Calculus – Multiple integrals – Double integrals over rectangles

Fubini's theorem

Theorem: If \(f\) is continuous on the rectangle

\[R=[a,b]\times[c,d]=\{(x,y)\,|\,a\leq x\leq b,\;c\leq y\leq d\},\]

then we have:

\[\iint\limits_Rf(x,y)\,dA=\int_a^b\int_c^df(x,y)\,dy\,dx=\int_c^2\int_a^bf(x,y)\,dx\,dy.\]

More generally, this is true if we assume that \(f\) is bounded on \(R\), \(f\) is discontinuous only on a finite number of smooth curves, and the iterated integrals exist.

Stewart §15.1, Example 5
Evaluate \(\displaystyle\iint\limits_R(x-3y^2)\,dA\), where \(R=[0,2]\times[1,2]\).

Solution: We have:

\[\iint\limits_R(x-3y^2)\,dA=\int_0^2\int_1^2(x-3y^2)\,dy\,dx=\int_0^2\bigg[xy-y^3\bigg]_{y=1}^2\,dx=\int_0^2(x-7)\,dx =\bigg[\frac{1}{2}x^2-7x\bigg]_{x=0}^2=-12\]

and

\[\iint\limits_R(x-3y^2)\,dA=\int_1^2\int_0^2(x-3y^2)\,dx\,dy=\int_1^2\bigg[\frac{1}{2}x^2-3xy^2\bigg]_{x=0}^2\,dy =\int_1^2\left(2-6y^2\right)\,dy=\bigg[2y-2y^3\bigg]_{y=1}^2=-12.\]

Stewart §15.1, Example 6
Evaluate \(\displaystyle\iint\limits_Ry\sin(xy)\,dA\), where \(R=[1,2]\times[0,\pi]\).

Solution: We have:

\[\iint\limits_Ry\sin(xy)\,dA=\int_0^{\pi}\int_1^2y\sin(xy)\,dx\,dy=\int_0^{\pi}\bigg[-\cos(xy)\bigg]_{x=1}^2\,dy =\int_0^{\pi}\left(\cos(x)-\cos(2x)\right)\,dx=\bigg[\sin(x)-\frac{1}{2}\sin(2x)\bigg]_{x=0}^{\pi}=0.\]

Here the other order is much more difficult. Using integration by parts we have

\[\int y\sin(xy)\,dy=-\frac{1}{x}\int y\,d\cos(xy)=-\frac{y}{x}\cos(xy)+\frac{1}{x}\int\cos(xy)\,dy=-\frac{y}{x}\cos(xy)+\frac{1}{x^2}\sin(xy)+C.\]

Hence we have:

\[\iint\limits_Ry\sin(xy)\,dA=\int_1^2\int_0^{\pi}y\sin(xy)\,dy\,dx=\int_1^2\bigg[-\frac{y}{x}\cos(xy)+\frac{1}{x^2}\sin(xy)\bigg]_{y=0}^{\pi} =\int_1^2\left(-\frac{\pi}{x}\cos(\pi x)+\frac{1}{x^2}\sin(\pi x)\right)\,dx.\]

Note that \(-\displaystyle\frac{1}{x}\sin(\pi x)\) is an antiderivative, since using the chain rule we have: \(\displaystyle\frac{d}{dx}\left(-\frac{1}{x}\sin(\pi x)\right)=-\frac{\pi}{x}\cos(\pi x)+\frac{1}{x^2}\sin(\pi x)\). Hence:

\[\iint\limits_Ry\sin(xy)\,dA=\int_1^2\int_0^{\pi}y\sin(xy)\,dy\,dx=\int_1^2\left(-\frac{\pi}{x}\cos(\pi x)+\frac{1}{x^2}\sin(\pi x)\right)\,dx =\bigg[-\frac{1}{x}\sin(\pi x)\bigg]_{x=1}^2=\sin(\pi)-\frac{1}{2}\sin(2\pi)=0.\]

Average value

The average value of a function \(f\) of one variable defined on an interval \([a,b]\) is \(\displaystyle\frac{1}{b-a}\int_a^bf(x)\,dx\).

Similarly, we have: the average value of a function \(f\) of two variables defined on a rectangle \(R\) is \(\displaystyle\frac{1}{\textrm{area}(R)}\iint\limits_Rf(x,y)\,dA\).

Catalan's constant

Earlier we considered Catalan's constant \(G\) which is defined as the sum of the series \(\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}\).

We obtained several (integral) representations for \(G\), such as: \(\displaystyle\int_0^1\frac{\arctan(x)}{x}\,dx=G\).

Now this can be used to prove a nice representation as a "double integral": \(\displaystyle\int_0^1\int_0^1\frac{dx\,dy}{1+x^2y^2}=G\). The proof is straightforward:

\[\int_0^1\int_0^1\frac{dx\,dy}{1+x^2y^2}=\int_0^1\int_0^1\frac{dy\,dx}{1+x^2y^2}=\int_0^1\bigg[\frac{\arctan(xy)}{x}\bigg]_{y=0}^1\,dx=\int_0^1\frac{\arctan(x)}{x}\,dx=G.\]

Alternatively, we might use \(\displaystyle\frac{1}{1-t}=\sum_{n=0}^{\infty}t^n\) for \(|t|<1\) to find that

\[\int_0^1\int_0^1\frac{dx\,dy}{1+x^2y^2}=\sum_{n=0}^{\infty}(-1)^n\int_0^1\int_0^1(xy)^{2n}\,dx\,dy =\sum_{n=0}^{\infty}(-1)^n\left(\int_0^1t^{2n}\,dt\right)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}=G.\]

Evaluating single integrals using double integrals

Sometimes "difficult" single integrals can be evaluated using "double integrals". A few examples:

1) If we deal with improper integrals as before, we use the fact that \(\displaystyle\frac{d}{dy}\left(\frac{\arctan(xy)}{x}\right)=\frac{1}{1+x^2y^2}\) to obtain that for \(0<\alpha<\beta\):

\begin{align*} \int_0^{\infty}\frac{\arctan(\beta x)-\arctan(\alpha x)}{x}\,dx&=\int_0^{\infty}\int_{\alpha}^{\beta}\frac{dy\,dx}{1+x^2y^2} =\int_{\alpha}^{\beta}\int_0^{\infty}\frac{dx\,dy}{1+x^2y^2}=\int_{\alpha}^{\beta}\bigg[\frac{\arctan(xy)}{y}\bigg]_{x=0}^{\infty}\,dy\\[2.5mm] &=\frac{1}{2}\pi\int_{\alpha}^{\beta}\frac{dy}{y}=\frac{1}{2}\pi\left(\ln(\beta)-\ln(\alpha)\right)=\frac{1}{2}\pi\ln\left(\frac{\beta}{\alpha}\right). \end{align*}

2) Similarly, using the fact that \(\displaystyle\frac{d}{dy}\left(\frac{e^{-xy}}{x}\right)=-e^{-xy}\) we obtain for \(0<\alpha<\beta\):

\[\int_0^{\infty}\frac{e^{-\alpha x}-e^{-\beta x}}{x}\,dx=\int_0^{\infty}\int_{\alpha}^{\beta}e^{-x}\,dy\,dx =\int_{\alpha}^{\beta}\int_0^{\infty}e^{-xy}\,dx\,dy=\int_{\alpha}^{\beta}\bigg[-\frac{e^{-xy}}{y}\bigg]_{x=0}^{\infty}\,dy =\int_{\alpha}^{\beta}\frac{dy}{y}=\ln(\beta)-\ln(\alpha)=\ln\left(\frac{\beta}{\alpha}\right).\]

3) Using integration by parts it can be shown that \(\displaystyle\int_0^{\infty}e^{-ax}\sin(bx)\,dx=\frac{b}{a^2+b^2}\) and \(\displaystyle\int_0^{\infty}e^{-ax}\cos(bx)\,dx=\frac{a}{a^2+b^2}\). For \(0<\alpha<\beta\) this leads to

\[\int_0^{\infty}\frac{(e^{-\alpha x}-e^{-\beta x})\sin(x)}{x}=\int_0^{\infty}\sin(x)\int_{\alpha}^{\beta}e^{-xy}\,dy\,dx =\int_{\alpha}^{\beta}\int_0^{\infty}e^{-xy}\sin(x)\,dx\,dy=\int_{\alpha}^{\beta}\frac{dy}{1+y^2}=\arctan(\beta)-\arctan(\alpha)\]

and

\begin{align*} \int_0^{\infty}\frac{(e^{-\alpha x}-e^{-\beta x})\cos(x)}{x}&=\int_0^{\infty}\cos(x)\int_{\alpha}^{\beta}e^{-xy}\,dy\,dx =\int_{\alpha}^{\beta}\int_0^{\infty}e^{-xy}\cos(x)\,dx\,dy=\int_{\alpha}^{\beta}\frac{y}{1+y^2}\,dy\\[2.5mm] &=\frac{1}{2}\ln(1+\beta^2)-\frac{1}{2}\ln(1+\alpha^2)=\frac{1}{2}\ln\left(\frac{1+\beta^2}{1+\alpha^2}\right). \end{align*}

Furthermore, using \(\displaystyle\int_0^1\cos(xy)\,dy=\frac{\sin(x)}{x}\) and \(\displaystyle\int_0^1\sin(xy)\,dy=\frac{1-\cos(x)}{x}\), we find for \(\alpha>0\) that

\begin{align*} \int_0^{\infty}\frac{e^{-\alpha x}\sin(x)}{x}\,dx&=\int_0^{\infty}e^{-\alpha x}\int_0^1\cos(xy)\,dy\,dx =\int_0^1\int_0^{\infty}e^{-\alpha x}\sin(xy)\,dx\,dy=\int_0^1\frac{\alpha}{\alpha^2+y^2}\,dy\\[2.5mm] &=\arctan\left(\frac{y}{\alpha}\right)\bigg|_0^1=\arctan\left(\frac{1}{\alpha}\right) \end{align*}

and

\begin{align*} \int_0^{\infty}\frac{e^{-\alpha x}(1-\cos(x))}{x}\,dx&=\int_0^{\infty}e^{-\alpha x}\int_0^1\sin(xy)\,dy\,dx =\int_0^1\int_0^{\infty}e^{-\alpha x}\sin(xy)\,dx\,dy=\int_0^1\frac{y}{\alpha^2+y^2}\,dy\\[2.5mm] &=\frac{1}{2}\ln(\alpha^2+y^2)\bigg|_0^1=\frac{1}{2}\ln(\alpha^2+1)-\ln(\alpha). \end{align*}

For \(\alpha=1\) this reads: \(\displaystyle\int_0^{\infty}\frac{e^{-x}\sin(x)}{x}\,dx=\frac{1}{4}\pi\) and \(\displaystyle\int_0^{\infty}\frac{e^{-x}(1-\cos(x))}{x}\,dx=\frac{1}{2}\ln(2)\).

Furthermore, we conclude that \(\displaystyle\int_0^{\infty}\frac{\sin(x)}{x}\,dx=\lim\limits_{\alpha\downarrow0}\arctan\left(\frac{1}{\alpha}\right)=\frac{1}{2}\pi\). This integral is known as the Dirichlet integral.

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Last modified on September 27, 2021