|
The Riemann integral of a function
is defined when  is bounded and
![$ I=[a,b]\subset \mathbf R$](http://images.planetmath.org:8080/cache/objects/2732/l2h/img4.png "$ I=[a,b]\subset \mathbf R$") is a compact interval. If  is any interval of  and  is bounded on every compact subset of  (for example  is continuous), then we can define the concept of improper integral of  on  by approximation of  with compact sets.
Let  with
 and
![$ b\in(-\infty,+\infty]$](http://images.planetmath.org:8080/cache/objects/2732/l2h/img15.png "$ b\in(-\infty,+\infty]$") and let  be a continuous function on  . Given  we know that the function  is Riemann integrable on ![$ [a,b]$](http://images.planetmath.org:8080/cache/objects/2732/l2h/img20.png "$ [a,b]$") ; hence we can define the improper integral of  on  as
Notice that the limit is taken in two variables. In the case when  is compact this is the usual Riemann integral on ![$ I=[a,b]$](http://images.planetmath.org:8080/cache/objects/2732/l2h/img25.png "$ I=[a,b]$") (because the integral function is continuous). So there is no ambiguity in using the same simbol for improper integrals and usual Riemann integrals (but we will see that there is an ambiguity when dealing with Lebesgue integrals). Similarly, in the case when the
interval  is semi-open, i.e. when  with  , the definition clearly reduces to
which is a limit in one variable.
Notice also that an improper integral can be infinite or can possibly not exist (even thought  is continuous).
The definition may be extended to the case when  is the union of a finite number of intervals
 by summing up the improper integrals on every interval:
In the case when
 (or more generally when  has some symmetry), one can define the symmetric improper integral as
This limit can exist in some cases when the improper integral (not symmetric) fails to exist.
Given ![$ I=(0,1]$](http://images.planetmath.org:8080/cache/objects/2732/l2h/img37.png "$ I=(0,1]$") ,
 one has:
Given
 ,
 one can check that the improper integral
 exists and is finite but the improper integral
 is infinite. In particular this function is not summable (in the sense of Lebesgue integrals) on the interval  .
The function
 has no improper integral in
 . But since
 one can easily check that the symmetric integral is zero:
| 
![$\displaystyle \int_{(0,1]} \frac 1 {\sqrt t}\, dt$](http://images.planetmath.org:8080/cache/objects/2732/l2h/img39.png "$\displaystyle \int_{(0,1]} \frac 1 {\sqrt t}\, dt$")
![$\displaystyle = \lim_{x\to 0^+} [2t^\frac 1 2]_x^1$](http://images.planetmath.org:8080/cache/objects/2732/l2h/img41.png "$\displaystyle = \lim_{x\to 0^+} [2t^\frac 1 2]_x^1$"){kind=small}
