Examples of Fubini's theorem

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Contents 1 Attention 2 Introduction 3 How to prove that :<math>\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy</math> does not converge 4 How to evaluate this integral 5 The moral of the story

Attention

Introduction

The iterated integral

<math>\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx</math>

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

<math>\int_0^1\int_0^1

\left|\frac{x^2-y^2}{(x^2+y^2)^2}\right|\,dy\,dx=\infty.</math>

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.

How to prove that :<math>\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy</math> does not converge

<math>\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</math>

<math>=\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</math>

How to evaluate this integral

The integral

<math>\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy</math>

can be evaluated via the trigonometric substitution

<math>y=x\tan(\theta),</math>

<math>dy=x\sec^2(\theta)\,d\theta,</math>

<math>x^2+y^2=x^2+x^2\tan^2(\theta)=x^2(1+\tan^2(\theta))

=x^2\sec^2(\theta).</math>

The bounds of integration can be found thus:

<math>0\leq y\leq 1,</math>

<math>0\leq x\tan(\theta)\leq 1,</math>

<math>0\leq\tan(\theta)\leq 1/x,</math>

<math>0\leq\theta\leq\arctan(1/x).</math>

The integral then becomes

<math>\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</math>

<math>=\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)}</math>

<math>=\frac{1}{x}\left[\sin(\theta)\cos(\theta)

\right]_{\theta:=0}^{\theta=\arctan(1/x)} =\frac{1}{x}\sin(\arctan(1/x))\cos(\arctan(1/x)).</math>

Now recall the trigonometric identities

<math>\sin(\arctan(1/x))=\frac{1}{\sqrt{1+x^2}}

\ \mbox{and}\ \cos(\arctan(1/x))=\frac{x}{\sqrt{1+x^2}}.</math>

The expression above then becomes

<math>\frac{1}{1+x^2}.</math>

This takes care of the "inside" integral with respect to y; now we do the "outside" integral with respect to x:

<math>\int_0^1\frac{1}{1+x^2}\,dx

=\left[\arctan(x)\right]_0^1

=\arctan(1)-\arctan(0)=\frac{\pi}{4}.</math>

Thus we have

<math>\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx=\frac{\pi}{4}</math>

and

<math>\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy=\frac{-\pi}{4}.</math>

The moral of the story

When

<math>\int_a^b\int_c^d \left|f(x,y)\right|\,dy\,dx=\infty</math>

then the two iterated integrals

<math>\int_a^b\int_c^d f(x,y)\,dy\,dx\ \mbox{and}\ \int_c^d\int_a^b f(x,y)\,dx\,dy</math>

may have different finite values.