 Solution

Remove the apparent singularity at $x=0$ by definingThis is jointly continuous: write $\varphi (t,x)=t\mathrm{sinc}(tx)$ using the unnormalized sinc function, whose value at zero is one. Its partial derivative in $t$ isalso at $x=0$, since there $\varphi (t,0)=t$. 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 $\varphi$, therefore makes $g$ continuously differentiable on $[0,\infty )$, with a right derivative at zero, andFor $t>0$, direct integration givesAt zero the same integral gives $\boxed{g_{+}^{\prime }(0)=1}$, so no division by zero is required. Also $g(0)=0$. The substitution rule $u=tx$ identifies $g(t)={\int }_0^t\sin u/udu$ for $t>0$, the sine integral.