm{/}(\gamma +1)$ by [F3], so $g_{\gamma }$ is integrable and the displayed value follows. The only choice assumption used is [A1], which supplies the underlying Lebesgue measure in [F1].