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Show $$\int_0^\infty \int_0^\infty \frac{(ax-by) {\rm e}^{-x} {\rm e}^{-y}}{(a^2 x + b^2 y + c x y)^{\frac{3}{2}}} \,dx \,dy = 0$$ for any $a,b,c > 0$.

I came upon the above double integral when simplifying an expression for a probability density function. How can I demonstrate that the integral is zero for any positive constants, $a$, $b$ and $c$?

I've verified the result numerically. At the moment it seems rather fascinating to me that the integral should always be zero given the presence of three free variables, and I would expect there to be a relatively simple derivation.

I've tried all sorts of approaches (integral substitutions including polar coordinates, differentiating under the integral, splitting the domain of integration into pieces) without seeming to make progress.

To answer Ali's comment, below I give my crude Matlab code for evaluating the double integral (with the trapezoidal rule) and one output. Regardless of what positive values of $a$,$b$,$c$ I put in, I get something close enough to zero for my liking.

Output of program:

double_int = getDoubleInt(2,3,4)

double_int =

Program code:

function double_int = getDoubleInt(a,b,c)

x_max = 25;
y_max = 25;
NN_x = 1000;
NN_y = 1000;

x_vec = logspace(log10(x_max/10^10),log10(x_max),NN_x);

y_vec = logspace(log10(y_max/10^10),log10(y_max),NN_y);

XX = x_vec'*ones(1,NN_y);

YY = ones(NN_x,1)*y_vec;

ZZ = (a*XX-b*YY)./(a^2*XX+b^2*YY+c*XX.*YY).^(3/2).*exp(-XX).*exp(-YY);

int_1 = zeros(size(x_vec));

for i = 1:NN_x

int_1(i) = trapz(y_vec,ZZ(i,:));


double_int = trapz(x_vec,int_1);
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Since MathJax is down: – user2468 Mar 13 '12 at 15:11
I'm still incredulous this could be true and haven't yet heard from other numerical tests, but at first glance I think you could interpret it as a function of $a,b,c$ and show it's constant... – anon Mar 13 '12 at 16:09
@djws: can yu show us a numerical test of this equality? – user17090 Mar 13 '12 at 16:40
For $a=b$, I believe the result. For $a\neq b$ this looks like a miracle (and I don't believe in miracles). – Fabian Mar 13 '12 at 20:41
@Fabian: I totally agree. Now, I'm doing a few numerical experiments with Mathematica, and so far I get zero for $a=.1$, $b=1$, and values of $c$ ranging from $.01$ to $5$. Every value I've tried gives equality. The caveat is that for some values Mathematica complains that the convergence of the numerical integration is too slow. – Martin Argerami Mar 13 '12 at 20:49
up vote 13 down vote accepted

The result is surprising because one would think that the combination with the exponentials would be too complicated to make the integral vanish for all $a,b,c$. However, the exponentials don't enter into it because the integral along every line with constant sum $x+y$ is zero, and the exponential factor is constant on these lines. Consider

$$ x=u+v\;, \\ y=u-v\;, $$

which, up to a constant factor from the Jacobian, transforms the integral into

$$\int_0^\infty\mathrm du\mathrm e^{-2u}\int_{-u}^u\mathrm dv\frac{a(u+v)-b(u-v)}{\left(a^2(u+v)+b^2(u-v)+c(u^2-v^2)\right)^{3/2}}\;.$$

The inner integral is readily performed; Wolfram|Alpha gives

$$\int_0^\infty\mathrm du\mathrm e^{-2u}\frac2{(a-b)^2+2cu}\left[\frac{a(u+v)+b(u-v)}{\sqrt{a^2(u+v)+b^2(u-v)+c(u^2-v^2)}}\right]_{v=-u}^{v=u}\;,$$

and the result is $0$ since the antiderivative takes the same value at $u$ and $-u$.

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very nice answer, thanks! – djws Mar 15 '12 at 21:17
@djws: You're welcome! – joriki Mar 15 '12 at 21:21
@joriki: +1 (+ bounty) thanks for helping us out ;-) – Fabian Mar 16 '12 at 8:25
@Fabian: My pleasure :-) – joriki Mar 16 '12 at 8:29

Using the substitution $$ r\frac{1+t}{1-t}=\frac1r\frac{1+s}{1-s} $$ maps $[-1,1]\mapsto[-1,1]$ and has proven useful in several situations similar to this. Some formulas pertinent to this substitution are $$ \frac{\mathrm{d}t}{1-t^2}=\frac{\mathrm{d}s}{1-s^2} $$ $$ \frac{r(1+t)+\frac1r(1-t)}{1-t^2}=\frac{\frac1r(1+s)+r(1-s)}{1-s^2} $$ $$ \frac{r(1+t)-(1-t)}{\sqrt{1-t^2}}=\frac{(1+s)-r(1-s)}{\sqrt{1-s^2}} $$ Change variables $u=x+y$ and $v=x-y$, then $v=ut$: $$ \begin{align} &\int_0^\infty\int_0^\infty\frac{(ax-by)\,\mathrm{e}^{-x}\,\mathrm{e}^{-y}}{(a^2 x + b^2 y + c x y)^{\frac{3}{2}}}\,\mathrm{d}x\,\mathrm{d}y\\[6pt] &=2\int_0^\infty\int_{-u}^u\frac{(a-b)u+(a+b)v}{\left(cu^2-cv^2+2(a^2+b^2)u+2(a^2-b^2)v\right)^{3/2}}\,\mathrm{e}^{-u}\,\mathrm{d}v\,\mathrm{d}u\\[6pt] &=2\int_0^\infty\int_{-1}^1\frac{(a-b)+(a+b)t}{\left(cu^2(1-t^2)+2u(a^2(1+t)+b^2(1-t))\right)^{3/2}}\,\mathrm{d}t\,\mathrm{e}^{-u}u^2\,\mathrm{d}u\tag{1} \end{align} $$ Using the substitution $\frac{a}{b}\frac{1+t}{1-t}=\frac{b}{a}\frac{1+s}{1-s}$ with $(1)$ yields $$ 2\int_0^\infty\int_{-1}^1\frac{(b-a)+(a+b)s}{\left(cu^2(1-s^2)+2u(b^2(1+s)+a^2(1-s))\right)^{3/2}}\,\mathrm{d}s\,\mathrm{e}^{-u}u^2\,\mathrm{d}u $$ Then substituting $w=-s$ yields $$ -2\int_0^\infty\int_{-1}^1\frac{(a-b)+(a+b)w}{\left(cu^2(1-s^2)+2u(a^2(1+w)+b^2(1-w))\right)^{3/2}}\,\mathrm{d}w\,\mathrm{e}^{-u}u^2\,\mathrm{d}u\tag{2} $$ Since $(1)$ and $(2)$ are equal, yet negatives, they are both $0$.

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Oh... my... God... – Norbert Mar 16 '12 at 22:01
I think you got the factors of $2$ wrong. That's why I didn't write them; I knew I'd get them wrong if I did :-) You have $x=(u+v)/2$ and $y=(u-v)/2$, so you should have factors of $1/2$, one from the numerator and one from the Jacobian if I'm not mistaken. – joriki Mar 17 '12 at 5:13
@joriki: I had originally included more steps, the second line was $$ \int_0^\infty\int_{-u}^u\frac{(a\frac{u+v}{2}-b\frac{u-v}{2})\,\mathrm{e}^{-u}}{‌​(a^2\frac{u+v}{2} + b^2\frac{u-v}{2} + c\frac{u^2-v^2}{4})^{3/2}}\tfrac12\mathrm{d}v\,\mathrm{d}u $$ I believe I have it correct. – robjohn Mar 17 '12 at 7:09
Ah, of course, sorry. See, I knew I'd get them wrong ;-) – joriki Mar 17 '12 at 9:09

Take $a=c=1$ and $b$ very small.

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Interesting that for $b>0$ the integral is $0$, but not for $b=0$. – GEdgar Mar 15 '12 at 22:10

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