Uniform sine integral bound and dirichlet value

The derivative of sine at zero makes $\sin u\mathrm{/}u\to 1$. MVT and $\mathrm{\mid }\cos u\mathrm{\mid }\le 1$ give $\mathrm{\mid }\sin u\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }$, so the extended quotient is continuous and bounded by one on $[0,1]$. It has a proper integral on every bounded interval. AC supplies the countable choice needed to identify these with Lebesgue integrals using F3 (and the compact integration interface F15).

For $B>A\ge 1$ and $\epsilon \ge 0$, put $w(u)=e^{-\epsilon u}\mathrm{/}u$. This is positive, decreasing, and continuously differentiable on $[A,B]$, with ${\int }_A^B\mathrm{\mid }w\mathrm{\prime }\mathrm{\mid }=w(A)-w(B)$. Integration by parts against $\sin u=(-\cos u)\mathrm{\prime }$ gives $\mid {\int }_A^{B}w(u)\sin udu\mid \le w(A)+w(B)+{\int }_A^B\mathrm{\mid }w\mathrm{\prime }\mathrm{\mid }=2w(A)\le 2\mathrm{/}A\mathrm{.}$ At $\epsilon =0$ this proves the Cauchy property of $S(T)$ as $T\to \mathrm{\infty }$ and the bound $\mathrm{\mid }S(T)\mathrm{\mid }\le 3$ for all $T\ge 0$. For positive damping it also bounds the infinite tail by $2\mathrm{/}A$.

Fix $\epsilon >0$. FTC gives $\sin u\mathrm{/}u={\int }_0^1\cos (su)ds$, including $u=0$. The double absolute integral of $e^{-\epsilon u}\cos (su)$ on $[0,\mathrm{\infty })\times [0,1]$ is at most $1\mathrm{/}\epsilon$, so Fubini applies. Differentiating $\frac{e^{-\epsilon u}(-\epsilon \cos (su)+s\sin (su))}{{\epsilon }^2+s^2}$ gives $e^{-\epsilon u}\cos (su)$; its limit at infinity is zero and its value at zero is $-\epsilon \mathrm{/}({\epsilon }^2+s^2)$. Consequently $J_{\epsilon }:={\int }_0^{\mathrm{\infty }}{e}^{-\epsilon u}\frac{\sin u}{u}du={\int }_0^1\frac{\epsilon }{{\epsilon }^2+s^2}ds=\arctan (1\mathrm{/}\epsilon )\mathrm{.}$

On $[0,A]$, dominated convergence gives convergence of the damped integral to $S(A)$ as $\epsilon \downarrow 0$. The two tails, damped and undamped, are each at most $2\mathrm{/}A$ by step 2.1. Thus, first taking $\epsilon \downarrow 0$ and then $A\to \mathrm{\infty }$, $J_{\epsilon }\to {\lim }_{T}S(T)$. The increasing inverse arctangent has limit $\pi \mathrm{/}2$ at infinity: its values are below $\pi \mathrm{/}2$, and for every $v<\pi \mathrm{/}2$ in its range, $y>\tan v$ implies $\arctan y>v$. Hence $S(T)\to \pi \mathrm{/}2$.

For $z=0$ the symmetric integral is zero. For $z\mathrm{\ne }0$, evenness in $t$ and substitution $u=\mathrm{\mid }z\mathrm{\mid }t$ give ${\int }_{-T}^T\frac{\sin (tz)}{t}dt=2\mathrm{sgn}(z)S(T\mathrm{\mid }z\mathrm{\mid })\mathrm{.}$ Its absolute value is at most six, and for each fixed nonzero $z$ its limit is $\pi \mathrm{sgn}(z)$. The uniform bound, but not uniform convergence in z, is asserted.