# Evaluating a double integral of a complicated rational function

Define the function $Q:\mathbb{C}^{2}\rightarrow\mathbb{C}$ to be the binary quadratic form,

$$Q{\left(z,w\right)}:=z^{2}+w^{2}.\tag{1a}$$

Also, define $P:\mathbb{C}^{4}\rightarrow\mathbb{C}$ to be the polynomial of degree $5$, in four variables,

\begin{align} P{\left(a,b,x,y\right)} &:=a\left[\left(a^{2}+1\right)\left(a^{2}+b^{2}+1\right)-4b^{2}\right]\\ &~~~~~+2\left[\left(a^{2}+1\right)\left(a^{2}+b^{2}+1\right)-2b^{2}\right]x\\ &~~~~~+a\left(a^{2}+b^{2}+1\right)x^{2}\\ &~~~~~+a\left(a^{2}+b^{2}+1\right)y^{2}.\tag{1b}\\ \end{align}

Note that $P{\left(a,b,x,y\right)}$ is obviously even in both $b$ and $y$.

Then, define the function $\mathcal{I}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ via the double integral

$$\mathcal{I}{\left(a,b\right)}:=\int_{-\infty}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}y\,\frac{2^{4}xy\,P{\left(a,b,x,y\right)}}{Q{\left(a+x,1-y\right)}\,Q{\left(a+x,1+y\right)}\,Q{\left(x,b-y\right)}\,Q{\left(x,b+y\right)}}.\tag{1c}$$

It's not hard to show then that $\mathcal{I}{\left(a,b\right)}$ is even in the second parameter $b$:

$$\mathcal{I}{\left(a,-b\right)}=\mathcal{I}{\left(a,b\right)};~~~\small{\left(a,b\right)\in\mathbb{R}^{2}}.$$

Problem: Given the pair of real parameters $\left(a,b\right)\in\mathbb{R}\times\mathbb{R}_{\ge0}$, find a closed form expression for the double integral $\mathcal{I}{\left(a,b\right)}$ in terms of elementary functions.

The obstacle in the way of solving this problem appears to be tedium more than anything else. Integrating the rational integrand in $(1c)$ over $x$ should in principle yield a piecewise rational function. Thus, subsequent integration over $y$ should lead to a function that is at the very least piecewise elementary, if not simpler.

However, attempting to solve the problem by brute force with partial fraction decompositions quickly leads to large numbers of cumbersome expressions, rending the integral quite unmanageable without a program such as Mathematica.

It is my hope that there is some cleverly efficient approach to this integral that I'm just not seeing at the moment. Any advice here would be welcome. Cheers!

• Fairly certain this integral is zero for all values of $a,b$. The nominator is odd in both $x$ and $y$ and you are integrating over a symmetrical domain. In addition the denominator is a product of even functions, hence it is even, and does not change the fact that the integral is zero. – N3buchadnezzar Apr 23 '17 at 12:29
• @N3buchadnezzar Thank you for your comment, and for teaching me a new word (I've never seen the word 'nominator' before and had to look it up ). =) Unfortunately, the numerator is not odd in $x$ like you say. And while the denominator is even in $y$, it isn't an even function of $x$. But it sure would be nice though if the integral was automatically zero, wouldn't it? ;) – David H Apr 23 '17 at 14:06
• I was too quick to say that it was symmetric in $x$, however that does not really matter as it is zero when integrating over $y$ =) – N3buchadnezzar Apr 23 '17 at 14:23
• @N3buchadnezzar The integration over $y$ would be zero if we were integrating over $-\infty<y<\infty$. But we are only integrating over $0\le y$. – David H Apr 23 '17 at 14:31
• @David H: This is nice stuff ;-). This is this kind of problem that is in principle solvable but in practice not unless you have infinite amount of time and patience on your disposal. My idea would be to fix $y$ and use Cauchy residue theorem to do the integration over $x$. The integrand diminishes fast at complex infinity so the the task boils down to extracting the poles which lie in the upper plane and computing the residues. Unfortunately there will be many cases depending on the relations between $y$, $a$ and $b$. – Przemo Jan 24 '18 at 18:54

This is going to be a partial answer to this question. We will integrate the integrand over $y$ first. In this variable the integrand is clearly a rational function and as such it can be decomposed into partial fractions. Having done this we find the anti-derivative of all relevant terms using the following identity: $$\int\frac{A+B y}{a+b y+y^2} dy = \frac{2(A-b B)}{\sqrt{4 a-b^2}} \arctan(\frac{b+2 y}{\sqrt{4 a-b^2}}) + \frac{B}{2} \log(a+b y+y^2)$$ All we have to do now is to evaluate the resulting expression at infinity and at zero. Using Mathematica and substituting particular numbers for $a$ and $b$ we have checked that the value at infinity vanishes and as such the integral over $y$ boils down to the negative value of our expression at zero. We evaluated this value in Mathematica and then tediously simplified it by hand. It appears that the resulting expression is very simple and neat. It reads: \begin{eqnarray} &&\int\limits_0^\infty \frac{2^{4}xy\,P{\left(a,b,x,y\right)}}{Q{\left(a+x,1-y\right)}\,Q{\left(a+x,1+y\right)}\,Q{\left(x,b-y\right)}\,Q{\left(x,b+y\right)}} dy=\\ &&\frac{2 \left(a^2-b+1\right) \left(\frac{a}{2}+x\right)+a b (b-1)}{b \left(\left(\frac{a}{2}+x\right)^2+\frac{1}{4} (1-b)^2\right)}\cdot\left(\arctan(\frac{b}{x})+\arctan(\frac{1}{x+a})\right)+\\ &&\frac{2 \left(a^2+b+1\right) \left(\frac{a}{2}+x\right)+a b (b+1)}{b \left(\left(\frac{a}{2}+x\right)^2+\frac{1}{4} (b+1)^2\right)}\cdot \left(\arctan(\frac{b}{x})-\arctan(\frac{1}{x+a})\right)+\\ &&\left( \frac{(1-b) \left(a^2-b+1\right)+2 a b \left(\frac{a}{2}+x\right)}{2 b \left(\left(\frac{a}{2}+x\right)^2+\frac{1}{4} (1-b)^2\right)}-\frac{(b+1) \left(a^2+b+1\right)-2 a b \left(\frac{a}{2}+x\right)}{2 b \left(\left(\frac{a}{2}+x\right)^2+\frac{1}{4} (b+1)^2\right)}\right)\cdot(\log(b^2+x^2) - \log(1+(a+x)^2)) \end{eqnarray} valid for $x,a,b \in{\mathbb R}$. We have checked this equality by randomly sampling values of $a$,$b$ and $x$ and varifying that the right hand side equals the left hand side to high numerical precision.