Improper Integrals and Proper Areas

A few years ago on the old AP Calculus discussion group a teachers asked a question about this improper integral:

\displaystyle \int_{0}^{\infty }{\frac{1}{1+{{x}^{2}}}dx}=\underset{b\to \infty }{\mathop{\lim }}\,\int_{0}^{b}{\frac{1}{1+{{x}^{2}}}dx}

=\underset{b\to \infty }{\mathop{\lim }}\,\left. \left( {{\tan }^{-1}}\left( x \right) \right) \right|_{0}^{b}

=\underset{b\to \infty }{\mathop{\lim }}\,\left( {{\tan }^{-1}}\left( b \right)-{{\tan }^{-1}}\left( 0 \right) \right)=\frac{\pi }{2}  

His (quite perceptive) student pointed out that the range of the inverse tangent function is arbitrarily restricted to the open interval \left( -\tfrac{\pi }{2},\tfrac{\pi }{2} \right). The student asked if some other range would affect the answer to this problem. The short answer is no, the result is the same. For example, if range were restricted to say \left( \tfrac{5\pi }{2},\tfrac{7\pi }{2} \right), then in the computation above:

\underset{b\to \infty }{\mathop{\lim }}\,\left( {{\tan }^{-1}}\left( b \right)-{{\tan }^{-1}}\left( 0 \right) \right)=\tfrac{7\pi }{2}-3\pi =\tfrac{\pi }{2}

The value is the same. While that is pretty straightforward, there are other things going on here which may be enlightening. The original indefinite integral represents the area in the first quadrant between the graph of y=\frac{1}{1+{{x}^{2}}} and the x-axis. Let’s consider the function that gives the area between the y-axis and the vertical line at various values of x.

A \displaystyle\left( x \right)=\int_{0}^{x}{\frac{1}{1+{{t}^{2}}}}\ dt

Pretending for the moment that we don’t know the antiderivative, we can use a calculator to graph the area function. Improper integral Of course we recognize this as the inverse tangent function, but what is more interesting is that whatever this function is, it seems to have a horizontal asymptote at y=\tfrac{\pi }{2}. The area is approaching a finite limit as x increases without bound.  The unbounded region has a finite area. The connection with improper integrals is obvious.

\displaystyle \underset{b\to \infty }{\mathop{\lim }}\,A\left( b \right)=\underset{b\to \infty }{\mathop{\lim }}\,\int_{0}^{b}{\frac{1}{1+{{x}^{2}}}dx=}\int_{0}^{\infty }{\frac{1}{1+{{x}^{2}}}dx}

Also, the improper integral is defined as the limit of the area function. This may give some insight as to why improper integrals are defined as they are.

Advertisements

One thought on “Improper Integrals and Proper Areas

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s