Improper and Parameter-Dependent Multiple Integrals: Examples and Counterexamples

Direct differentiation gives $f(x,y)={\mathrm{\partial }}_y(y\mathrm{/}(x^2+y^2))=-{\mathrm{\partial }}_x(x\mathrm{/}(x^2+y^2))$.

Integrating the first identity in $y$ gives $1\mathrm{/}(1+x^2)$, and integrating the second in $x$ gives $-1\mathrm{/}(1+y^2)$; [L4] and [L1] therefore give the two iterated values $\pi \mathrm{/}4$ and $-\pi \mathrm{/}4$.

On $[a,u]\times [b,v]\subset (0,1{)}^2$, step 1.1 gives the double integral $\arctan (u\mathrm{/}v)-\arctan (a\mathrm{/}v)-\arctan (u\mathrm{/}b)+\arctan (a\mathrm{/}b)$. Taking $u_j=v_j=1-1\mathrm{/}(j+2)$ and either $(a_j,b_j)=((j+2{)}^{-2},(j+2{)}^{-1})$ or the swapped pair produces nested compact Jordan exhaustions with limits $-\pi \mathrm{/}4$ and $\pi \mathrm{/}4$. By [L3], and hence by [L2], no exhaustion-independent signed improper integral exists.

A fixed coordinate $x\in [0,1]$ belongs to at most one prime grid, because a fraction $k\mathrm{/}p$ with $1\le k<p$ is reduced and its prime denominator is unique. Thus each horizontal and vertical section of $E$ is finite; covering its finitely many points by intervals of arbitrarily small total length gives content zero, so [L4] makes every section of $f$ integrable with value zero. By [L1], both iterated integrals are zero.

By [L5] and [L6], primes are unbounded, so in every nonempty open rectangle in the unit square a sufficiently fine prime grid supplies a point of $E$. By [L3], the same rectangle contains a point with irrational first coordinate, which is outside $E$. Hence both $E$ and its complement are dense in the square.

Every nondegenerate grid cell therefore has supremum $1$ and infimum $0$, so every Darboux gap equals the area of the unit square, namely $1$. By [L2], $f$ is not double Riemann integrable.

Step 1.1 gives equal iterated integrals although step 2.1 gives no double integral, refuting the Statement.

By [L3], [L5], [L6], and [L7], $(-e^{-u}{)}^{\mathrm{\prime }}=e^{-u}$ and ${\int }_0^R{e}^{-u}du=1-e^{-R}\to 1$. If $t\mathrm{\ne }0$, the substitution $u=t^{2}x$ in [L2] therefore gives $F(t)=t{\int }_0^{\mathrm{\infty }}{e}^{-u}du=t$; at $t=0$, the integrand is identically zero, so [L4] gives $F(0)=0$. Thus $F(t)=t$ for every real $t$.

By [L3], ${\mathrm{\partial }}_{t}f(x,t)=(3t^2-2x{t}^4)e^{-t^{2}x}$, so ${\mathrm{\partial }}_{t}f(x,0)=0$ for every $x\ge 0$ and its improper integral is $0$.

Step 1.1 gives $F^{\mathrm{\prime }}(0)=1$, while step 1.2 gives ${\int }_0^{\mathrm{\infty }}{\mathrm{\partial }}_{t}f(x,0)dx=0$. Restricting the witness to the open domain $(0,\mathrm{\infty })$ changes none of these integrals; there $f$ and ${\mathrm{\partial }}_{t}f$ are continuous and the zero base slice is absolutely integrable. Thus every hypothesis of [L1] except a parameter-uniform integrable derivative bound holds, differentiation under the sign fails, and no such dominator can exist near $t=0$.

The substitution $u=\sqrt{a}x$ is licensed by [L4], and [L1] gives $F(a)=a^{-1\mathrm{/}2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-u^2}du=\sqrt{\pi }{a}^{-1\mathrm{/}2}$.

Let $C\subset (0,\mathrm{\infty })$ be compact and put $m:=\min C>0$. Then $\mathrm{\mid }{\mathrm{\partial }}_a{e}^{-a{x}^2}\mathrm{\mid }=x^2{e}^{-a{x}^2}\le x^2{e}^{-m{x}^2}$ for $a\in C$. Applying [L5] with the variable $x^2$ shows this is eventually at most $x^{-2}$, so [L6] and [L7] make both tails integrable; continuity handles the compact middle interval. The slice at $a=1$ is absolutely integrable by [L1]. Thus every hypothesis of [L2] holds on the open parameter interval $(0,\mathrm{\infty })$ and gives $F^{\mathrm{\prime }}(a)=-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^2{e}^{-a{x}^2}dx$.

Differentiating the explicit formula in step 1.1 with [L3] gives $F^{\mathrm{\prime }}(a)=-(\sqrt{\pi }\mathrm{/}2)a^{-3\mathrm{/}2}$, which combined with step 1.2 gives the displayed second-moment identity.