On $0<x<1$, the geometric expansion gives summands $h_n(x)=x^{n+\beta -1}-x^{n-\beta }$. Their integrals areBoth powers are integrable because $0<\beta <1$. If $\beta <1/2$, each $h_n$ is nonnegative, since its first exponent is smaller. Their integrals form a convergent nonnegative series: the $n=0$ term is finite and subsequent terms are bounded by a constant times $n^{-2}$. The proved series-integration rule applies.

If $\beta >1/2$, apply precisely that rule to $-h_n$, which are nonnegative, and negate the result. At $\beta =1/2$ all summands and the integrand are zero. Thus in every case the integral is absolutely integrable andThe common sign of the summands is the reason no unproved theorem for integrating arbitrary signed series is needed.

For $x>0$, the geometric series givesEvery summand is nonnegative and has integral $(n+\alpha {)}^{-2}$ by part (a). This numerical series converges by comparison with ${\sum }_{n\ge 1}{n}^{-2}$. The series-integration rule proved above therefore yieldsThe value at the single endpoint zero does not affect this Lebesgue integral.

For $\alpha >0$, integration by parts on $[0,R]$ givesThe exponential remainder tends to zero. Since the integrand is nonnegative, the monotone convergence theorem applied to truncations identifies the improper integral with the Lebesgue integral. ThereforeIts finite value proves integrability as well as evaluating it.

Put $p=4272943$. Since $p\equiv 7(\mathrm{mod}8)$, the supplementary laws give $(-1/p)=-1$ and $(2/p)=1$. Therefore $(-286/p)=-(143/p)$.

Both $143$ and $p$ are three modulo four. Reciprocity for the Jacobi symbol, followed by the supplied remainder, givesAgain both denominators are three modulo four, soNow $103\equiv 7(\mathrm{mod}8)$ and $40=8\cdot 5$. Consequentlybecause the nonzero squares modulo five are one and four. Thus $(103/143)=1$, $(143/p)=-1$, and finallyThe congruence is soluble, with exactly two distinct solutions modulo this prime: a nonzero root and its negative. The computation uses a prime denominator in its final step; a positive Jacobi symbol with composite denominator would not by itself imply solubility.

Factor $a={\prod }_i{q}_i$ and $b={\prod }_j{p}_j$, listing prime factors with repetitions. Coprimality ensures $q_i\ne p_j$. By the definition of the Jacobi symbol and multiplicativity of the Legendre symbol,For odd integers $r,s$,since the difference is $(r-1)(s-1)/2$, an even integer. Inductively the two sums in the exponent are congruent to $(a-1)/2$ and $(b-1)/2$ modulo two. HenceIf either number is one, the empty product convention proves the same identity. This extends the prime law to arbitrary coprime positive odd denominators.