How to decide convergence of IntegralsLooking for the recurrence relation for certain trigonometric integralsEvaluating integrals with trigonometric functionHow to find the indefinite Integral of ln|sec|?Convergence of Riemann sums for improper integralsShowing convergence/divergence of generalized integralsCan definite integrals be indeterminate?Whats wrong in this approach to evaluate $int_0^fracpi2 fracsin xcos xdxsin x+cos x$Is there a non-trivial definite integral that values to $fracepi$?Evaluate Integrals Using Convergence TheoremsExamples of use of integrals to show convergence of a limit
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How to decide convergence of Integrals
Looking for the recurrence relation for certain trigonometric integralsEvaluating integrals with trigonometric functionHow to find the indefinite Integral of ln|sec|?Convergence of Riemann sums for improper integralsShowing convergence/divergence of generalized integralsCan definite integrals be indeterminate?Whats wrong in this approach to evaluate $int_0^fracpi2 fracsin xcos xdxsin x+cos x$Is there a non-trivial definite integral that values to $fracepi$?Evaluate Integrals Using Convergence TheoremsExamples of use of integrals to show convergence of a limit
$begingroup$
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 2 hours ago
Umesh shankarUmesh shankar
3,04431220
$endgroup$
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 2 hours ago
Umesh shankarUmesh shankar
3,04431220
$endgroup$
$begingroup$
Comparison test is definite a way to go.
$endgroup$
– Sangchul Lee
2 hours ago
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 2 hours ago
Umesh shankarUmesh shankar
3,04431220
$endgroup$
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 2 hours ago
Umesh shankarUmesh shankar
3,04431220
asked 2 hours ago
Umesh shankarUmesh shankar
3,04431220
asked 2 hours ago
Umesh shankarUmesh shankar
3,04431220
asked 2 hours ago
Umesh shankarUmesh shankar
3,04431220
asked 2 hours ago
Umesh shankarUmesh shankar
3,04431220
3,04431220
$begingroup$
Comparison test is definite a way to go.
$endgroup$
– Sangchul Lee
2 hours ago
add a comment |
$begingroup$
Comparison test is definite a way to go.
$endgroup$
– Sangchul Lee
2 hours ago
$begingroup$
Comparison test is definite a way to go.
$endgroup$
– Sangchul Lee
2 hours ago
$begingroup$
Comparison test is definite a way to go.
$endgroup$
– Sangchul Lee
2 hours ago
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 2 hours ago
John Wayland BalesJohn Wayland Bales
15k21238
$endgroup$
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
2 hours ago
$begingroup$
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
1 hour ago
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 1 hour ago
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 2 hours ago
John Wayland BalesJohn Wayland Bales
15k21238
$endgroup$
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
2 hours ago
$begingroup$
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
1 hour ago
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 2 hours ago
John Wayland BalesJohn Wayland Bales
15k21238
$endgroup$
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
2 hours ago
$begingroup$
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
1 hour ago
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 2 hours ago
John Wayland BalesJohn Wayland Bales
15k21238
$endgroup$
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 2 hours ago
John Wayland BalesJohn Wayland Bales
15k21238
edited 2 hours ago
answered 2 hours ago
John Wayland BalesJohn Wayland Bales
15k21238
answered 2 hours ago
John Wayland BalesJohn Wayland Bales
15k21238
answered 2 hours ago
John Wayland BalesJohn Wayland Bales
15k21238
15k21238
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
2 hours ago
$begingroup$
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
1 hour ago
add a comment |
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
2 hours ago
$begingroup$
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
1 hour ago
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
2 hours ago
$begingroup$
So does it mean if limit is finite at the end points, integral exists?
$endgroup$
– Umesh shankar
2 hours ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
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 1 hour ago
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 1 hour ago
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 1 hour ago
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 1 hour ago
user626368user626368
193
edited 1 hour ago
answered 1 hour ago
user626368user626368
193
answered 1 hour ago
user626368user626368
193
answered 1 hour ago
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
2 hours ago