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Source context: published paper, printed p. 412 (PDF p. 2), the integrals in Theorem 1. The elementary bounds below are supplied by this compilation to make the finite calculation in Theorem 2 fully reproducible; they are not stated as a separate lemma in the paper.

Statement

For $z>0$ and $a>1$, put

$$h=1-a^{-2}>0,\qquad E=\exp\left(-\frac z2(a+a^{-1})\right).$$

Then

$$K_1(z,a)\le\frac{E}{zh},\qquad K_2(z,a)\le\frac Ez\left(\frac ah+\frac{2}{zh^2}\right). \tag{1}$$

These bound the complete integrals from $a$ to infinity, including every tail value.

Full proof

For $u\ge0$, direct subtraction gives

$$\frac1{a+u}-\left(\frac1a-\frac{u}{a^2}\right) =\frac{u^2}{a^2(a+u)}\ge0.$$

Consequently, with $c=z/2$,

$$(a+u)+(a+u)^{-1}\ge a+a^{-1}+hu, \quad e^{-c((a+u)+(a+u)^{-1})}\le E e^{-chu}.$$

Substitute $t=a+u$ in the definition of $K_\nu$. Since $\int_0^\infty e^{-vu}\,du=v^{-1}$ and $\int_0^\infty u e^{-vu}\,du=v^{-2}$ for $v>0$,

$$\begin{aligned} K_1(z,a)&\le\frac E2\int_0^\infty e^{-chu}\,du =\frac E{zh},\\ K_2(z,a)&\le\frac E2\int_0^\infty(a+u)e^{-chu}\,du =\frac Ez\left(\frac ah+\frac2{zh^2}\right). \end{aligned}$$

All multipliers of $K_1,K_2$ in the numerical specialization of Theorem 1 are positive, including $\log(17/(2\pi))$. Therefore substituting (1) gives an upper bound on its error constant. The exact certificate verifies $a=A'>1$ and all required signs before making that substitution.