Can someone explain this integration trick for log-sine integrals?
$begingroup$
I was working on this rather challenging log-sin integral:
$$
int_{0}^{2pi}x^{2}ln^{2}left(2sinleft(x over 2right)right),{rm d}x = {13pi^{5} over 45}
$$
The upper limit is a waiver from the norm of $frac{pi}{2}$. Anyway, when integrating log-sin integrals one can often times use the famous identity
$$displaystyle -ln(2sin(x/2))=sum_{k=1}^{infty}frac{cos(kx)}{k}$$
by switching sum and integral, integrating, then evaluating the sums. However, this is only valid when $xneq pm pi,pm2pi,ldots$. So, just for the heck of it I decided to do something I figured would not be viable, but done it anyway. I integrated $displaystyle int_{0}^{2pi}frac{x^{2}cos^{2}(kx)}{k^{2}}$, evaluated the resulting sums and arrived at $displaystyle frac{41{pi}^{5}}{180}$.
Now, take the integral $displaystyle int_{0}^{pi}x^{2}ln^{2}(2cos(x/2))dx=frac{11{pi}^{5}}{180}$. This one can be done by using Cauchy's cosine formula, differentiating, and so on. Anyway, note that when I add the two results, the correct result for the integral at hand is obtained.
$$displaystyle frac{41{pi}^{5}}{180}+frac{11{pi}^{5}}{180}=frac{13{pi}^{5}}{45}.$$
By quandary/query is why does this happen to work?. A fluke?. Does that cosine sum represent something that when evaluated and added to the above log-cos integral happens to be equivalent to the log-sin integral in question?.
See what I am trying to explain?. I done something I knew I was not supposed to, but it happened to work out. If viable, what would $displaystyle int_{0}^{2pi}frac{x^{2}cos^{2}(kx)}{k^{2}}$ represent such that when it is added to $displaystyle int_{0}^{pi}x^{2}ln^{2}(2cos(x/2))dx=frac{11{pi}^{5}}{180}$ equals $displaystyle int_{0}^{2pi}x^{2}ln^{2}(2sin(x/2))dx=frac{13{pi}^{5}}{45}$.
I knew I could not just square the cosine and multiply by $x^2$. But,sometimes it can be done due to Parseval, i.e., $displaystyle ln^{2}(2sin(x/2))=sum_{n=1}^{infty}sum_{k=1}^{infty}frac{cos(nx)cos(kx)}{nk}$. Then, because $displaystyle int_{0}^{pi}cos(nx)cos(kx)dx=0, ; nneq k$, one can integrate $displaystyle int_{0}^{pi}frac{cos^{2}(kx)}{k^{2}}$, then sum.
This works for $2pi$ as well, but that $x^2$ term tends to throw a wrench in things But I do not think I can do this in my case because of the $x^2$ term. Or can I?. I was surprised when I saw this. I thought perhaps I stumbled onto a cool way to evaluate this, but I am not entirely sure what I have.
Or, if anyone has their own clever method?.
EDIT: I managed to evaluate this integral by considering the identity
$displaystyle ln(2sin(x/2))=ln(1-e^{ix})+frac{i}{2}(pi -x)$
definite-integrals
edited Dec 7 '18 at 16:57
Klangen
1,73711334
asked May 2 '13 at 20:05
CodyCody
8,1211976
$endgroup$
|
show 1 more comment
$begingroup$
I was working on this rather challenging log-sin integral:
$$
int_{0}^{2pi}x^{2}ln^{2}left(2sinleft(x over 2right)right),{rm d}x = {13pi^{5} over 45}
$$
The upper limit is a waiver from the norm of $frac{pi}{2}$. Anyway, when integrating log-sin integrals one can often times use the famous identity
$$displaystyle -ln(2sin(x/2))=sum_{k=1}^{infty}frac{cos(kx)}{k}$$
by switching sum and integral, integrating, then evaluating the sums. However, this is only valid when $xneq pm pi,pm2pi,ldots$. So, just for the heck of it I decided to do something I figured would not be viable, but done it anyway. I integrated $displaystyle int_{0}^{2pi}frac{x^{2}cos^{2}(kx)}{k^{2}}$, evaluated the resulting sums and arrived at $displaystyle frac{41{pi}^{5}}{180}$.
Now, take the integral $displaystyle int_{0}^{pi}x^{2}ln^{2}(2cos(x/2))dx=frac{11{pi}^{5}}{180}$. This one can be done by using Cauchy's cosine formula, differentiating, and so on. Anyway, note that when I add the two results, the correct result for the integral at hand is obtained.
$$displaystyle frac{41{pi}^{5}}{180}+frac{11{pi}^{5}}{180}=frac{13{pi}^{5}}{45}.$$
By quandary/query is why does this happen to work?. A fluke?. Does that cosine sum represent something that when evaluated and added to the above log-cos integral happens to be equivalent to the log-sin integral in question?.
See what I am trying to explain?. I done something I knew I was not supposed to, but it happened to work out. If viable, what would $displaystyle int_{0}^{2pi}frac{x^{2}cos^{2}(kx)}{k^{2}}$ represent such that when it is added to $displaystyle int_{0}^{pi}x^{2}ln^{2}(2cos(x/2))dx=frac{11{pi}^{5}}{180}$ equals $displaystyle int_{0}^{2pi}x^{2}ln^{2}(2sin(x/2))dx=frac{13{pi}^{5}}{45}$.
I knew I could not just square the cosine and multiply by $x^2$. But,sometimes it can be done due to Parseval, i.e., $displaystyle ln^{2}(2sin(x/2))=sum_{n=1}^{infty}sum_{k=1}^{infty}frac{cos(nx)cos(kx)}{nk}$. Then, because $displaystyle int_{0}^{pi}cos(nx)cos(kx)dx=0, ; nneq k$, one can integrate $displaystyle int_{0}^{pi}frac{cos^{2}(kx)}{k^{2}}$, then sum.
This works for $2pi$ as well, but that $x^2$ term tends to throw a wrench in things But I do not think I can do this in my case because of the $x^2$ term. Or can I?. I was surprised when I saw this. I thought perhaps I stumbled onto a cool way to evaluate this, but I am not entirely sure what I have.
Or, if anyone has their own clever method?.
EDIT: I managed to evaluate this integral by considering the identity
$displaystyle ln(2sin(x/2))=ln(1-e^{ix})+frac{i}{2}(pi -x)$
definite-integrals
edited Dec 7 '18 at 16:57
Klangen
1,73711334
asked May 2 '13 at 20:05
CodyCody
8,1211976
$endgroup$
$begingroup$
The problem with the identity $displaystyle -ln(2sin(x/2))=sum_{k=1}^{infty}frac{cos(kx)}{k}$ for $x=npi$ arises only when $(4m-2)pile nle 4mpi,$ where $min Bbb Z$. So the values $x=pmpi,pm 5pi, dots$ are OK. Note that $0$ is in the list of problematic values, too.
$endgroup$
– John Bentin
May 4 '13 at 22:32
$begingroup$
Looks like a fluke to me.
$endgroup$
– joriki
May 6 '13 at 7:15
$begingroup$
Thanks joriki. I had a feeling it may be, but I wanted to see what you all thought.
$endgroup$
– Cody
May 6 '13 at 18:03
$begingroup$
John Benton, slight correction. The "good" negative multiples of pi are -3*pi, -7*pi, etc, i.e. whenever sin (x/2) is +1.
$endgroup$
– Oscar Lanzi
Feb 21 '16 at 13:41
$begingroup$
It ought not to cause a great deal of trouble that the range of integration spans a value at which the expansion is not valid. It's only not valid by reason of the sine going to zero & therefore the logarithm going to ∞. functions that go to ∞ at the end of some region are quite routinely integrated over that region & found to have finite (though strictly speaking improper) integrals. If the ∞ is in the midst of the region, then it's just two such integrals juxtaposed. If you take the usual precautions for evaluating such integrals, then what you are doing is not extraordinary.
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 19:21
|
show 1 more comment
$begingroup$
I was working on this rather challenging log-sin integral:
$$
int_{0}^{2pi}x^{2}ln^{2}left(2sinleft(x over 2right)right),{rm d}x = {13pi^{5} over 45}
$$
The upper limit is a waiver from the norm of $frac{pi}{2}$. Anyway, when integrating log-sin integrals one can often times use the famous identity
$$displaystyle -ln(2sin(x/2))=sum_{k=1}^{infty}frac{cos(kx)}{k}$$
by switching sum and integral, integrating, then evaluating the sums. However, this is only valid when $xneq pm pi,pm2pi,ldots$. So, just for the heck of it I decided to do something I figured would not be viable, but done it anyway. I integrated $displaystyle int_{0}^{2pi}frac{x^{2}cos^{2}(kx)}{k^{2}}$, evaluated the resulting sums and arrived at $displaystyle frac{41{pi}^{5}}{180}$.
Now, take the integral $displaystyle int_{0}^{pi}x^{2}ln^{2}(2cos(x/2))dx=frac{11{pi}^{5}}{180}$. This one can be done by using Cauchy's cosine formula, differentiating, and so on. Anyway, note that when I add the two results, the correct result for the integral at hand is obtained.
$$displaystyle frac{41{pi}^{5}}{180}+frac{11{pi}^{5}}{180}=frac{13{pi}^{5}}{45}.$$
By quandary/query is why does this happen to work?. A fluke?. Does that cosine sum represent something that when evaluated and added to the above log-cos integral happens to be equivalent to the log-sin integral in question?.
See what I am trying to explain?. I done something I knew I was not supposed to, but it happened to work out. If viable, what would $displaystyle int_{0}^{2pi}frac{x^{2}cos^{2}(kx)}{k^{2}}$ represent such that when it is added to $displaystyle int_{0}^{pi}x^{2}ln^{2}(2cos(x/2))dx=frac{11{pi}^{5}}{180}$ equals $displaystyle int_{0}^{2pi}x^{2}ln^{2}(2sin(x/2))dx=frac{13{pi}^{5}}{45}$.
I knew I could not just square the cosine and multiply by $x^2$. But,sometimes it can be done due to Parseval, i.e., $displaystyle ln^{2}(2sin(x/2))=sum_{n=1}^{infty}sum_{k=1}^{infty}frac{cos(nx)cos(kx)}{nk}$. Then, because $displaystyle int_{0}^{pi}cos(nx)cos(kx)dx=0, ; nneq k$, one can integrate $displaystyle int_{0}^{pi}frac{cos^{2}(kx)}{k^{2}}$, then sum.
This works for $2pi$ as well, but that $x^2$ term tends to throw a wrench in things But I do not think I can do this in my case because of the $x^2$ term. Or can I?. I was surprised when I saw this. I thought perhaps I stumbled onto a cool way to evaluate this, but I am not entirely sure what I have.
Or, if anyone has their own clever method?.
EDIT: I managed to evaluate this integral by considering the identity
$displaystyle ln(2sin(x/2))=ln(1-e^{ix})+frac{i}{2}(pi -x)$
definite-integrals
edited Dec 7 '18 at 16:57
Klangen
1,73711334
asked May 2 '13 at 20:05
CodyCody
8,1211976
$endgroup$
I was working on this rather challenging log-sin integral:
$$
int_{0}^{2pi}x^{2}ln^{2}left(2sinleft(x over 2right)right),{rm d}x = {13pi^{5} over 45}
$$
The upper limit is a waiver from the norm of $frac{pi}{2}$. Anyway, when integrating log-sin integrals one can often times use the famous identity
$$displaystyle -ln(2sin(x/2))=sum_{k=1}^{infty}frac{cos(kx)}{k}$$
by switching sum and integral, integrating, then evaluating the sums. However, this is only valid when $xneq pm pi,pm2pi,ldots$. So, just for the heck of it I decided to do something I figured would not be viable, but done it anyway. I integrated $displaystyle int_{0}^{2pi}frac{x^{2}cos^{2}(kx)}{k^{2}}$, evaluated the resulting sums and arrived at $displaystyle frac{41{pi}^{5}}{180}$.
Now, take the integral $displaystyle int_{0}^{pi}x^{2}ln^{2}(2cos(x/2))dx=frac{11{pi}^{5}}{180}$. This one can be done by using Cauchy's cosine formula, differentiating, and so on. Anyway, note that when I add the two results, the correct result for the integral at hand is obtained.
$$displaystyle frac{41{pi}^{5}}{180}+frac{11{pi}^{5}}{180}=frac{13{pi}^{5}}{45}.$$
By quandary/query is why does this happen to work?. A fluke?. Does that cosine sum represent something that when evaluated and added to the above log-cos integral happens to be equivalent to the log-sin integral in question?.
See what I am trying to explain?. I done something I knew I was not supposed to, but it happened to work out. If viable, what would $displaystyle int_{0}^{2pi}frac{x^{2}cos^{2}(kx)}{k^{2}}$ represent such that when it is added to $displaystyle int_{0}^{pi}x^{2}ln^{2}(2cos(x/2))dx=frac{11{pi}^{5}}{180}$ equals $displaystyle int_{0}^{2pi}x^{2}ln^{2}(2sin(x/2))dx=frac{13{pi}^{5}}{45}$.
I knew I could not just square the cosine and multiply by $x^2$. But,sometimes it can be done due to Parseval, i.e., $displaystyle ln^{2}(2sin(x/2))=sum_{n=1}^{infty}sum_{k=1}^{infty}frac{cos(nx)cos(kx)}{nk}$. Then, because $displaystyle int_{0}^{pi}cos(nx)cos(kx)dx=0, ; nneq k$, one can integrate $displaystyle int_{0}^{pi}frac{cos^{2}(kx)}{k^{2}}$, then sum.
This works for $2pi$ as well, but that $x^2$ term tends to throw a wrench in things But I do not think I can do this in my case because of the $x^2$ term. Or can I?. I was surprised when I saw this. I thought perhaps I stumbled onto a cool way to evaluate this, but I am not entirely sure what I have.
Or, if anyone has their own clever method?.
EDIT: I managed to evaluate this integral by considering the identity
$displaystyle ln(2sin(x/2))=ln(1-e^{ix})+frac{i}{2}(pi -x)$
definite-integrals
definite-integrals
edited Dec 7 '18 at 16:57
Klangen
1,73711334
asked May 2 '13 at 20:05
CodyCody
8,1211976
edited Dec 7 '18 at 16:57
Klangen
1,73711334
asked May 2 '13 at 20:05
CodyCody
8,1211976
edited Dec 7 '18 at 16:57
Klangen
1,73711334
edited Dec 7 '18 at 16:57
Klangen
1,73711334
edited Dec 7 '18 at 16:57
Klangen
1,73711334
1,73711334
asked May 2 '13 at 20:05
CodyCody
8,1211976
asked May 2 '13 at 20:05
CodyCody
8,1211976
asked May 2 '13 at 20:05
CodyCody
8,1211976
8,1211976
$begingroup$
The problem with the identity $displaystyle -ln(2sin(x/2))=sum_{k=1}^{infty}frac{cos(kx)}{k}$ for $x=npi$ arises only when $(4m-2)pile nle 4mpi,$ where $min Bbb Z$. So the values $x=pmpi,pm 5pi, dots$ are OK. Note that $0$ is in the list of problematic values, too.
$endgroup$
– John Bentin
May 4 '13 at 22:32
$begingroup$
Looks like a fluke to me.
$endgroup$
– joriki
May 6 '13 at 7:15
$begingroup$
Thanks joriki. I had a feeling it may be, but I wanted to see what you all thought.
$endgroup$
– Cody
May 6 '13 at 18:03
$begingroup$
John Benton, slight correction. The "good" negative multiples of pi are -3*pi, -7*pi, etc, i.e. whenever sin (x/2) is +1.
$endgroup$
– Oscar Lanzi
Feb 21 '16 at 13:41
$begingroup$
It ought not to cause a great deal of trouble that the range of integration spans a value at which the expansion is not valid. It's only not valid by reason of the sine going to zero & therefore the logarithm going to ∞. functions that go to ∞ at the end of some region are quite routinely integrated over that region & found to have finite (though strictly speaking improper) integrals. If the ∞ is in the midst of the region, then it's just two such integrals juxtaposed. If you take the usual precautions for evaluating such integrals, then what you are doing is not extraordinary.
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 19:21
|
show 1 more comment
$begingroup$
The problem with the identity $displaystyle -ln(2sin(x/2))=sum_{k=1}^{infty}frac{cos(kx)}{k}$ for $x=npi$ arises only when $(4m-2)pile nle 4mpi,$ where $min Bbb Z$. So the values $x=pmpi,pm 5pi, dots$ are OK. Note that $0$ is in the list of problematic values, too.
$endgroup$
– John Bentin
May 4 '13 at 22:32
$begingroup$
Looks like a fluke to me.
$endgroup$
– joriki
May 6 '13 at 7:15
$begingroup$
Thanks joriki. I had a feeling it may be, but I wanted to see what you all thought.
$endgroup$
– Cody
May 6 '13 at 18:03
$begingroup$
John Benton, slight correction. The "good" negative multiples of pi are -3*pi, -7*pi, etc, i.e. whenever sin (x/2) is +1.
$endgroup$
– Oscar Lanzi
Feb 21 '16 at 13:41
$begingroup$
It ought not to cause a great deal of trouble that the range of integration spans a value at which the expansion is not valid. It's only not valid by reason of the sine going to zero & therefore the logarithm going to ∞. functions that go to ∞ at the end of some region are quite routinely integrated over that region & found to have finite (though strictly speaking improper) integrals. If the ∞ is in the midst of the region, then it's just two such integrals juxtaposed. If you take the usual precautions for evaluating such integrals, then what you are doing is not extraordinary.
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 19:21
$begingroup$
The problem with the identity $displaystyle -ln(2sin(x/2))=sum_{k=1}^{infty}frac{cos(kx)}{k}$ for $x=npi$ arises only when $(4m-2)pile nle 4mpi,$ where $min Bbb Z$. So the values $x=pmpi,pm 5pi, dots$ are OK. Note that $0$ is in the list of problematic values, too.
$endgroup$
– John Bentin
May 4 '13 at 22:32
$begingroup$
The problem with the identity $displaystyle -ln(2sin(x/2))=sum_{k=1}^{infty}frac{cos(kx)}{k}$ for $x=npi$ arises only when $(4m-2)pile nle 4mpi,$ where $min Bbb Z$. So the values $x=pmpi,pm 5pi, dots$ are OK. Note that $0$ is in the list of problematic values, too.
$endgroup$
– John Bentin
May 4 '13 at 22:32
$begingroup$
Looks like a fluke to me.
$endgroup$
– joriki
May 6 '13 at 7:15
$begingroup$
Looks like a fluke to me.
$endgroup$
– joriki
May 6 '13 at 7:15
$begingroup$
Thanks joriki. I had a feeling it may be, but I wanted to see what you all thought.
$endgroup$
– Cody
May 6 '13 at 18:03
$begingroup$
Thanks joriki. I had a feeling it may be, but I wanted to see what you all thought.
$endgroup$
– Cody
May 6 '13 at 18:03
$begingroup$
John Benton, slight correction. The "good" negative multiples of pi are -3*pi, -7*pi, etc, i.e. whenever sin (x/2) is +1.
$endgroup$
– Oscar Lanzi
Feb 21 '16 at 13:41
$begingroup$
John Benton, slight correction. The "good" negative multiples of pi are -3*pi, -7*pi, etc, i.e. whenever sin (x/2) is +1.
$endgroup$
– Oscar Lanzi
Feb 21 '16 at 13:41
$begingroup$
It ought not to cause a great deal of trouble that the range of integration spans a value at which the expansion is not valid. It's only not valid by reason of the sine going to zero & therefore the logarithm going to ∞. functions that go to ∞ at the end of some region are quite routinely integrated over that region & found to have finite (though strictly speaking improper) integrals. If the ∞ is in the midst of the region, then it's just two such integrals juxtaposed. If you take the usual precautions for evaluating such integrals, then what you are doing is not extraordinary.
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 19:21
$begingroup$
It ought not to cause a great deal of trouble that the range of integration spans a value at which the expansion is not valid. It's only not valid by reason of the sine going to zero & therefore the logarithm going to ∞. functions that go to ∞ at the end of some region are quite routinely integrated over that region & found to have finite (though strictly speaking improper) integrals. If the ∞ is in the midst of the region, then it's just two such integrals juxtaposed. If you take the usual precautions for evaluating such integrals, then what you are doing is not extraordinary.
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 19:21
|
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$begingroup$
The problem with the identity $displaystyle -ln(2sin(x/2))=sum_{k=1}^{infty}frac{cos(kx)}{k}$ for $x=npi$ arises only when $(4m-2)pile nle 4mpi,$ where $min Bbb Z$. So the values $x=pmpi,pm 5pi, dots$ are OK. Note that $0$ is in the list of problematic values, too.
$endgroup$
– John Bentin
May 4 '13 at 22:32
$begingroup$
Looks like a fluke to me.
$endgroup$
– joriki
May 6 '13 at 7:15
$begingroup$
Thanks joriki. I had a feeling it may be, but I wanted to see what you all thought.
$endgroup$
– Cody
May 6 '13 at 18:03
$begingroup$
John Benton, slight correction. The "good" negative multiples of pi are -3*pi, -7*pi, etc, i.e. whenever sin (x/2) is +1.
$endgroup$
– Oscar Lanzi
Feb 21 '16 at 13:41
$begingroup$
It ought not to cause a great deal of trouble that the range of integration spans a value at which the expansion is not valid. It's only not valid by reason of the sine going to zero & therefore the logarithm going to ∞. functions that go to ∞ at the end of some region are quite routinely integrated over that region & found to have finite (though strictly speaking improper) integrals. If the ∞ is in the midst of the region, then it's just two such integrals juxtaposed. If you take the usual precautions for evaluating such integrals, then what you are doing is not extraordinary.
$endgroup$
– AmbretteOrrisey
Dec 7 '18 at 19:21