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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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    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.2011-03-10

2 Answers 2

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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 =)2011-03-14
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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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    @Viktor: No worries =)2011-03-14