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How to decide convergence of Integrals

5

$begingroup$

I got this doubt while evaluating the integrals:

$$I=int_0^fracpi2ln(sin x)sin xdx$$ and

$$J=int_0^fracpi4csc xdx$$

Now even though the integrand $f(x)=ln(sin x)sin x$ is not defined at $x=0$ which is the lower limit, still it has a finite answer.

But integrand in $J$ is not defined at $x=0$ and integral is infinite.

So how to identify without explicitly evaluating?

integration algebra-precalculus convergence

share|cite|improve this question

asked Mar 24 at 17:31

Umesh shankarUmesh shankar

3,09231220

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Comparison test is definite a way to go.
  $endgroup$
  – Sangchul Lee
  Mar 24 at 17:57
  

  
  
  
  

add a comment | 

5

$begingroup$

I got this doubt while evaluating the integrals:

$$I=int_0^fracpi2ln(sin x)sin xdx$$ and

$$J=int_0^fracpi4csc xdx$$

Now even though the integrand $f(x)=ln(sin x)sin x$ is not defined at $x=0$ which is the lower limit, still it has a finite answer.

But integrand in $J$ is not defined at $x=0$ and integral is infinite.

So how to identify without explicitly evaluating?

integration algebra-precalculus convergence

share|cite|improve this question

asked Mar 24 at 17:31

Umesh shankarUmesh shankar

3,09231220

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Comparison test is definite a way to go.
  $endgroup$
  – Sangchul Lee
  Mar 24 at 17:57
  

  
  
  
  

add a comment | 

$begingroup$

I got this doubt while evaluating the integrals:

$$I=int_0^fracpi2ln(sin x)sin xdx$$ and

$$J=int_0^fracpi4csc xdx$$

Now even though the integrand $f(x)=ln(sin x)sin x$ is not defined at $x=0$ which is the lower limit, still it has a finite answer.

But integrand in $J$ is not defined at $x=0$ and integral is infinite.

So how to identify without explicitly evaluating?

integration algebra-precalculus convergence

share|cite|improve this question

asked Mar 24 at 17:31

Umesh shankarUmesh shankar

3,09231220

$endgroup$

share|cite|improve this question

asked Mar 24 at 17:31

Umesh shankarUmesh shankar

3,09231220

share|cite|improve this question

asked Mar 24 at 17:31

Umesh shankarUmesh shankar

3,09231220

asked Mar 24 at 17:31

Umesh shankarUmesh shankar

3,09231220

asked Mar 24 at 17:31

Umesh shankarUmesh shankar

3,09231220

2 Answers
2

active

oldest

votes

6

$begingroup$

Notice, however, that

begineqnarray
lim_xto0^+ln(sin x)sin x&=&lim_xto0^+fracln(sin x)csc x\
&=&lim_xto0^+fraccot x(-csc xcot x)\
&=&lim_xto0^+(-sin x)\
&=&0
endeqnarray

Here is the graph of $y=ln(sin x)sin x$

share|cite|improve this answer

edited Mar 24 at 18:03

answered Mar 24 at 17:54

John Wayland BalesJohn Wayland Bales

15.4k21238

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So does it mean if limit is finite at the end points, integral exists?
  $endgroup$
  – Umesh shankar
  Mar 24 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In this case, the discontinuity at $x=0$ was a removable discontinuity. But such is not the case at $x=0$ for $csc x$ because of the vertical asymptote. However, even in the case where there is a vertical asymptote the integral may still converge. For example, $int_0^1dfrac1sqrtx,dx$.
  $endgroup$
  – John Wayland Bales
  Mar 24 at 18:20
  
  

  
  
  
  

add a comment | 

0

$begingroup$

For the second integral $J$, the integrand goes to $infty$ as $xto 0$. Roughly $frac1sin x = frac1x$. From the examples $int_0^1 x^-a dx$ we know that $a=1$ is divergent, albeit borderline so. Since
$$
sin x = x - frac13!x^3 pm ...
$$
we have $sin x leq x$ for $x>0$ small (from elementary trigonometry, or since the tail of a (edit: convergent) alternating series with terms of decreasing value is dominated by any previous term). Thus $frac1sin x geq frac1x$ and the integral $J$ is divergent.

share|cite|improve this answer

edited Mar 24 at 18:54

answered Mar 24 at 18:46

user626368user626368

193

$endgroup$

add a comment | 

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6

$begingroup$

Notice, however, that

begineqnarray
lim_xto0^+ln(sin x)sin x&=&lim_xto0^+fracln(sin x)csc x\
&=&lim_xto0^+fraccot x(-csc xcot x)\
&=&lim_xto0^+(-sin x)\
&=&0
endeqnarray

Here is the graph of $y=ln(sin x)sin x$

share|cite|improve this answer

edited Mar 24 at 18:03

answered Mar 24 at 17:54

John Wayland BalesJohn Wayland Bales

15.4k21238

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So does it mean if limit is finite at the end points, integral exists?
  $endgroup$
  – Umesh shankar
  Mar 24 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In this case, the discontinuity at $x=0$ was a removable discontinuity. But such is not the case at $x=0$ for $csc x$ because of the vertical asymptote. However, even in the case where there is a vertical asymptote the integral may still converge. For example, $int_0^1dfrac1sqrtx,dx$.
  $endgroup$
  – John Wayland Bales
  Mar 24 at 18:20
  
  

  
  
  
  

add a comment | 

6

$begingroup$

Notice, however, that

begineqnarray
lim_xto0^+ln(sin x)sin x&=&lim_xto0^+fracln(sin x)csc x\
&=&lim_xto0^+fraccot x(-csc xcot x)\
&=&lim_xto0^+(-sin x)\
&=&0
endeqnarray

Here is the graph of $y=ln(sin x)sin x$

share|cite|improve this answer

edited Mar 24 at 18:03

answered Mar 24 at 17:54

John Wayland BalesJohn Wayland Bales

15.4k21238

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So does it mean if limit is finite at the end points, integral exists?
  $endgroup$
  – Umesh shankar
  Mar 24 at 18:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In this case, the discontinuity at $x=0$ was a removable discontinuity. But such is not the case at $x=0$ for $csc x$ because of the vertical asymptote. However, even in the case where there is a vertical asymptote the integral may still converge. For example, $int_0^1dfrac1sqrtx,dx$.
  $endgroup$
  – John Wayland Bales
  Mar 24 at 18:20
  
  

  
  
  
  

add a comment | 

$begingroup$

Notice, however, that

begineqnarray
lim_xto0^+ln(sin x)sin x&=&lim_xto0^+fracln(sin x)csc x\
&=&lim_xto0^+fraccot x(-csc xcot x)\
&=&lim_xto0^+(-sin x)\
&=&0
endeqnarray

Here is the graph of $y=ln(sin x)sin x$

share|cite|improve this answer

edited Mar 24 at 18:03

answered Mar 24 at 17:54

John Wayland BalesJohn Wayland Bales

15.4k21238

$endgroup$

share|cite|improve this answer

edited Mar 24 at 18:03

answered Mar 24 at 17:54

John Wayland BalesJohn Wayland Bales

15.4k21238

answered Mar 24 at 17:54

John Wayland BalesJohn Wayland Bales

15.4k21238

answered Mar 24 at 17:54

John Wayland BalesJohn Wayland Bales

15.4k21238

0

$begingroup$

For the second integral $J$, the integrand goes to $infty$ as $xto 0$. Roughly $frac1sin x = frac1x$. From the examples $int_0^1 x^-a dx$ we know that $a=1$ is divergent, albeit borderline so. Since
$$
sin x = x - frac13!x^3 pm ...
$$
we have $sin x leq x$ for $x>0$ small (from elementary trigonometry, or since the tail of a (edit: convergent) alternating series with terms of decreasing value is dominated by any previous term). Thus $frac1sin x geq frac1x$ and the integral $J$ is divergent.

share|cite|improve this answer

edited Mar 24 at 18:54

answered Mar 24 at 18:46

user626368user626368

193

$endgroup$

add a comment | 

0

$begingroup$

For the second integral $J$, the integrand goes to $infty$ as $xto 0$. Roughly $frac1sin x = frac1x$. From the examples $int_0^1 x^-a dx$ we know that $a=1$ is divergent, albeit borderline so. Since
$$
sin x = x - frac13!x^3 pm ...
$$
we have $sin x leq x$ for $x>0$ small (from elementary trigonometry, or since the tail of a (edit: convergent) alternating series with terms of decreasing value is dominated by any previous term). Thus $frac1sin x geq frac1x$ and the integral $J$ is divergent.

share|cite|improve this answer

edited Mar 24 at 18:54

answered Mar 24 at 18:46

user626368user626368

193

$endgroup$

add a comment | 

$begingroup$

For the second integral $J$, the integrand goes to $infty$ as $xto 0$. Roughly $frac1sin x = frac1x$. From the examples $int_0^1 x^-a dx$ we know that $a=1$ is divergent, albeit borderline so. Since
$$
sin x = x - frac13!x^3 pm ...
$$
we have $sin x leq x$ for $x>0$ small (from elementary trigonometry, or since the tail of a (edit: convergent) alternating series with terms of decreasing value is dominated by any previous term). Thus $frac1sin x geq frac1x$ and the integral $J$ is divergent.

share|cite|improve this answer

edited Mar 24 at 18:54

answered Mar 24 at 18:46

user626368user626368

193

$endgroup$

share|cite|improve this answer

edited Mar 24 at 18:54

answered Mar 24 at 18:46

user626368user626368

193

answered Mar 24 at 18:46

user626368user626368

193

answered Mar 24 at 18:46

user626368user626368

193