A naive absolute bound on the original derivative would be $1/x$, which is not integrable on $[1,\infty )$. Instead make the change of variable $s=tx$, valid for $t>0$:The integral is absolutely convergent for every positive lower endpoint. Its integrand is continuous there, so the fundamental theorem of calculus givesThe product rule now yieldsThis proves differentiability and the equation without an unjustified differentiation of a conditionally convergent derivative integral.

For $0<x<1$, expand $(1-x{)}^{-2}={\sum }_{k\ge 1}k{x}^{k-1}$. Multiplying by $(\log x{)}^2$ gives nonnegative terms, so the monotone convergence theorem permits termwise integration. Put $x=e^{-t}$ to obtainwhere scaling followed by two integrations by parts evaluates the last integral. HenceThe finite value also establishes integrability at both endpoints.

Apply the monotone convergence theorem to $s_N={\sum }_{k=1}^N|a_k|$. Its limit $s$ satisfiesThus $s$ is finite almost everywhere, which proves absolute convergence of ${\sum }_k{a}_k$ almost everywhere. Define its sum $a$ there, and set $a=0$ on the exceptional null set. Then $|a|\le s$, so $a$ is integrable. The partial sums of ${\sum }_k{a}_k$ are also dominated by $s$; dominated convergence givesThe series of integrals converges absolutely, since $|\int a_k|\le \int |a_k|$.

Part (ii), with exponent $30$, shows that a prime $q$ fails to divide the sum precisely when $q-1$ divides $30$. The positive divisors are $1,2,3,5,6,10,15,30$. Adding one gives $2,3,4,6,7,11,16,31$, of which the primes areFor these primes the sum is $-1$ modulo $q$; for every other prime it is zero.

Let $R$ be the sum of the nonzero quadratic residues. Each has two square roots by part (iii), sowhere part (ii) applies because $p>3$ makes $p-1$ not divide $2$. Since $2$ is invertible modulo $p$, $R\equiv 0$.

The sum of all nonzero classes is $p(p-1)/2\equiv 0(\mathrm{mod}p)$. Subtracting $R$ shows that the sum of quadratic nonresidues is also zero. Consequently both requested sums vanish modulo $p$.

The Legendre symbol equal to $1$ says that $a$ is a nonzero square modulo $p$. Choose one root $r\ne 0$. Then $-r$ is another root, distinct because $p$ is odd. Any solution $x$ satisfiesin the field ${\mathbb{F}}_p$, so $x=r$ or $x=-r$. Thus there are

For $n\ge 1$, the zero residue contributes zero. Write $S={\sum }_{x\in {\mathbb{F}}_p^{\times }}{x}^n$. Multiplication by any nonzero $\alpha$ permutes these residues, so $S={\alpha }^{n}S$ in ${\mathbb{F}}_p$. If $(p-1)\nmid n$, choose a primitive root $g$; then $g^n\ne 1$, and $(1-g^n)S=0$ forces $S=0$.

If $(p-1)\mid n$, every nonzero $x$ has $x^n=1$, so $S=p-1\equiv -1(\mathrm{mod}p)$. ThereforeThe source also lists $n=0$, although its premise assumes $n\ge 1$. For that separate case, the integer sum has $p$ terms each equal to one, so it is zero modulo $p$.

A primitive root modulo $p$ is a residue class generating the cyclic multiplicative group $(\mathbb{Z}/p\mathbb{Z}{)}^{\times }$, so its order is $p-1$. If $g$ is one generator, all generators are $g^j$ with $\gcd (j,p-1)=1$. Thus the number iswhere $\phi$ is Euler's totient function. This includes $p=2$, with $\phi (1)=1$.