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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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    $g(x, y)$ 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.2011-03-10
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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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    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 $\sqrt{24}i$.2011-03-10
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    1. Sorry. (a,b)=(u,v). I fixed that already.2011-03-10
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    When you change variables, you should multiply the *absolute value* of the determinant.2011-03-10
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    It should be $y=\frac{u}{12}-\frac{v}{3}$ This will fix the factor $4$.2011-03-10
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    Thanks Byron! Thanks Ross! Haven't done that for awhile.2011-03-10
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    @Byron: but the negative sign on y means we integrate from $+\infty$ to $-\infty$ and I think switching the direction of integration introduces a negative sign.2011-03-10
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    @Ross You're probably right, I'm not used to "oriented integrals" if that is the right term. In probability, we take the absolute value of the Jacobian to do change of variables. In any case, the result of integrating a positive function ought to be positive.2011-03-10
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    I was taught $\int_1^0dx=-1$, not $1$ supported by the fundamental theorem of calculus.2011-03-10
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    @Ross, @Byron 1. Jacobian does have sign. 2. We indeed swap limits when substituting $v=x-3y$.2011-03-10
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    The fact that it comes out imaginary means you should say there is no solution. You can't have an integral of a non-negative function (needed for Tonelli and specified) be imaginary, as it is presumably real.2011-03-11
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    @Ross I doesn't come out imaginary. Jacobian sign and limit swap cancel each other. I updated the solution some time ago.2011-03-11
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    @Viktor: Wonderful solution, thank you. I chose Ross's answer only because he left more of the details for me to figure out, but I appreciate your response =)2011-03-14
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    @Bey. Sorry for revealing the whole thing. I am quite new to the site, so I didn't figure out the spirit of it from the very beginning. Now I am getting it. :)2011-03-14
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    @Viktor: No worries =)2011-03-14