A direct derivative bound by $1/x$ would not be integrable, so the earlier theorem cannot be applied that way. Instead, for $t>0$ make the legitimate substitution $s=tx$ in the absolutely convergent integral:The tail integral has continuous integrand away from zero and converges absolutely. The fundamental theorem of calculus therefore gives derivative $-\cos t/t^2$ with respect to its lower endpoint. The product rule yieldsThis establishes differentiability carefully without an invalid domination argument.

For $0<x<1$, expand $(1-x{)}^{-2}={\sum }_{k\ge 1}k{x}^{k-1}$. All terms after multiplication by $(\log x{)}^2$ are nonnegative, so monotone convergence permits integration term by term. Substituting $x=e^{-t}$ and integrating by parts twice givesConsequently the integral is ${\sum }_{k\ge 1}2/k^2$, and the supplied sum gives $\boxed{{\pi }^2/3}$.

Apply monotone convergence to $A_n={\sum }_{k=1}^n|a_k|$. Its limit $A={\sum }_{k\ge 1}|a_k|$ has integral ${\sum }_k\int |a_k|<\infty$, so $A<\infty$ almost everywhere. Thus the original series converges absolutely almost everywhere. Its partial sums are bounded in absolute value by the integrable function $A$, and their limit $s$ satisfies $|s|\le A$. Dominated convergence givesAssigning any value on the null exceptional set defines an integrable representative of $s$.

By (ii), failure of divisibility is equivalent to $q-1\mid 30$. The positive divisors of $30$ are $1,2,3,5,6,10,15,30$; adding one gives $2,3,4,6,7,11,16,31$. Retaining only primes yields

Each nonzero quadratic residue is attained twice as $x^2$ for $1\le x\le p-1$. Therefore twice the sum of those residues is ${\sum }_{x=1}^{p-1}{x}^2$, which is zero modulo $p$ by (ii), since $p-1>2$. As two is invertible modulo $p$, the sum of the residues is zero. The sum of all nonzero residues is $p(p-1)/2\equiv 0$, so subtracting gives the same result for the nonresidues. Hence $\boxed{\text{both indicated sums are }0(\mathrm{mod}p)}$.