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For each positive integer $n$, consider the following intgral: $$\int_0^\infty\int_0^\infty\frac{(x^4-y^2)x^{2n+1}}{(x^4+y^2)[(1+x^2)^2+y^2]^{n+1}}dxdy.$$ I want to know if there is any easy way to see that it's positive. If there is no easy proof to show that it's positive, I am satisfied if I know that it is positive. I don't need to know the exact value of the integral. I try to write the integral as $$\int_0^\infty\int_0^\infty\frac{x^{2n+1}}{[(1+x^2)^2+y^2]^{n+1}}dxdy- \int_0^\infty\int_0^\infty\frac{2y^2x^{2n+1}}{(x^4+y^2)[(1+x^2)^2+y^2]^{n+1}}dxdy,$$ but it seems to me that it doesn't help much.

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It is somewhat visually confusing to have both $r$ and $\tau$ as variables... –  Qiaochu Yuan Jun 22 '12 at 8:20
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If you change the variable $r$ to $u$ via $u=r^2$, you have the more symmetric $\int_0^\infty\int_0^\infty\frac{(u^2-\tau^2)u^{n}}{2(u^2+\tau^2)[(1+u)^2+\tau^2‌​]^{n+1}}dud\tau.$ –  alex.jordan Jun 22 '12 at 8:51
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Numerical calculation suggests that it is positive with sharp decay for $n \geq 1$. I am still working to devise a proof, though unsuccessful so far. –  sos440 Jun 22 '12 at 9:27
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For $n=1$ the integral is $-\frac{1}{4}\pi\ln(2)+\frac{3}{16}\pi=0.044652\ldots$ (by computer). –  vesszabo Jun 22 '12 at 10:29
    
Actually, exact computation of the integral up to $n = 20$ suggests that it is of the form $$r_n \pi - \frac{n+1}{2^{n+2}} \pi \log 2,$$ where $r_n \in \mathbb{Q}^{+}$ decreases exponentially. But still I have no idea how to derive positivity. –  sos440 Jun 22 '12 at 11:41
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up vote 3 down vote accepted

Start with the substitution suggested by @alex.jordan: $u=\sqrt{x}, v=y$. Then think of $(u,v)$ as Cartesian coordinates with the integral over the first quadrant. Change to polar coordinates $u=r\cos\theta, v=r\sin\theta$ to obtain $$ \hbox{your integral } = I_n = \frac12\int_0^\infty r^{n+1} dr \int_0^{\pi/2} \frac{\cos^n\theta(\cos^2\theta-\sin^2\theta)}{(r^2+2r\cos\theta+1)^{n+1}}d\theta.$$ Split $$J_n=\int_0^{\pi/2} \frac{\cos^n\theta(\cos^2\theta-\sin^2\theta)}{(r^2+2r\cos\theta+1)^{n+1}}d\theta$$ at $\theta=\pi/4$ to obtain integrands that are obviously positive and negative. In the subinterval $\theta\in[\pi/4,\pi/2]$ change the variable to $\theta^\prime=\pi/2-\theta$ - the domain for the second integral is now $\theta^\prime\in[0,\pi/4]$ - and then change the name of $\theta^\prime$ to $\theta$. Recombine the two integrals to obtain $$ J_n=\int_0^{\pi/4} (h_n(r,\cos\theta)-h_n(r,\sin\theta))(\cos^2\theta-\sin^2\theta)d\theta$$ where $$h_n(r,z)=\frac{z^n}{(r^2+2rz+1)^{n+1}}.$$ For $n\geq 2$, this is an increasing function of $z$ on $[0,1]$ which is sufficient to show that the integrand in the second representation of $J_n$ is positive. I think this deals with everything but the case $n=1$.

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Very smart idea!I think this idea deals with everything including the case n=1. –  y zhao Jun 22 '12 at 13:13
    
@user12477: Your idea is brilliant! Thank you very much! –  Paul Jun 22 '12 at 14:05
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It's an idea only, but too long in the comment (I run out of the space.)(Sorry.) Following the idea of @alex.jordan, introduce the new variables $X:=\sqrt{x}, Y:=\sqrt{y}$ by steps. Then we obtain $$ \int_0^{\infty}\int_0^{\infty}\frac{(X^2-Y)X^n}{4(X^2+Y)((1+X)^2+Y)^{n+1}\sqrt{Y}}\,dXdY $$ Now we use the transformation $X^2+Y=u,(1+X)^2+Y=v$. Then the abs'value of Jacobi-det is $\frac{1}{2}$ and we get $$ \int_0^{\infty}\int_1^{\infty} \frac{\left(\frac{1}{2}-v+\frac{1}{2}u^2-uv+\frac{1}{2}v^2\right)\left(-\frac{1}{2}-\frac{1}{2}u+\frac{1}{2}v\right)^n}{8uv^{n+1}\sqrt{-\frac{1}{4}+\frac{1}{2}u+\frac{1}{2}v-\frac{1}{4}u^2+\frac{1}{2}uv-\frac{1}{4}v^2}}\,dvdu. $$ The advantage of the last integral, that the terms containing $n$ in the exponent are simple. But I have to think further ...

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