[CALCULUS] Fubini and iterated integrals

This page turns a double integral over a rectangle into two one-variable integrals. The Riemann-sum definition says what the integral means but is not a practical way to evaluate it. Instead, fix one coordinate, integrate across the resulting slice and then add the slices with a second integral.

When slice endpoints depend on the outer variable, regions bounded by curves extend the method. A change of variables reshapes slices that remain awkward.

[DEFINITION]

Double integral over a rectangle

Let $f$ be bounded on $R=[a,b]\times[c,d]$ . If there is a number $I$ such that, for every $\varepsilon>0$ , some $\delta>0$ makes every tagged partition with $\lVert P\rVert<\delta$ satisfy

\left| \sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*,y_{ij}^*)\,\Delta A_{ij}-I \right|<\varepsilon,

then $f$ is Riemann integrable over $R$ and

\iint_R f(x,y)\,dA = \lim \sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*,y_{ij}^*)\,\Delta A_{ij}.

Add the slices

Let

R=[a,b]\times[c,d].

Fixing $x=x_0$ leaves a slice in the $y$ -direction. Its signed cross-sectional area is the one-variable integral

C_{\mathrm{vert}}(x_0)=\int_c^d f(x_0,y)\,dy,

provided this integral exists. Adding these areas as $x$ varies gives the double integral. Fixing $y=y_0$ instead gives slices in the other direction.

[DEFINITION]

Cross-section integral

For a function $f$ on $R=[a,b]\times[c,d]$ , when the one-variable integrals exist, the vertical and horizontal cross-section integrals are

C_{\mathrm{vert}}(x)=\int_c^d f(x,y)\,dy, \qquad C_{\mathrm{horiz}}(y)=\int_a^b f(x,y)\,dx.

They measure the signed areas of slices obtained by fixing $x$ or $y$ .

When $f\ge0$ , these are ordinary cross-sectional areas. When $f$ takes both signs, negative parts are subtracted.

[1]:Fubini's theorem on a rectangle

[THEOREM]

Fubini's theorem on a rectangle

If $f$ is continuous on

R=[a,b]\times[c,d],

then

\begin{aligned} \iint_R f(x,y)\,dA &=\int_a^b\left(\int_c^d f(x,y)\,dy\right)dx\\ &=\int_c^d\left(\int_a^b f(x,y)\,dx\right)dy. \end{aligned}

The inner integral computes one cross-section and the outer integral adds the cross-sections. In

\int_a^b\int_c^d f(x,y)\,dy\,dx,

the $dy$ says that $y$ is integrated first and the $dx$ says that $x$ is integrated second. Limits must belong to the variable whose differential they accompany.

The iterated formula remains valid for a bounded piecewise-continuous $f$ whose discontinuities lie on finitely many piecewise-smooth curves. This includes zero extensions whose only discontinuities lie on the region boundary.

Unbounded or improper integrals need stronger safeguards. If $f$ is measurable and $\iint_R|f|\,dA<\infty$ , Fubini's theorem gives the two iterated integrals and permits their order to be exchanged. If $f\ge0$ is measurable, Tonelli's theorem gives the same equality with values in $[0,\infty]$ , even when the integral is infinite. Similarly, if nonnegative functions increase pointwise to a limit, monotone convergence justifies passing the limit through the integral, as in the Basel sum.

An improper calculation should therefore first use bounded truncations (or a nonnegative convergence theorem). It must not subtract two separately divergent integrals.

[METHOD]

Choose an order of integration

To compute a double integral over a rectangle:

inspect the integrand for a direct antiderivative in $x$ or $y$ ;
write the corresponding inner integral first;
evaluate the inner integral while treating the other variable as constant;
evaluate the remaining one-variable integral;
use the other order as a check when it is comparably simple.

Under Fubini's hypotheses the two orders give the same value, but they need not be equally convenient.

[EXAMPLE]

Computing in both orders

Let

R=[0,1]\times[0,2], \qquad f(x,y)=xy^2.

Integrating with respect to $x$ first gives

\begin{aligned} \iint_R xy^2\,dA &=\int_0^2\int_0^1xy^2\,dx\,dy\\ &=\int_0^2\left[\frac{x^2y^2}{2}\right]_{0}^{1}dy\\ &=\int_0^2\frac{y^2}{2}\,dy =\frac43. \end{aligned}

Integrating with respect to $y$ first gives

\begin{aligned} \iint_R xy^2\,dA &=\int_0^1\int_0^2xy^2\,dy\,dx\\ &=\int_0^1\left[\frac{xy^3}{3}\right]_{0}^{2}dx\\ &=\int_0^1\frac{8x}{3}\,dx =\frac43. \end{aligned}

[routine]

[1]:Choose an order of integration

[EXAMPLE]

Choosing the useful order

Consider

\iint_R xe^{xy}\,dA, \qquad R=[2,4]\times[1,3].

Integrating in $y$ first uses the factor $x$ already present:

\begin{aligned} \iint_R xe^{xy}\,dA &=\int_2^4\int_1^3xe^{xy}\,dy\,dx\\ &=\int_2^4\left[e^{xy}\right]_{1}^{3}dx\\ &=\int_2^4(e^{3x}-e^x)\,dx\\ &=\left[\frac{e^{3x}}3-e^x\right]_2^4\\ &=\frac{e^{12}}3-e^4-\frac{e^6}{3}+e^2. \end{aligned}

The reverse order leaves $xe^{xy}$ as the inner integrand in $x$ , which needs integration by parts.

[exam]

[1]:Choose an order of integration

Products separate over rectangles

When the integrand is a function of $x$ times a function of $y$ , the double integral over a rectangle is the product of two single integrals.

[PROPOSITION]

Product separation

If $g$ is integrable on $[a,b]$ and $h$ is integrable on $[c,d]$ , then $f(x,y)=g(x)h(y)$ is integrable on $R=[a,b]\times[c,d]$ and

\iint_R g(x)h(y)\,dA = \left(\int_a^b g(x)\,dx\right) \left(\int_c^d h(y)\,dy\right).

The factor that does not involve the inner variable pulls out of the inner integral.

[1]:Fubini's theorem on a rectangle
[2]:A separated product

[EXAMPLE]

A separated product

For $R=[-1,1]\times[0,\pi/4]$ ,

\begin{aligned} \iint_R x^2\sin y\,dA &=\left(\int_{-1}^{1}x^2\,dx\right) \left(\int_0^{\pi/4}\sin y\,dy\right)\\ &=\frac23\left(1-\frac1{\sqrt2}\right). \end{aligned}

[routine]

[1]:Product separation

Why the hypotheses matter

Fubini's theorem is about a genuine double integral. It does not say that two improper iterated integrals must agree.

[EXAMPLE]

Different improper iterated integrals

Define on $[0,1]^2$

f(x,y)= \begin{cases} \dfrac1{x^2},&0<y\le x\le1,\\[6pt] -\dfrac1{y^2},&0<x<y\le1,\\[6pt] 0,&x=0\text{ or }y=0. \end{cases}

The values on the axes remove the undefined endpoint without changing any integral. For fixed $x>0$ , split the improper $y$ -integral at $y=x$ :

\begin{aligned} \int_0^1 f(x,y)\,dy &=\int_0^x\frac1{x^2}\,dy \;+\int_x^1-\frac1{y^2}\,dy\\ &=\frac1x+\left(1-\frac1x\right) =1. \end{aligned}

Thus one iterated integral gives

\int_0^1\int_0^1f(x,y)\,dy\,dx=1.

For fixed $y>0$ , split the improper $x$ -integral at $x=y$ :

\begin{aligned} \int_0^1f(x,y)\,dx &=\int_0^y-\frac1{y^2}\,dx \;+\int_y^1\frac1{x^2}\,dx\\ &=-\frac1y+\left(\frac1y-1\right) =-1. \end{aligned}

The other order gives

\int_0^1\int_0^1f(x,y)\,dx\,dy=-1.

There is no contradiction. The positive and negative parts each have infinite integral near the origin, so $f$ is not integrable over the square in the sense Fubini requires. The two calculations are different improper cancellations, not two values of one double integral.

[challenge]

[1]:Fubini's theorem on a rectangle

Fubini and iterated integrals

Double integral over a rectangle

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