# Improper Integrals and Proper Areas $\displaystyle \int_{0}^{\infty }{\frac{1}{1+{{x}^{2}}}dx}=\underset{b\to \infty }{\mathop{\lim }}\,\int_{0}^{b}{\frac{1}{1+{{x}^{2}}}dx}$ $=\underset{b\to \infty }{\mathop{\lim }}\,\left. \left( {{\tan }^{-1}}\left( x \right) \right) \right|_{0}^{b}$ $=\underset{b\to \infty }{\mathop{\lim }}\,\left( {{\tan }^{-1}}\left( b \right)-{{\tan }^{-1}}\left( 0 \right) \right)=\frac{\pi }{2}$

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. 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.

1. scottreedvb says: