# Fubini's theorem problem

Let $f$ be a non-negative measurable function in $\mathbb{R}$. Suppose that $$\iint_{\mathbb{R}^2}{f(4x)f(x-3y)\,dxdy}=2\,.$$ Calculate $\int_{-\infty}^{\infty}{f(x)\,dx}$.

My first thought was to change the order of integration so that I integrate in $y$ first, since there's only a single $y$ in the integrand... but I'm not sure how/if that even helps me.

Then the more I thought about it, the less clear it was to me that Fubini's theorem even applies as it's written. Somehow I need a function of two variables. So should I set $g(x,y) = f(4x)f(x-3y)$ and do something with that? At least Fubini's theorem applies for $g(x,y)$, since we know it's integrable on $\mathbb{R}^2$. .... Maybe?

I'm just pretty lost on this, so any help you could offer would be great. Thanks!

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$g(x, y)$ is integrable on $\mathbb{R}^2$: $f$ is non-negative, so $|g|=g$. You know the integral of $g$ over $\mathbb{R}^2$ and it is finite. –  Vitaly Lorman Mar 10 '11 at 17:25
if both functions are integrable you can use Tonelli's theorem which is a simpler version of Fubini's theorem. I think for Tonelli to apply you don't have to know that $g(x,y)$ is integrable, you can just exchange the integration order. –  Rudy the Reindeer Mar 10 '11 at 17:45

Your thought is a good one. If you integrate in $y$ first you can change variable to $u=x-3y$. You are considering $x$ a constant, so you get a decoupled product of integrals of $f$ over the real line.

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Thanks so much. Chose your answer because you left more of it up to me =) –  Bey Mar 14 '11 at 20:39

I think both Vitali and Matt are right. As soon as $G(x,y)$ is integrable on $\mathbb{R}^2$, $$\iint_{\mathbb{R}^2}{G(x,y)\,dxdy}=\int_{\mathbb{R}}dx\,\int_{\mathbb{R}}{G(x,y)\,dy}=\int_{\mathbb{R}}dy\,\int_{\mathbb{R}}{G(x,y)\,dx}$$

So you can substitute: $$u=4x$$ $$v=x-3y$$ $$x=\frac{u}{4}$$ $$y=\frac{u}{12}-\frac{v}{3}$$ Jacobian $$J=\det D(x(u,v),y(u,v))=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}=\begin{vmatrix}\frac{1}{4} & 0\\\\\frac{1}{12} & -\frac{1}{3}\end{vmatrix}=-\frac{1}{12}$$

Then $$\iint_{\mathbb{R}^2}{f(4x)\,f(x-3y)\,dxdy}=-\frac{1}{12}\int_{-\infty}^{\infty}\,du\,\int_{\infty}^{-\infty}{f(u)\,f(v)\,dv}=\frac{1}{12}(\int_{-\infty}^{\infty}{f(z)\,dz})^2=2$$ Thus giving us $$\int_{-\infty}^{\infty}{f(z)\,dz}=2\sqrt 6$$

Please tell me if I $\mathbb{F}\bigoplus$ something up.

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Where did $a$ and $b$ come from? I think you are out a factor 4 as in the next to last line as $\int_{-\infty}^{\infty}f(4x)dx=\frac{1}{4}\int_{-\infty}^{\infty}f(x)dx$ and the same for 3 in the $y$ direction. Then before taking the square root, you multiply by 48(should be 12) instead of dividing, getting $\sqrt{24}i$. –  Ross Millikan Mar 10 '11 at 21:56
1. Sorry. (a,b)=(u,v). I fixed that already. –  Viktor Mar 10 '11 at 22:01
When you change variables, you should multiply the absolute value of the determinant. –  Byron Schmuland Mar 10 '11 at 22:02
It should be $y=\frac{u}{12}-\frac{v}{3}$ This will fix the factor $4$. –  Ross Millikan Mar 10 '11 at 22:04
Thanks Byron! Thanks Ross! Haven't done that for awhile. –  Viktor Mar 10 '11 at 22:09