Ucale

# Convergence of Improper Integral

### Convergence of Improper Integral of First kind

Definition 1.   If f(x)  is a bounded and integrable  function on $\displaystyle [a,\infty )$, then the improper integral of f(x) on $\displaystyle [a,\infty )$ is denoted by $\displaystyle \int _{ a }^{ \infty }{ f\left( x \right) } dx$ and is defined as

$\displaystyle \int _{ a }^{ \infty }{ f\left( x \right) } dx=\lim _{ k\longrightarrow \infty }{ \int _{ a }^{ k }{ f\left( x \right) } } dx$

provided that the limit on right hand side exists.

The improper integral  $\displaystyle \int _{ a }^{ \infty }{ f\left( x \right) } dx$ is said to be convergent, if $\displaystyle \lim _{ k\longrightarrow \infty }{ \int _{ a }^{ k }{ f\left( x \right) } } dx$ is either $\displaystyle +\infty \quad and\quad -\infty$, then the integral is said to be divergent.

Definition 2.  Let f(x) be a bounded integrable function defined on $\displaystyle (\infty ,b]$ Then the improper integral of f(x) on $\displaystyle (\infty ,b]$ is denoted by $\displaystyle \int _{ -\infty }^{ b }{ f\left( x \right) } dx$ and is defined as

$\displaystyle \int _{ -\infty }^{ b }{ f\left( x \right) } dx=\lim _{ k\longrightarrow \infty }{ \int _{ -\infty }^{ b }{ f\left( x \right) } dx }$

provided that the limit on right hand side exists.

The improper integral  $\displaystyle \int _{ -\infty }^{ b }{ f\left( x \right) } dx$ is said to be convergent, if $\displaystyle \lim _{ k\longrightarrow \infty }{ \int _{ -\infty }^{ b }{ f\left( x \right) } dx }$ exists finitely and this limit is called the value of the improper integral $\displaystyle \int _{ -\infty }^{ b }{ f\left( x \right) } dx.\quad If\quad \lim _{ k\longrightarrow \infty }{ \int _{ -\infty }^{ b }{ f\left( x \right) } dx }$ is either $\displaystyle +\infty \quad or\quad -\infty$ then the integral is said to be divergent.

Definition 3.  Let f(x) be a bounded integrable function defined on $\displaystyle \left( -\infty ,\infty \right)$. Then the improper integral of f(x) on $\displaystyle \left( -\infty ,\infty \right)$ is denoted by $\displaystyle \int _{ -\infty }^{ \infty }{ f\left( x \right) } dx$ and is defined as

$\displaystyle \int _{ -\infty }^{ \infty }{ f\left( x \right) } dx=\lim _{ k\longrightarrow -\infty }{ \int _{ k }^{ c }{ f\left( x \right) } } dx+\lim _{ l\longrightarrow \infty }{ \int _{ c }^{ l }{ f\left( x \right) } } dx$

provided that both the limits on right hand side exist , where c is any real number satisfying k<c<1.

The improper integral $\displaystyle \int _{ -\infty }^{ \infty }{ f\left( x \right) } dx$ is said to be convergent,  if both the limits on the right- hand side exist finitely and are independent of each other. The improper integral $\displaystyle \int _{ -\infty }^{ \infty }{ f\left( x \right) } dx$ is said to be divergent if the right hand side is $\displaystyle +\infty \quad or\quad -\infty$

### Example

(i) $\displaystyle \int _{ 0 }^{ \infty }{ { e }^{ -x } } dx$

Solution:

$\displaystyle We\quad have\quad \int _{ 0 }^{ \infty }{ { e }^{ -x } } dx\\ =\lim _{ k\longrightarrow \infty }{ { \left[ -{ e }^{ -x } \right] }_{ 0 }^{ k } } \\ =\lim _{ k\longrightarrow \infty }{ { \left( -{ e }^{ -k }+{ e }^{ 0 } \right) } } \\ =\lim _{ k\longrightarrow \infty }{ { \left( 1-{ e }^{ -k } \right) } } \\ =1-0\qquad \left[ \because \lim _{ k\longrightarrow \infty }{ { \left( -{ e }^{ -k } \right) } } ={ e }^{ -\infty }=0 \right] \\ =1$

Thus $\displaystyle \lim _{ k\longrightarrow \infty }{ \int _{ 0 }^{ k }{ { e }^{ -x } } } dx$ exists and is finite. Hence, the given integral is convergent and $\displaystyle \int _{ 0 }^{ \infty }{ { e }^{ -x } } dx=1$

(ii)  $\displaystyle \int _{ 0 }^{ \infty }{ \frac { 1 }{ x } } dx$

Solution:

$\displaystyle \int _{ 0 }^{ \infty }{ \frac { 1 }{ x } } dx\\ =\lim _{ k\longrightarrow \infty }{ \int _{ 1 }^{ k }{ \frac { 1 }{ x } } } dx\\ =\lim _{ k\longrightarrow \infty }{ { \left[ \log { \left| x \right| } \right] }_{ 1 }^{ k } } \\ =\lim _{ k\longrightarrow \infty }{ { \left[ \log { \left| k \right| } -\log { 1 } \right] } } \\ =\lim _{ k\longrightarrow \infty }{ { \left[ \log { \left| k \right| } -0 \right] } } \\ =\infty$

So, $\displaystyle \int _{ 0 }^{ \infty }{ \frac { 1 }{ x } } dx$ is a divergent integral.

### Convergence of Improper Integral of Second  kind

In this section we shall discussed the convergence of improper integral of second kind.

Recall that a definite integral $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx$ is an improper integral of second kind if its range of integration is finite and f(x) has at least one point of infinite discontinuity in [a,b]

Definition 1.  Let f(x) be a bounded function defined on (a,b] such that a is the only point of infinite discontinuity of f(x) i.e. $\displaystyle f\left( x \right)\longrightarrow \infty \quad as\quad x\longrightarrow a$. Then the improper integral of f(x) on (a,b] is denoted by $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx$ and is defined as

$\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\lim _{ \epsilon \longrightarrow 0 }{ \int _{ a+\epsilon }^{ b }{ f\left( x \right) } } dx$

provided that the limit on right hand side exists finitely. If l denotes the limit on right hand side, then the improper integral $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx$ is said to be converge to l, when l is finite. If  If $\displaystyle l=+\infty \quad or\quad l=-\infty$ , then the integral is said to be  a divergent integral.

Definition 2.   Let f(x) be bounded function defined on [a,b) such that b is only point of infinite discontinuity of f(x) i.e. $\displaystyle f\left( x \right)\longrightarrow \infty \quad as\quad x\longrightarrow b$ Then the improper integral of f(x) on [a,b) is denoted by $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx$ and is defined as

$\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\lim _{ \epsilon \longrightarrow 0 }{ \int _{ a }^{ b-\epsilon }{ f\left( x \right) } } dx$

provided that both the  limits on the right hand side exist.

Definition 3.   Let f(x) be bounded function defined on [a,b]-{c} ,$\displaystyle c\quad \epsilon \quad \left[ a,b \right]$ and c is the only point of infinite discontinuity of f(x) i.e. $\displaystyle f\left( c \right)\longrightarrow \infty$ and is defined as

$\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\lim _{ \epsilon \longrightarrow 0 }{ \int _{ a }^{ c-\epsilon }{ f\left( x \right) } } dx+\lim _{ \delta \longrightarrow 0 }{ \int _{ c+\delta }^{ }{ f\left( x \right) } }dx$

provided that the limit on right hand side exists finitely. The improper integral $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx$ is said to be convergent if both the limits on the right hand side exists finitely. If either of the two or both the limits on RHS are $\displaystyle \pm \infty$ then the integral is said to be divergent.

### Example

(i) $\displaystyle \int _{ 0 }^{ 4 }{ \frac { 1 }{ \sqrt { x } } } dx$

Solution:

$\displaystyle Let\quad f\left( x \right) =\int _{ 0 }^{ 4 }{ \frac { 1 }{ \sqrt { x } } } dx.\quad Then,\quad \lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } =\infty$ So, 0 is the only point of infinite discontinuity of f(x) on [0,4)

$\displaystyle \therefore \int _{ 0 }^{ 4 }{ \frac { 1 }{ \sqrt { x } } } dx=\lim _{ \epsilon \longrightarrow 0 }{ \int _{ 0+\epsilon }^{ 4 }{ \frac { 1 }{ \sqrt { x } } } } dx\\ =\lim _{ \epsilon \longrightarrow 0 }{ { \left[ \frac { { x }^{ { 1 }/{ 2 } } }{ { 1 }/{ 2 } } \right] }_{ \epsilon }^{ 4 } } \\ =\lim _{ \epsilon \longrightarrow 0 }{ { \left[ 2\sqrt { x } \right] }_{ \epsilon }^{ 4 } } \\ =2\lim _{ \epsilon \longrightarrow 0 }{ { \left[ 2-\sqrt { \epsilon } \right] }_{ \epsilon }^{ 4 } } \\ =2\left( 2-0 \right) \\ =4$

Hence, the given integral is convergent and its value is 4.

(ii) $\displaystyle \int _{ 0 }^{ 1 }{ \frac { 1 }{ { x }^{ 2 } } } dx$

Solution:

$\displaystyle Let\quad f\left( x \right) =\int _{ 0 }^{ 1 }{ \frac { 1 }{ { x }^{ 2 } } } dx\\ Then,\lim _{ x\longrightarrow { 0 }^{ + } }{ f\left( x \right) } =\lim _{ x\longrightarrow { 0 }^{ + } }{ \frac { 1 }{ { x }^{ 2 } } } =\infty$

So, 0 is the only point of discontinuity of f(x) on (0,1].

$\displaystyle \therefore \int _{ 0 }^{ 1 }{ \frac { 1 }{ { x }^{ 2 } } } dx=\lim _{ \epsilon \longrightarrow 0 }{ \int _{ 0+\epsilon }^{ 1 }{ \frac { 1 }{ { x }^{ 2 } } } } dx\\ =\lim _{ \epsilon \longrightarrow 0 }{ { \left[ -\frac { 1 }{ { x }^{ 2 } } \right] }_{ \epsilon }^{ 1 } } \\ =\lim _{ \epsilon \longrightarrow 0 }{ { \left[ -\frac { 1 }{ 1 } +\frac { 1 }{ \epsilon } \right] } } \\ =-1+\infty \\ =\infty$

So, the given integral is divergent.

April 18, 2019
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