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$begingroup$
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
asked yesterday
Umesh shankarUmesh shankar
3,05931220
$endgroup$
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
asked yesterday
Umesh shankarUmesh shankar
3,05931220
$endgroup$
$begingroup$
Comparison test is definite a way to go.
$endgroup$
– Sangchul Lee
yesterday
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
asked yesterday
Umesh shankarUmesh shankar
3,05931220
$endgroup$
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
integration algebra-precalculus convergence
asked yesterday
Umesh shankarUmesh shankar
3,05931220
asked yesterday
Umesh shankarUmesh shankar
3,05931220
asked yesterday
Umesh shankarUmesh shankar
3,05931220
asked yesterday
Umesh shankarUmesh shankar
3,05931220
asked yesterday
Umesh shankarUmesh shankar
3,05931220
3,05931220
$begingroup$
Comparison test is definite a way to go.
$endgroup$
– Sangchul Lee
yesterday
add a comment |
$begingroup$
Comparison test is definite a way to go.
$endgroup$
– Sangchul Lee
yesterday
$begingroup$
Comparison test is definite a way to go.
$endgroup$
– Sangchul Lee
yesterday
$begingroup$
Comparison test is definite a way to go.
$endgroup$
– Sangchul Lee
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $y=ln(sin x)sin x$
answered yesterday
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
yesterday
$begingroup$
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
yesterday
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.
answered yesterday
user626368user626368
193
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $y=ln(sin x)sin x$
answered yesterday
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
yesterday
$begingroup$
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
yesterday
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 $y=ln(sin x)sin x$
answered yesterday
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
yesterday
$begingroup$
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
yesterday
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 $y=ln(sin x)sin x$
answered yesterday
John Wayland BalesJohn Wayland Bales
15.1k21238
$endgroup$
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 $y=ln(sin x)sin x$
answered yesterday
John Wayland BalesJohn Wayland Bales
15.1k21238
edited yesterday
answered yesterday
John Wayland BalesJohn Wayland Bales
15.1k21238
answered yesterday
John Wayland BalesJohn Wayland Bales
15.1k21238
answered yesterday
John Wayland BalesJohn Wayland Bales
15.1k21238
15.1k21238
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
yesterday
$begingroup$
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
yesterday
add a comment |
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
yesterday
$begingroup$
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
yesterday
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
yesterday
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
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.
answered yesterday
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.
answered yesterday
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.
answered yesterday
user626368user626368
193
$endgroup$
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.
answered yesterday
user626368user626368
193
edited yesterday
answered yesterday
user626368user626368
193
answered yesterday
user626368user626368
193
answered yesterday
user626368user626368
193
193
add a comment |
add a comment |
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$begingroup$
Comparison test is definite a way to go.
$endgroup$
– Sangchul Lee
yesterday