Fubini's theorem

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In mathematical analysis, Fubini's theorem, named in honor of Guido Fubini, states that if

$\int_{A\times B} |f(x,y)|\,d(x,y)<\infty,$

the integral being taken with respect to a product measure on the space over $A\times B$ , where A and B are σ-finite measure spaces, then

$\int_A\left(\int_B f(x,y)\,dy\right)\,dx=\int_B\left(\int_A f(x,y)\,dx\right)\,dy=\int_{A\times B} f(x,y)\,d(x,y),$

the first two integrals being iterated integrals, and the third being an integral with respect to a product measure. Also,

$\int_A f(x)\, dx \int_B g(y)\, dy = \int_{A\times B} f(x)g(y)\,d(x,y)$

the third integral being with respect to a product measure.

If the above integral of the absolute value is not finite, then the two iterated integrals may actually have different values. See below for an illustration of this possibility.

1 Tonelli's theorem
2 Applications
- 2.1 Rearranging a conditionally convergent iterated integral
3 See also
4 References

[edit] Tonelli's theorem

Tonelli's theorem (named after Leonida Tonelli) is a predecessor of Fubini's theorem. The conclusion of Tonelli's theorem is identical to that of Fubini's theorem, but the assumptions are different. Tonelli's theorem states that a product measure integral can be evaluated by way of an iterated integral for nonnegative measurable functions, regardless of whether they have finite integral.

In fact, the existence of the first integral above (the integral of the absolute value), is guaranteed by Tonelli's theorem (see below).

[edit] Applications

One of the most beautiful applications of Fubini's theorem is the evaluation of the Gaussian integral which is the basis for much of probability theory:

$\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}.$

To see how Fubini's theorem is used to prove this, see Gaussian integral.

Another nice use of Tonelli's theorem is to apply it to $| f (x, y) |$ for a complex valued function $f$ .

It is useful to note that if

$\varphi(x)=\int |f(x,y)|\,dy$ and $\int \varphi(x) dx <\infty,$

then

$\int |f(x,y)| \, d(x,y) < \infty.$

This is often a useful way to check the conditions of Fubini's theorem.

[edit] Rearranging a conditionally convergent iterated integral

The iterated integral

$\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx$

does not converge absolutely (i.e. the integral of the absolute value is not finite):

$\int_0^1\int_0^1 \left|\frac{x^2-y^2}{(x^2+y^2)^2}\right|\,dy\,dx=\infty.$

Fubini's theorem tells us that if the integral of the absolute value is finite, then the order of integration does not matter; if we integrate first with respect to x and then with respect to y, we get the same result as if we integrate first with respect to y and then with respect to x. The assumption that the integral of the absolute value is finite is "Lebesgue integrability". That the assumption of Lebesgue integrability in Fubini's theorem cannot be dropped can be seen by examining this particular iterated integral. Clearly putting "dx dy" in place of "dy dx" has the effect of multiplying the value of the integral by −1 because of the "antisymmetry" of the function being integrated. Therefore, unless the value of the integral is zero, putting "dx dy" in place of "dy dx" actually changes the value of the integral. That is indeed what happens in this case.

[edit] Proof

One way to do this without using Fubini's theorem is as follows:

$\int_0^1\int_0^1 \left|\frac{x^2-y^2}{(x^2+y^2)^2}\right|\,dx\,dy=\int_0^1\left[\int_0^y \frac{y^2-x^2}{(x^2+y^2)^2}\,dx+\int_y^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\right]\,dy$

$=\int_0^1\left(\frac{1}{2y}+\frac{1}{2y}-\frac{1}{y^2+1}\right)\,dy=\int_0^1 \frac{1}{y}\,dy-\int_0^1\frac{1}{1+y^2}\,dy.$

[edit] Evaluation

The integral

$\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy$

can be evaluated via the trigonometric substitution

y = x tan(θ),

$dy=x\sec^2(\theta)\,d\theta,$

x 2 + y 2 = x 2 + x 2 tan 2 (θ) = x 2 (1 + tan 2 (θ)) = x 2 sec 2 (θ).

The bounds of integration can be found thus:

$0\leq y\leq 1,$

$0\leq x\tan(\theta)\leq 1,$

$0\leq\tan(\theta)\leq 1/x,$

$0\leq\theta\leq\arctan(1/x).$

The integral then becomes

$\int_0^{\arctan(1/x)} \frac{x^2(1-\tan^2(\theta))}{(x^2\sec^2(\theta))^2} x\sec^2(\theta)\,d\theta =\frac{1}{x}\int_0^{\arctan(1/x)} \frac{1-\tan^2(\theta)}{\sec^2(\theta)}\,d\theta$

$=\frac{1}{x}\int_0^{\arctan(1/x)} \cos^2(\theta)-\sin^2(\theta)\,d\theta =\frac{1}{x}\int_0^{\arctan(1/x)} \cos(2\theta)\,d\theta =\frac{1}{x}\left[\frac{\sin(2\theta)}{2} \right]_{\theta:=0}^{\theta=\arctan(1/x)}$