Integral $\int_{a}^{b}$ f(x) dx requires the function f(x) to be bounded in the interval [a, b]. An integral is an improper integral if either the interval of integration is not finite or if the function to integrate is not continuous.

If the interval of integration is not finite, then the improper integral is of type 1 and if the function to integrate is not continuous then the interval of improper integral is of type 2. Improper integrals cannot be computed using a normal Riemann integral.

There are two types in improper integrals: Type-1 and Type-2.

Type- 1 Improper integral
In type-1 integral, one or both of the limits of integration will be infinity and the interval of integration will be over an infinite interval.
Given below are the following cases of type- 1 integral.

1. If $\int_{l}^{k}$ f(x) dx exists for all k $\geq$ l then
$\int_{l}^{\infty}$  f(x) dx = $\lim_{}_{k\rightarrow \infty}\int_{l}^{k}$ f(x) dx
provided the limit exists as a finite number.

2. If $\int_{k}^{n}$ f(x) dx exists for all k $\leq$ n then
$\int_{-\infty}^{n}$  f(x) dx = $\lim_{}_{k\rightarrow -\infty}\int_{k}^{n}$ f(x) dx
provided the limit exists as a finite number.

3. If both $\int_{l}^{\infty}$f(x) dx and $\int_{-\infty}^{n}$ f(x)dx converge then
$\int_{-\infty}^{\infty}$ f(x) dx = $\int_{-\infty}^{l}$ + $\int_{l}^{\infty}$ f(x) dx

Type- 2 Improper integral
In type -2 integral, the interval [a, b] will be unbounded. (a = -$\infty$ and b = $\infty$ ) and  given below are the cases of  type -2 integral.

1. If f is continuous on [a, b) and not continuous at b then
$\int_{a}^{b}$  f(x) dx = $\lim_{t \to b^{-}} \int_{a}^{t}$  f(x) dx
provided the limit exists as a finite number.

2. If f is continuous on (a, b] and not continuous at a then
$\int_{a}^{b}$  f(x) dx = $\lim_{t \to a^{+}}\int_{t}^{b}$  f(x) dx
provided the limit exists as a finite number.

3. If f is not continuous at c where a < c < b and both $\int_{a}^{c}$  f(x) dx and
 $\int_{c}^{b}$  f(x) dx converge,
then
$\int_{a}^{b}$  f(x) dx  = $\int_{a}^{c}$  f(x) dx  + $\int_{c}^{b}$  f(x) dx
Consider a function f(x) having a type-1 or type-2 behavior on the interval [a, b] then we have two cases to be considered.

case-1 : If the limit exists and is a finite number, then the improper integral is said to be convergent.

case-2 : If the limit doesn't exist (infinite) then the improper integral will diverge.
If an integral is split into two improper integrals then the interval is said to be convergent only if both the integrals converge else it is said to be divergent.

Solved Examples

Question 1: Evaluate  $\int_{1}^{\infty}$ $\frac{dx}{x^{2}}$
Solution:
 
The given integral is an improper integral of type-1

We evaluate  this by finding  $\lim_{}_{t\rightarrow \infty}\int_{1}^{t}$ $\frac{dx}{x^{2}}$

Now integrate $\int_{1}^{t}$ $\frac{dx}{x^{2}}$

= ( 1 - $\frac{1}{t}$)

$\lim_{}_{t\rightarrow \infty}( 1 - $$\frac{1}{t}$$)$  = 1    

Therefore $\int_{1}^{\infty}$ $\frac{dx}{x^{2}}$ = 1
 

Question 2: Evaluate  $\int_{0}^{\frac{\pi}{2}}$ sec xdx
Solution:
 
The given integral is an improper integral of type-2 as sec x  will not be continuous at $\frac{\pi}{2}$

Evaluate by finding   $\lim_{}_{t\rightarrow \frac{\pi}{2}-}\int_{0}^{t}$ sec xdx

$\int_{0}^{t}$ sec xdx = ln|Sec x + tan x|$_{0}^{t}$

= ln |sec t + tan t|

$\lim_{}_{t\rightarrow \frac{\pi}{2}-}$ (ln|sec t + tan t|) = $\infty$

Therefore $\int_{0}^{\frac{\pi}{2}}$ sec xdx diverges.
 

If f and g are two continuous functions with f(x) $\geq$ g(x) $\geq$ 0 for x $\geq$ a then

1. If $\int_{a}^{\infty}$ f(x) dx is convergent, then $\int_{a}^{\infty}$ g(x) dx is also convergent.

2.  If $\int_{a}^{\infty}$ g(x) dx is divergent, then $\int_{a}^{\infty}$ f(x) dx is also divergent.

If f(x) is larger than g(x) then the area under f(x) should also be larger than the area under g(x).
Note that $\int_{1}^{\infty}$ $\frac{dx}{x^{p}}$ converges if p > 1, diverges otherwise.

Solved Examples

Question 1: Examine the given integral $\int_{1}^{\infty}\frac{1}{x}$ dx and examine whether it is convergent or divergent.
Solution:
 
Convert the integral to a limit.

$\int_{1}^{\infty}\frac{1}{x}$dx = $\lim_{t \to \infty}\int_{1}^{t}\frac{1}{x}$dx

Integrating the given integral

$\int_{1}^{\infty}\frac{1}{x}$dx = $\lim_{t \to \infty}ln(x)_{1}^{t}$

= $\lim_{t \to \infty}[ln(t)-ln(1)]$

= $\infty$

As the limit is infinite the given integral is said to be divergent.
 

Question 2: Find $\int_{0}^{\pi}Sec^{2}$ xdx
Solution:
 
Given integral is an improper integral of type -2 and $sec^{2}$x is not continuous at $\frac{\pi}{2}$

We can write it as
$\int_{0}^{\pi}$ $sec^{2}$xdx =  $\int_{0 }^{\frac{\pi}{2}}$ $sec^{2}$xdx + $\int_{\frac{\pi}{2}}^{\pi}$ $sec^{2}$xdx

First evaluate $\int_{0 }^{\frac{\pi}{2}}$ $sec^{2}$xdx

$\int_{0}^{\frac{\pi}{2}}$ $sec^{2}$xdx =  $\lim_{t \to \frac{\pi}{2}-}$ $\int_{0}^{t}$ $sec^{2}$xdx

$\int_{0}^{t}$ $sec^{2}$xdx = tan t - tan 0

= tan t

Therefore
$\int_{0}^{\frac{\pi}{2}}$$sec^{2}$xdx =  $\lim_{t \to \frac{\pi}{2}-}$(tan t )
= $\infty$

As $\int_{0}^{\frac{\pi}{2}}$$sec^{2}$xdx diverges the other intergral $\int_{0}^{\pi}$$sec^{2}$xdx will also diverge.

Therefore the given integral $\int_{0}^{\pi}Sec^{2}$ xdx is divergent.