= Solution
Remove the apparent singularity at $x=0$ by defining
$$
\phi(t,x)=\begin{cases}\sin(tx)/x,&x>0,\\t,&x=0.\end{cases}
$$
This is jointly continuous: write $\phi(t,x)=t\,\operatorname{sinc}(tx)$ using the <unnormalized sinc function>, whose value at zero is one. Its <partial derivative> in $t$ is
$$
\partial_t\phi(t,x)=\cos(tx)
$$
also at $x=0$, since there $\phi(t,0)=t$. This <partial derivative> is jointly continuous on the whole strip.
The value at the single endpoint $x=0$ does not change a <Lebesgue integral>. Part (b), whose <dominated convergence> proof applies to $\phi$, therefore makes $g$ continuously differentiable on $[0,\infty)$, with a <right derivative> at zero, and
$$
g'(t)=\int_0^1\cos(tx)\,dx.
$$
For $t>0$, direct <integration> gives
$$
\boxed{g'(t)=\frac{\sin t}{t},\qquad tg'(t)=\sin t.}
$$
At zero the same integral gives $\boxed{g'_+(0)=1}$, so no division by zero is required. Also $g(0)=0$. The <substitution rule> $u=tx$ identifies $g(t)=\int_0^t\sin u/u\,du$ for $t>0$, the <sine integral>.
Back to article page
