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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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