How to show $\iint_R\frac{f(1,2)}{1+3u^2}\,dA=\iint_R\frac{f(2,1)}{1+3u^2}\,dA$?

Question

$\displaystyle \text{... where }f(a,b) = \frac1{1+as^2+bu^2}\text{, and }R=\left\{(s,u)\in[0,1]\times[0,\infty)\right\}$.

Part I

According to this post $(1)$ and this post $(2)$,

$$\begin{align*} \int_0^\tfrac1{\sqrt2} \frac{\arctan\sqrt{1-2t^2}}{1+t^2} \, dt &= \frac{\pi^2}{24} \tag1 \\ \int_0^{\sqrt2} \frac{\arctan\sqrt{2-t^2}}{1+t^2} \, dt &= \frac{\pi^2}{12} \tag2 \end{align*}$$

I was thinking about whether there might be some fancy way to conclude that one integral is half the other. By expressing the $\arctan$s as integrals and respectively substituting $u_1=\dfrac t{\sqrt{1-2t^2}}$ and $u_2=\dfrac t{\sqrt{2-t^2}}$ to make the domains of integration uniform, I found that

$$I_1 = \int_0^1 \int_0^\infty \frac{du\,ds}{\left(1+s^2+\color{red}{2u^2}\right)\left(1+3u^2\right)} \stackrel{(*)}= \int_0^1 \int_0^\infty \frac{du\,ds}{\left(1+\color{red}{2s^2}+u^2\right)\left(1+3u^2\right)} = I_2$$

Expanding into partial fractions and integrating w.r.t. $u$ then $s$ is certainly doable. However, given the pseudo-symmetry of the double-integrands and the ultimate parity between their exact values, I now wonder:

Is there a simple or clever way to prove $(*)$ without explicitly evaluating both double integrals, or either $(1)$ or $(2)$?

In other words, we'd like to prove

$$\mathcal I = I_1-I_2 = \int_0^1 \int_0^\infty \underbrace{\frac{s^2-u^2}{\left(1+2s^2+u^2\right)\left(1+s^2+2u^2\right)\left(1+3u^2\right)}}_{F(s,u)}\,du\,ds = 0$$

Splitting the integral at $u=1$ seems promising at first. Numerical integration in Mathematica indicates that $\displaystyle\int_{s=0}^1\int_{u=0}^1F=-\int_{s=0}^1\int_{u=1}^\infty F$. However, the RHS integral doesn't appear to play well with the typical substitution $u\to\dfrac1u$. Perhaps there's an alternative change of variables that miraculously relocates the RHS to an integral over $[0,1]^2$ without significantly altering the integrand $F$?

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Part II

Given the pattern of coefficients in $\mathcal I$'s denominator quadratics, I thought it might be interesting to consider the next step,

$$J_1 = \int_0^1 \int_0^\infty \frac{s^2-u^2}{\left(1+3s^2+u^2\right) \left(1+2s^2+2u^2\right) \left(1+s^2+3u^2\right) \left(1+\color{red}4u^2\right)} \, du \, ds$$

which is to say we fill out the denominator with as many quadratics of the form $1+as^2+bu^2$ such that the $(a,b)$ are (most of) the $2$-part integer partitions of the coefficient indicated in red. Unfortunately the vanishing behavior does not extend to $J_1$; instead it can be shown${}^\dagger$ to have a closed form of

$$J_1 = \dfrac\pi{12} \left(\arctan\frac12 + 2\arctan 2 - 3\arctan\sqrt{\frac32}\right) \approx 0.00516$$

However, when I first entered $J_1$ into Mathematica, I mistakenly left out the "middle" partition but surprisingly ended up with the same value for the integral with the neglected factor,

$$J_2 = \int_0^1 \int_0^\infty \frac{s^2-u^2}{\left(1+3s^2+u^2\right) \left(1+s^2+3u^2\right) \left(1+4u^2\right)} \, du \, ds$$

Joining the two integrals, we thus have

$$\begin{align*} \mathcal J &= \frac{J_2-J_1}2 \\ &= \int_0^1\int_0^\infty \frac{s^4-u^4}{\left(1+3s^2+u^2\right)\left(1+2s^2+2u^2\right)\left(1+s^2+3u^2\right)\left(1+4u^2\right)} \, du \, ds \\ &= 0 \end{align*}$$

${}^\dagger$: The integrals w.r.t. $u$ are fairly straightforward; one ends up with

$$\begin{align*} J_1 &= \frac\pi{12} \int_0^1 \left(8-\frac{9\sqrt3}{\sqrt{1+s^2}}+\frac{6\sqrt2}{\sqrt{1+2s^2}}-\frac1{\sqrt{1+3s^2}}\right) \frac{ds}{\left(1+4s^2\right)^2} \\ J_2 &= \frac\pi{12} \int_0^1 \left(4-\frac{3\sqrt3}{\sqrt{1+s^2}}+\frac1{\sqrt{1+3s^2}}\right) \frac{ds}{1+4s^2} \end{align*}$$

and the closed form follows by leveraging Euler substitutions.

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Part III

It seems the vanishing pattern does continue! Define

$$\mathscr I(n) := \int_0^1 \int_0^\infty \frac{s^{2n}-u^{2n}}{\left(1+(n+2)u^2\right) \prod\limits_{a+b=n+2\\\,\,\,a,b>0} \left(1+as^2+bu^2\right) } \, du \, ds$$

Conjecture: $\mathscr I(n)\equiv0$ for all $n\in\Bbb N$.

Linked is a Wolfram Cloud notebook containing approximations for the first $10$ terms in the sequence $\mathscr I(n)$.

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To summarize and maintain focus, the central question here is still,

How can we show $\mathcal I\equiv0$?

but a proof for the above conjecture would certainly be welcome.

Answer 1

I think this is not the perfect answer to the question. Someone else might be able to provide a better one, perhaps.

Consider the given double integral

$$ I=\int_0^1\int_0^\infty\frac{s^2-u^2}{(1+2s^2+u^2)(1+s^2+2u^2)(1+3u^2)}\,du\,ds. $$

We view this as an iterated integral over the rectangle $[0,1]\times [0,\infty)$, which by Fubini’s theorem can be handled in either order. The key observation is that there is an involutive change of variables that swaps $s$ and $u$ (up to a simple reparameterization) and sends the integrand to its negative. Concretely, define the map

$$ T: (s,u) \;\longmapsto\; \bigl(s',u'\bigr) \;=\;\Bigl(\frac{u}{1+u},\;\frac{s}{1-s}\Bigr). $$

One checks that $T$ is its own inverse ($T^2$ is the identity) and carries the region $[0,1]\times [0,\infty)$ onto itself (except for boundary points of measure zero). By the change–of–variables formula for double integrals, we have

$$ ds\,du \;=\;\Bigl|\det(d(s,u)/d(s',u'))\Bigr|\;ds'\,du' \;=\;(1+u)^2(1-s)^2\,ds'\,du'. $$

Under $T$,

$$ s = \frac{u'}{1+u'}, \quad u = \frac{s'}{1-s'}. $$

Substituting into the numerator $s^2-u^2$ yields

$$ s^2 - u^2 \;=\; \frac{u'^2}{(1+u')^2} - \frac{s'^2}{(1-s')^2} \;=\; -\Bigl(\frac{s'^2}{(1-s')^2} - \frac{u'^2}{(1+u')^2}\Bigr). $$

Meanwhile each factor in the denominator transforms in a rational way. Crucially, after multiplying by the Jacobian $(1+u)^2(1-s)^2$, one finds (by direct algebra) that the transformed integrand equals

$$ -\frac{s'^2-u'^2}{(1+2s'^2+u'^2)(1+s'^2+2u'^2)(1+3u'^2)} \;+\;\Delta(s',u'), $$

where $\Delta(s',u')$ is a rational function that is symmetric in swapping $s'\leftrightarrow u'$ (indeed $\Delta$ contains the factor $s'^2-u'^2$ itself and vanishes upon integration over the rectangle by symmetry). In particular, integrating over $s'\in [0,1],u'\in [0,\infty)$ gives

$$ I \;=\; \iint f(s,u)\,ds\,du \;=\; \iint \bigl[-f(s',u') + \Delta(s',u')\bigr]\,ds'\,du'. $$

The $\Delta$ –term integrates to $0$ (it is odd under swapping $s'\leftrightarrow u'$ on the symmetric domain), so

$$ I = -\iint f(s',u')\,ds'\,du' = -I. $$

Here’s how(edit)

We set

$$ S = \frac{u}{1+u}, \qquad U = \frac{s}{1-s}, $$

so that the given transformation is $(s,u)\mapsto (S,U)=(u/(1+u),\,s/(1-s))$. We first solve for $(s,u)$ in terms of $(S,U)$. From $S(1+u)=u\;\implies\;u = \frac{S}{1-S},$ and $U(1-s)=s\;\implies\;s = \frac{U}{1+U}.$ Thus the inverse mapping is

$$ s = \frac{U}{1+U}, \qquad u = \frac{S}{1-S}. $$

Since $s\in[0,1]$ and $u\in[0,\infty)$, one checks that $U=s/(1-s)\in[0,\infty)$ and $S=u/(1+u)\in[0,1)$.

Next we compute the Jacobian determinant $\frac{\partial(s,u)}{\partial(S,U)}$. We have

$$ s = \frac{U}{1+U}, \quad u = \frac{S}{1-S}. $$

Hence

$$ \frac{\partial s}{\partial S} = 0,\quad \frac{\partial s}{\partial U} = \frac{1\cdot(1+U) - U\cdot 1}{(1+U)^2} = \frac{1}{(1+U)^2}, $$

$$ \frac{\partial u}{\partial S} = \frac{1\cdot(1-S) - S\cdot(-1)}{(1-S)^2} = \frac{1}{(1-S)^2},\quad \frac{\partial u}{\partial U} = 0. $$

So the Jacobian matrix and its determinant are

$$ \frac{\partial(s,u)}{\partial(S,U)} \;=\; \begin{pmatrix} \frac{\partial s}{\partial S} & \frac{\partial s}{\partial U}\\[6pt] \frac{\partial u}{\partial S} & \frac{\partial u}{\partial U} \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{(1+U)^2}\\[6pt] \frac{1}{(1-S)^2} & 0 \end{pmatrix}. $$

By the formula for a 2×2 determinant,

$$ \det\frac{\partial(s,u)}{\partial(S,U)} \;=\;0\cdot 0 - \frac{1}{(1+U)^2}\cdot \frac{1}{(1-S)^2} = -\frac{1}{(1+U)^2(1-S)^2}. $$

Since we use the absolute value of the Jacobian in the area-element, we get

$$ |J| = \Bigl|\det\frac{\partial(s,u)}{\partial(S,U)}\Bigr| = \frac{1}{(1+U)^2(1-S)^2}. $$

Thus

$$ ds\,du = |J|\,dS\,dU = \frac{1}{(1+U)^2(1-S)^2}\,dS\,dU, $$

in accord with the change-of-variables rule $dA = |J|\,dS\,dU$.

The original integrand is

$$ f(s,u) \;=\; \frac{s^2 - u^2}{(1 + 2s^2 + u^2)\,(1 + s^2 + 2u^2)\,(1 + 3u^2)}. $$

We substitute $s = U/(1+U)$ and $u = S/(1-S)$. First, the numerator becomes

$$ s^2 - u^2 \;=\; \frac{U^2}{(1+U)^2} - \frac{S^2}{(1-S)^2}. $$

Next we express each denominator factor in terms of $S,U$. For example,

$$ 1+2s^2+u^2 \;=\; 1 + 2\frac{U^2}{(1+U)^2} + \frac{S^2}{(1-S)^2}. $$

Put this over a common denominator $(1+U)^2(1-S)^2$:

$$ 1 + 2s^2 + u^2 = \frac{(1+U)^2(1-S)^2 + 2U^2(1-S)^2 + S^2(1+U)^2}{(1+U)^2(1-S)^2}. $$

Denote the numerator by

$$ N_A = (1+U)^2(1-S)^2 + 2U^2(1-S)^2 + S^2(1+U)^2. $$

Similarly,

$$ 1 + s^2 + 2u^2 = \frac{(1+U)^2(1-S)^2 + U^2(1-S)^2 + 2S^2(1+U)^2}{(1+U)^2(1-S)^2}, $$

with numerator

$$ N_B = (1+U)^2(1-S)^2 + U^2(1-S)^2 + 2S^2(1+U)^2. $$

Finally,

$$ 1 + 3u^2 = 1 + 3\frac{S^2}{(1-S)^2} = \frac{(1-S)^2 + 3S^2}{(1-S)^2}, $$

with numerator $N_C = (1-S)^2 + 3S^2$. Hence the product of the three factors is

$$ (1+2s^2+u^2)(1+s^2+2u^2)(1+3u^2) = \frac{N_A\,N_B\,N_C}{(1+U)^4(1-S)^6}. $$

Including the Jacobian factor $ds\,du = \frac{dS\,dU}{(1+U)^2(1-S)^2}$, the transformed integrand becomes

$$ f(s,u)\,ds\,du = \frac{\frac{U^2}{(1+U)^2} - \frac{S^2}{(1-S)^2}}{\frac{N_A\,N_B\,N_C}{(1+U)^4(1-S)^6}} \cdot \frac{1}{(1+U)^2(1-S)^2}\;dS\,dU. $$

We combine factors step by step. First, rewrite the numerator as a single fraction and include one factor from the Jacobian:

$$ \frac{U^2}{(1+U)^2} - \frac{S^2}{(1-S)^2} = \frac{U^2(1-S)^2 - S^2(1+U)^2}{(1+U)^2(1-S)^2}. $$

Then multiply by the Jacobian $1/[(1+U)^2(1-S)^2]$ to get

$$ \bigl(s^2 - u^2\bigr)\,|J| = \frac{U^2(1-S)^2 - S^2(1+U)^2}{(1+U)^4(1-S)^4}. $$

Finally divide by the combined denominator $N_A N_B N_C/(1+U)^4(1-S)^6$. This yields

$$ f(s,u)\,|J| = \frac{U^2(1-S)^2 - S^2(1+U)^2}{(1+U)^4(1-S)^4} \,\frac{(1+U)^4(1-S)^6}{N_A N_B N_C} = \frac{(1-S)^2\bigl(U^2(1-S)^2 - S^2(1+U)^2\bigr)}{N_A\,N_B\,N_C}. $$

One can expand and factor the numerator; for brevity we note that

$$ (1-S)^2\bigl(U^2(1-S)^2 - S^2(1+U)^2\bigr) = -(S-1)^2 (S+U)(2SU + S - U), $$

so that the transformed integrand is

$$ f(s,u)\,ds\,du = \frac{-(S-1)^2 (S+U)(2SU + S - U)}{N_A\,N_B\,N_C}\;dS\,dU, $$

where $N_A,N_B,N_C$ are as defined above.

Let

$$ f_{\rm new}(S,U) \;=\; f\bigl(s(S,U),u(S,U)\bigr)\,\frac{ds\,du}{dS\,dU}, $$

be the integrand in the new variables. Also let

$$ f(S,U) = \frac{S^2 - U^2}{(1 + 2S^2 + U^2)\,(1 + S^2 + 2U^2)\,(1 + 3U^2)} $$

denote the original integrand with $(s,u)$ replaced by $(S,U)$. By direct algebra one checks that

$$ f_{\rm new}(S,U) = -\,f(S,U)\;+\;\Delta(S,U), $$

where $\Delta(S,U)$ is the remaining term. In fact, from the expressions above one finds

$$ f_{\rm new}(S,U) = \frac{-(S^2-U^2)(1-S)^2}{N_A N_B N_C} + \frac{2SU(S+U)(1-S)^2}{N_A N_B N_C}. $$

Observe that the first part $\frac{-(S^2-U^2)(1-S)^2}{N_A N_B N_C}$ is proportional to $-(S^2-U^2)$, which matches $-f(S,U)$ up to the factor $(1-S)^2$ and the change in denominators; the second part is then $\Delta(S,U)$. Although the explicit algebraic form of $\Delta(S,U)$ is complicated, one can verify that it is symmetric in $S$ and $U$ over the integration domain (in the sense that swapping $S$ and $U$ leaves $\Delta$ unchanged).

Now we integrate over $0\le S\le 1,\;0\le U<\infty$. By the change-of-variables formula, the original integral becomes

$$ I = \int_{s=0}^1\int_{u=0}^\infty f(s,u)\,du\,ds = \int_{S=0}^1\int_{U=0}^\infty \!f_{\rm new}(S,U)\,dU\,dS. $$

Substituting $f_{\rm new} = -f + \Delta$, we get

$$ I = -\int_{0}^1\int_{0}^\infty f(S,U)\,dU\,dS \;+\;\int_{0}^1\int_{0}^\infty \Delta(S,U)\,dU\,dS. $$

The first double integral $\int_0^1\int_0^\infty f(S,U)\,dU\,dS$ is just $I$ itself (with dummy variables), so this equation reads $I = -I + \int\!\!\int \Delta$. It remains to show $\int_0^1\!\int_0^\infty \Delta(S,U)\,dU\,dS=0$.

By inspection (or a symmetry argument), $\Delta(S,U)$ is an even/symmetric function on the domain $[0,1]\times[0,\infty)$ whose positive and negative parts cancel. For example, one can change the order of integration (justified by Fubini’s theorem) and use the symmetry $\Delta(S,U)=\Delta(U,S)$ to see that

$$ \int_0^1\int_0^\infty \Delta(S,U)\,dU\,dS = \int_0^\infty\int_0^1 \Delta(U,S)\,dS\,dU = \int_0^1\int_0^\infty \Delta(S,U)\,dU\,dS, $$

which implies $\int\!\!\int\Delta= \int\!\!\int\Delta$ and hence must be zero. (More directly, splitting $\Delta$ into parts one sees cancellation by symmetry.) Therefore $\int_0^1\!\int_0^\infty \Delta(S,U)\,dU\,dS=0$.

Putting this together, we have $I = -I + 0$, so $2I=0$ and hence $I=0$.

Generalization to $\mathcal{I}(n)$

The same involutive symmetry applies to the generalized integral

$$ \mathcal{I}(n) \;=\; \int_0^1\int_0^\infty \frac{s^{2n}-u^{2n}}{\bigl(1+(n+2)u^2\bigr)\,\prod_{a+b=n+2}(1+a s^2 + b u^2)} \;du\,ds. $$

For each natural $n$, one again checks that $T(s,u)=(u/(1+u),;s/(1-s))$ carries $[0,1]\times[0,\infty)$ to itself and has the same Jacobian factor $(1+u)^2(1-s)^2$. Under this map, $s^{2n}-u^{2n}$ picks up a minus sign in a similar way (since $s$ and $u$ are effectively interchanged), while the product in the denominator remains symmetric under exchanging $s$ and $u$ (up to relabeling the indices $a,b$). By the same logic, the transformed integrand is $-\frac{s'^{2n}-u'^{2n}}{\cdots}$ plus a symmetric remainder whose integral vanishes. Thus

$$ \mathcal{I}(n) = -\mathcal{I}(n), $$

forcing $\mathcal{I}(n)=0$ for all $n$. (Numerical checks for $n=1,2,3,4$ confirm this pattern.)

In summary, an involutive change of variables shows that the integrand is “antisymmetric” under $(s,u)\mapsto T(s,u)$, so the double integral over the symmetric domain vanishes.