An improper double integral can often be computed similarly to an improper integral of one variable. The first iteration of the following improper integral is conducted just as if it were a proper integral. One then evaluates an improper integral of a single variable by taking appropriate limits. Evaluate the following improper integral as an iterated integral. Integrate infinity ? infinity Integrate infinity ? infinity 1/(x^2+16)(y^2+4) dx dy Integrate infinity ? infinity Integrate infinity ? infinity 1/(x^2+16)(y^2+4) dx dy =

Question

15-01-2020
Mathematics

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An improper double integral can often be computed similarly to an improper integral of one variable. The first iteration of the following improper integral is conducted just as if it were a proper integral. One then evaluates an improper integral of a single variable by taking appropriate limits. Evaluate the following improper integral as an iterated integral. Integrate infinity ? infinity Integrate infinity ? infinity 1/(x^2+16)(y^2+4) dx dy Integrate infinity ? infinity Integrate infinity ? infinity 1/(x^2+16)(y^2+4) dx dy =

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erturkmemmedli erturkmemmedli · Answer 1 · 2020-01-16T09:44:01+01:00

Answer:

[tex]\frac{\pi^2}{8}[/tex]

Step-by-step explanation:

We are given the following improper integral:

[tex]\int\limits^{\infty}_{-\infty}\int\limits^{\infty}_{-\infty}\frac{1}{(x^2+16)(y^2+4)} \:dxdy[/tex]

Above double integral can be written as follows:

[tex]\int\limits^{\infty}_{-\infty}\frac{1}{x^2+16} \:dx\int\limits^{\infty}_{-\infty}\frac{1}{y^2+4}\:dy[/tex]

When we have improper integral with two infinite bounds, we are separating them into two parts.

So let's calculate them one by one.

First,

[tex]\int\limits^{\infty}_{-\infty}\frac{1}{x^2+16} \:dx=\int\limits^{0}_{-\infty}\frac{1}{x^2+16} \:dx+\int\limits^{\infty}_0}\frac{1}{x^2+16} \:dx=\\\\= \lim_{n \to -\infty} \int\limits^{0}_{n}\frac{1}{x^2+16} \:dx + \lim_{m \to \infty}\int\limits^{m}_0}\frac{1}{x^2+16} \:dx[/tex]

[tex]x = 4 \tan \theta\\dx = 4 \sec^2\theta\:d\theta\\\theta=\arctan(\frac{x}{4})[/tex]

[tex]\lim_{n \to -\infty} \int\limits^{0}_{n}\frac{1}{x^2+16} \:dx = \lim_{n \to -\infty} \int\limits^{0}_{n}\frac{4 \sec^2\theta\:d\theta}{16\tan^2\theta+16} = \lim_{n \to -\infty} \frac{1}{4}\int\limits^{0}_{n}\frac{\sec^2\theta\:d\theta}{\tan^2\theta+1}=\\\\=\lim_{n \to -\infty} \frac{1}{4}\int\limits^{0}_{n}\frac{\sec^2\theta\:d\theta}{\sec^2\theta}=\lim_{n \to -\infty} \frac{1}{4}\theta=\lim_{n \to -\infty} \frac{1}{4}\arctan(\frac{x}{4})|^0_n=[/tex]

[tex]=\lim_{n \to -\infty} \frac{1}{4}(\arctan(\frac{0}{4})-\arctan(\frac{n}{4}))= \frac{1}{4}(0-(-\frac{\pi}{2} ))=\frac{\pi}{8}[/tex]

Since,

[tex]\lim_{m \to \infty}\int\limits^{m}_0}\frac{1}{x^2+16}\:dx=\lim_{n \to -\infty} \int\limits^{0}_{n}\frac{1}{x^2+16} \:dx=\frac{\pi}{8}[/tex]

[tex]\int\limits^{\infty}_{-\infty}\frac{1}{x^2+16} \:dx= \lim_{n \to -\infty} \int\limits^{0}_{n}\frac{1}{x^2+16} \:dx + \lim_{m \to \infty}\int\limits^{m}_0}\frac{1}{x^2+16} \:dx=\frac{\pi}{8}+\frac{\pi}{8}=\frac{\pi}{4}[/tex]

Second,

[tex]\int\limits^{\infty}_{-\infty}\frac{1}{y^2+4} \:dy=\int\limits^{0}_{-\infty}\frac{1}{y^2+4}} \:d4+\int\limits^{\infty}_0}\frac{1}{y^2+4} \:dy=\\\\= \lim_{n \to -\infty} \int\limits^{0}_{n}\frac{1}{y^2+4} \:dy + \lim_{m \to \infty}\int\limits^{m}_0}\frac{1}{y^2+4} \:dy[/tex]

[tex]y = 2 \tan \theta\\dy = 2 \sec^2\theta\:d\theta\\\theta=\arctan(\frac{y}{2})[/tex]

[tex]\lim_{n \to -\infty} \int\limits^{0}_{n}\frac{1}{y^2+4} \:dy = \lim_{n \to -\infty} \int\limits^{0}_{n}\frac{2 \sec^2\theta\:d\theta}{4\tan^2\theta+4} = \lim_{n \to -\infty} \frac{1}{2}\int\limits^{0}_{n}\frac{\sec^2\theta\:d\theta}{\tan^2\theta+1}=\\\\=\lim_{n \to -\infty} \frac{1}{2}\int\limits^{0}_{n}\frac{\sec^2\theta\:d\theta}{\sec^2\theta}=\lim_{n \to -\infty} \frac{1}{2}\theta=\lim_{n \to -\infty} \frac{1}{2}\arctan(\frac{y}{2})|^0_n=[/tex]

[tex]=\lim_{n \to -\infty} \frac{1}{2}(\arctan(\frac{0}{2})-\arctan(\frac{n}{2}))= \frac{1}{2}(0-(-\frac{\pi}{2} ))=\frac{\pi}{4}[/tex]

Since,

[tex]\lim_{m \to \infty}\int\limits^{m}_0}\frac{1}{y^2+4}\:dy=\lim_{n \to -\infty} \int\limits^{0}_{n}\frac{1}{y^2+4}\:dy=\frac{\pi}{4}[/tex]

[tex]\int\limits^{\infty}_{-\infty}\frac{1}{y^2+4} \:dy= \lim_{n \to -\infty} \int\limits^{0}_{n}\frac{1}{y^2+4} \:dy + \lim_{m \to \infty}\int\limits^{m}_0}\frac{1}{y^2+4} \:dy=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}[/tex]

Hence,

[tex]\int\limits^{\infty}_{-\infty}\int\limits^{\infty}_{-\infty}\frac{1}{(x^2+16)(y^2+4)} \:dxdy = \int\limits^{\infty}_{-\infty}\frac{1}{x^2+16} \:dx\int\limits^{\infty}_{-\infty}\frac{1}{y^2+4}\:dy = \frac{\pi}{4}*\frac{\pi}{2}=\frac{\pi^2}{8}[/tex]