The function $h$ is measurable and nonnegative. In first-quadrant polar coordinates, including the Jacobian, its integral isThe radial factor is $1/2$ by the supplied fact and the angular integrand is at most one, giving finiteness. The nonnegative change-of-variables formula is valid before finiteness is known, and this bound proves $h\in L^1$.

Fubini may now factor the Cartesian integral. Substituting $u=x^2$, $v=y^2$ in the respective factors givesBoth are finite: their powers at zero exceed $-1$, and the exponential controls infinity. Comparing with the polar expression proveswhich is exactly the product of integrals requested.

The function $g$ is measurable, and for fixed positive $y$ both terms are integrable over $0<x<1$ by linear substitution and integrability of $f$. Those substitutions giveThe strict sign follows from positivity of $f$ on a nonempty interval, since $a>1$. Similarly each section in $y$ is absolutely integrable for fixed $x>0$, andIf $g$ were absolutely integrable on the stated rectangle, Fubini would make the two finite iterated integrals equal. The first would be strictly negative and the second strictly positive, which is impossible. Therefore $\boxed{g\notin L^1((0,1)\times (1,\infty ))}$. The one-dimensional section integrability does not rescue the missing product-space hypothesis.

For product measure spaces, Tonelli's theorem states that for a nonnegative measurable function the product-space integral equals either iterated integral, allowing the value $+\infty$. Fubini's theorem states that if a measurable function has finite integral of its absolute value, then its sections are integrable almost everywhere, both iterated integrals exist and equal the product-space integral. The absolute-integrability hypothesis is what permits this conclusion for a signed function. These statements apply to the Lebesgue product spaces used here, which are sigma-finite.

The integrand is nonnegative and measurable. It is bounded by $x$ near zero and by $x{e}^{-x}$ for $x>0$, so its integral is finite. For every $x>0$ the geometric expansion gives $x/(e^x+1)={\sum }_{n\ge 1}(-1{)}^{n-1}x{e}^{-nx}$. To justify termwise integration, note thatTonelli gives an integrable sum of absolute values; dominated convergence then applies to the signed partial sums. ConsequentlyIf evaluated further, this is $(1-2^{-1})\zeta (2)={\pi }^2/12$.

Suppose there are only finitely many primes $p_1,\dots ,p_r\equiv 3(\mathrm{mod}8)$, and let $N$ be their product. This is odd, so $N^2+2\equiv 3(\mathrm{mod}8)$. No listed prime divides this number, since reduction modulo any $p_i$ gives two. Every prime divisor $q$ is odd and does not divide $N$, and satisfies $N^2\equiv -2(\mathrm{mod}q)$. Thus $(-2/q)=1$, forcing $q\equiv 1$ or $3(\mathrm{mod}8)$ by (v).

A product of prime factors all congruent to one modulo eight could not be three modulo eight. More precisely, the product of factors congruent to three is three modulo eight exactly when their total exponent is odd. Thus at least one prime divisor $q\equiv 3(\mathrm{mod}8)$ occurs. It is not in the alleged complete list, a contradiction. Hence there are infinitely many primes congruent to three modulo eight.

Euler's criterion gives $(-1/p)=(-1{)}^{(p-1)/2}$. The Legendre symbol is multiplicative, so multiplying this sign by the result of (iv) yields

For $a=2$ the displayed multiples are $2,4,\dots ,p-1$, already least positive residue classes. Those greater than $p/2$ correspond to $j>p/4$. Thus their number is $m=(p-1)/2-\lfloor p/4\rfloor$. For $p\equiv 1,3,5,7(\mathrm{mod}8)$, its parity is respectively even, odd, odd, even. Gauss's lemma therefore gives

Gauss's lemma for quadratic residues states: if $p\nmid a$ and $m$ is the number of least positive residue classes among $a,2a,\dots ,((p-1)/2)a$ that exceed $p/2$, then $\boxed{(a/p)=(-1{)}^m}$.

Euler's criterion states that for $p\nmid a$,For a multiple of $p$, both the Legendre symbol and that positive power vanish modulo $p$, so the congruence extends to that case as well.