Editing Improper integral (section)

==Examples==
The original definition of the [[Riemann integral]] does not apply to a function such as <math>1/{x^2}</math> on the interval {{closed-open|1, ∞}}, because in this case the domain of integration is [[bounded set|unbounded]]. However, the Riemann integral can often be extended by [[continuous function|continuity]], by defining the improper integral instead as a [[Limit (mathematics)|limit]]

:<math>\int_1^\infty \frac{dx}{x^2}=\lim_{b\to\infty} \int_1^b\frac{dx}{x^2} = \lim_{b\to\infty} \left(-\frac{1}{b} + \frac{1}{1}\right) = 1. </math>

The narrow definition of the Riemann integral also does not cover the function <math display="inline">1/\sqrt{x}</math> on the interval {{closed-closed|0, 1}}. The problem here is that the integrand is [[bounded function|unbounded]] in the domain of integration. In other words, ''the definition of the Riemann integral requires that both the domain of integration and the integrand be bounded''. However, the improper integral does exist if understood as the limit

:<math>\int_0^1 \frac{dx}{\sqrt{x}}=\lim_{a\to 0^+}\int_a^1\frac{dx}{\sqrt{x}} = \lim_{a\to 0^+} \left(2 - 2\sqrt{a}\right)=2.</math>

[[File:Improper integral.svg|right|thumb|The improper integral<br/><math>\int_{0}^{\infty} \frac{dx}{(x+1)\sqrt{x}} = \pi</math><br/> has unbounded intervals for both domain and range.]]Sometimes integrals may have two singularities where they are improper. Consider, for example, the function {{math|1/((''x'' + 1){{sqrt|''x''}})}} integrated from 0 to {{math|∞}} (shown right). At the lower bound of the integration domain, as {{mvar|x}} goes to 0 the function goes to {{math|∞}}, and the upper bound is itself {{math|∞}}, though the function goes to 0. Thus this is a doubly improper integral. Integrated, say, from 1 to 3, an ordinary Riemann sum suffices to produce a result of {{pi}}/6. To integrate from 1 to {{math|∞}}, a Riemann sum is not possible. However, any finite upper bound, say {{mvar|t}} (with {{math|''t'' > 1}}), gives a well-defined result, {{math|2 arctan({{sqrt|''t''}}) − π/2}}. This has a finite limit as {{mvar|t}} goes to infinity, namely {{pi}}/2. Similarly, the integral from 1/3 to 1 allows a Riemann sum as well, coincidentally again producing {{pi}}/6. Replacing 1/3 by an arbitrary positive value {{mvar|s}} (with {{math|''s'' < 1}}) is equally safe, giving {{math|π/2 − 2 arctan({{sqrt|''s''}})}}. This, too, has a finite limit as {{mvar|s}} goes to zero, namely {{pi}}/2. Combining the limits of the two fragments, the result of this improper integral is
:<math>\begin{align}
 \int_{0}^{\infty} \frac{dx}{(1+x)\sqrt{x}} &{} = \lim_{s \to 0^+} \int_s^1 \frac{dx}{(1+x)\sqrt{x}}
   + \lim_{t \to \infty} \int_1^t \frac{dx}{(1+x) \sqrt{x}} \\
  &{} = \lim_{s \to 0^+} \left(\frac{\pi}{2} - 2 \arctan{\sqrt{s}} \right)
   + \lim_{t \to \infty} \left(2 \arctan{\sqrt{t}} - \frac{\pi}{2} \right) \\
  &{} = \frac{\pi}{2} + \left(\pi - \frac{\pi}{2} \right) \\
  &{} = \pi .
\end{align}</math>
This process does not guarantee success; a limit might fail to exist, or might be infinite. For example, over the bounded interval from 0 to 1 the integral of {{math|1/''x''}} does not converge; and over the unbounded interval from 1 to {{math|∞}} the integral of {{math|1/{{sqrt|''x''}}}} does not converge.

[[File:Improper integral unbounded internally.svg|right|thumb|The improper integral<br/><math>\int_{-1}^{1} \frac{dx}{\sqrt[3]{x^2}} = 6</math><br/> converges, since both left and right limits exist, though the integrand is unbounded near an interior point.]]
It might also happen that an integrand is unbounded near an interior point, in which case the integral must be split at that point. For the integral as a whole to converge, the limit integrals on both sides must exist and must be bounded. For example:
:<math>\begin{align}
 \int_{-1}^{1} \frac{dx}{\sqrt[3]{x^2}} &{} = \lim_{s \to 0^-} \int_{-1}^{s} \frac{dx}{\sqrt[3]{x^2}}
   + \lim_{t \to 0^+} \int_t^1 \frac{dx}{\sqrt[3]{x^2}} \\
  &{} = \lim_{s \to 0^-} 3\left(1-\sqrt[3]{s}\right) + \lim_{t \to 0^+} 3\left(1-\sqrt[3]{t}\right) \\
  &{} = 3 + 3 \\
  &{} = 6.
\end{align}</math>
But the similar integral
:<math>\int_{-1}^{1} \frac{dx}{x}</math>
cannot be assigned a value in this way, as the integrals above and below zero in the integral domain do not independently converge. (However, see [[Cauchy principal value]].)