Question

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 =

Answer 1

`(\pi^2)/(8)`

Explanation:

We are given the following improper integral:

`\int\limits^(\infty)_(-\infty)\int\limits^(\infty)_(-\infty)(1)/((x^2+16)(y^2+4)) \:dxdy`

Above double integral can be written as follows:

`\int\limits^(\infty)_(-\infty)(1)/(x^2+16) \:dx\int\limits^(\infty)_(-\infty)(1)/(y^2+4)\:dy`

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,

`\int\limits^(\infty)_(-\infty)(1)/(x^2+16) \:dx=\int\limits^(0)_(-\infty)(1)/(x^2+16) \:dx+\int\limits^(\infty)_0}(1)/(x^2+16) \:dx=\\\\= \lim_(n \to -\infty) \int\limits^(0)_(n)(1)/(x^2+16) \:dx + \lim_(m \to \infty)\int\limits^(m)_0}(1)/(x^2+16) \:dx`

`x = 4 \tan \theta\\dx = 4 \sec^2\theta\:d\theta\\\theta=\arctan((x)/(4))`

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

`=\lim_(n \to -\infty) (1)/(4)(\arctan((0)/(4))-\arctan((n)/(4)))= (1)/(4)(0-(-(\pi)/(2) ))=(\pi)/(8)`

Since,

`\lim_(m \to \infty)\int\limits^(m)_0}(1)/(x^2+16)\:dx=\lim_(n \to -\infty) \int\limits^(0)_(n)(1)/(x^2+16) \:dx=(\pi)/(8)`

`\int\limits^(\infty)_(-\infty)(1)/(x^2+16) \:dx= \lim_(n \to -\infty) \int\limits^(0)_(n)(1)/(x^2+16) \:dx + \lim_(m \to \infty)\int\limits^(m)_0}(1)/(x^2+16) \:dx=(\pi)/(8)+(\pi)/(8)=(\pi)/(4)`

Second,

`\int\limits^(\infty)_(-\infty)(1)/(y^2+4) \:dy=\int\limits^(0)_(-\infty)(1)/(y^2+4)} \:d4+\int\limits^(\infty)_0}(1)/(y^2+4) \:dy=\\\\= \lim_(n \to -\infty) \int\limits^(0)_(n)(1)/(y^2+4) \:dy + \lim_(m \to \infty)\int\limits^(m)_0}(1)/(y^2+4) \:dy`

`y = 2 \tan \theta\\dy = 2 \sec^2\theta\:d\theta\\\theta=\arctan((y)/(2))`

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

`=\lim_(n \to -\infty) (1)/(2)(\arctan((0)/(2))-\arctan((n)/(2)))= (1)/(2)(0-(-(\pi)/(2) ))=(\pi)/(4)`

Since,

`\lim_(m \to \infty)\int\limits^(m)_0}(1)/(y^2+4)\:dy=\lim_(n \to -\infty) \int\limits^(0)_(n)(1)/(y^2+4)\:dy=(\pi)/(4)`

`\int\limits^(\infty)_(-\infty)(1)/(y^2+4) \:dy= \lim_(n \to -\infty) \int\limits^(0)_(n)(1)/(y^2+4) \:dy + \lim_(m \to \infty)\int\limits^(m)_0}(1)/(y^2+4) \:dy=(\pi)/(4)+(\pi)/(4)=(\pi)/(2)`

Hence,

`\int\limits^(\infty)_(-\infty)\int\limits^(\infty)_(-\infty)(1)/((x^2+16)(y^2+4)) \:dxdy = \int\limits^(\infty)_(-\infty)(1)/(x^2+16) \:dx\int\limits^(\infty)_(-\infty)(1)/(y^2+4)\:dy = (\pi)/(4)*(\pi)/(2)=(\pi^2)/(8)`