An integral inequality with cosine
I tried to prove that $$int_{a}^{b}frac{|cos (x)|}{x}dxleq frac{2}{pi}logleft(frac{b}{a}right)+O(1).$$ Of course $O(1)$ as a function of $b$, i.e. a bounded function of $b$. $a$ is considered constant with respect to $b$.
I found this inequality in a paper. There, it is written that we can be prove it by splitting the integral in integrals of length $2pi$ and by using the basic equality $$frac{1}{2pi}int_0^{2pi}|cos(x)|dx=frac{2}{pi}.$$ I tried to follow the aforementioned hint, but I got nowhere. I suppose that the term $log(b/a)$ pops out from the integral $int_{a}^{b}x^{-1}dx$, but I could not make this integral appear and obtain the desired inequality at the same time.
Do you have any hint? I suspect that it must be easy, but I am probably missing something now. I hope it is not a duplicate. Thanks in advance for your help!
inequality integral-inequality
asked Nov 28 '18 at 2:03
Sachpazis Stelios
765315
add a comment |
I tried to prove that $$int_{a}^{b}frac{|cos (x)|}{x}dxleq frac{2}{pi}logleft(frac{b}{a}right)+O(1).$$ Of course $O(1)$ as a function of $b$, i.e. a bounded function of $b$. $a$ is considered constant with respect to $b$.
I found this inequality in a paper. There, it is written that we can be prove it by splitting the integral in integrals of length $2pi$ and by using the basic equality $$frac{1}{2pi}int_0^{2pi}|cos(x)|dx=frac{2}{pi}.$$ I tried to follow the aforementioned hint, but I got nowhere. I suppose that the term $log(b/a)$ pops out from the integral $int_{a}^{b}x^{-1}dx$, but I could not make this integral appear and obtain the desired inequality at the same time.
Do you have any hint? I suspect that it must be easy, but I am probably missing something now. I hope it is not a duplicate. Thanks in advance for your help!
inequality integral-inequality
asked Nov 28 '18 at 2:03
Sachpazis Stelios
765315
I suspect you do a Taylor expansion for this problem.
– user23793
Nov 28 '18 at 2:09
@user23793 And what do you do after that actually?
– Sachpazis Stelios
Nov 28 '18 at 2:11
What are the conditions on $a$ and $b$? When you say "prove it by splitting the integral in integrals of length $2pi$", the length of $[a,b]$ may be less than $2pi$, in which case you may have to consider some cases. Also, could you clarify what you mean in the sentence I quoted?
– user23793
Nov 28 '18 at 2:34
@user23793 Usually the length of the interval is large, a lot greater than $2pi.$ There are no other conditions on $a$ and $b$. We just have to know that both sides are treated as functions of $b$, while $a$ is considered to be constant in terms of $a$. I also edited the question about this clarification.
– Sachpazis Stelios
Nov 28 '18 at 2:39
The reason I ask is that there might be a singularity at $0$ if $0in [a,b]$. It seems that there is some missing information.
– user23793
Nov 28 '18 at 2:44
add a comment |
I tried to prove that $$int_{a}^{b}frac{|cos (x)|}{x}dxleq frac{2}{pi}logleft(frac{b}{a}right)+O(1).$$ Of course $O(1)$ as a function of $b$, i.e. a bounded function of $b$. $a$ is considered constant with respect to $b$.
I found this inequality in a paper. There, it is written that we can be prove it by splitting the integral in integrals of length $2pi$ and by using the basic equality $$frac{1}{2pi}int_0^{2pi}|cos(x)|dx=frac{2}{pi}.$$ I tried to follow the aforementioned hint, but I got nowhere. I suppose that the term $log(b/a)$ pops out from the integral $int_{a}^{b}x^{-1}dx$, but I could not make this integral appear and obtain the desired inequality at the same time.
Do you have any hint? I suspect that it must be easy, but I am probably missing something now. I hope it is not a duplicate. Thanks in advance for your help!
inequality integral-inequality
asked Nov 28 '18 at 2:03
Sachpazis Stelios
765315
I tried to prove that $$int_{a}^{b}frac{|cos (x)|}{x}dxleq frac{2}{pi}logleft(frac{b}{a}right)+O(1).$$ Of course $O(1)$ as a function of $b$, i.e. a bounded function of $b$. $a$ is considered constant with respect to $b$.
I found this inequality in a paper. There, it is written that we can be prove it by splitting the integral in integrals of length $2pi$ and by using the basic equality $$frac{1}{2pi}int_0^{2pi}|cos(x)|dx=frac{2}{pi}.$$ I tried to follow the aforementioned hint, but I got nowhere. I suppose that the term $log(b/a)$ pops out from the integral $int_{a}^{b}x^{-1}dx$, but I could not make this integral appear and obtain the desired inequality at the same time.
Do you have any hint? I suspect that it must be easy, but I am probably missing something now. I hope it is not a duplicate. Thanks in advance for your help!
inequality integral-inequality
inequality integral-inequality
asked Nov 28 '18 at 2:03
Sachpazis Stelios
765315
asked Nov 28 '18 at 2:03
Sachpazis Stelios
765315
edited Nov 28 '18 at 2:40
asked Nov 28 '18 at 2:03
Sachpazis Stelios
765315
asked Nov 28 '18 at 2:03
Sachpazis Stelios
765315
asked Nov 28 '18 at 2:03
Sachpazis Stelios
765315
765315
I suspect you do a Taylor expansion for this problem.
– user23793
Nov 28 '18 at 2:09
@user23793 And what do you do after that actually?
– Sachpazis Stelios
Nov 28 '18 at 2:11
What are the conditions on $a$ and $b$? When you say "prove it by splitting the integral in integrals of length $2pi$", the length of $[a,b]$ may be less than $2pi$, in which case you may have to consider some cases. Also, could you clarify what you mean in the sentence I quoted?
– user23793
Nov 28 '18 at 2:34
@user23793 Usually the length of the interval is large, a lot greater than $2pi.$ There are no other conditions on $a$ and $b$. We just have to know that both sides are treated as functions of $b$, while $a$ is considered to be constant in terms of $a$. I also edited the question about this clarification.
– Sachpazis Stelios
Nov 28 '18 at 2:39
The reason I ask is that there might be a singularity at $0$ if $0in [a,b]$. It seems that there is some missing information.
– user23793
Nov 28 '18 at 2:44
add a comment |
I suspect you do a Taylor expansion for this problem.
– user23793
Nov 28 '18 at 2:09
@user23793 And what do you do after that actually?
– Sachpazis Stelios
Nov 28 '18 at 2:11
What are the conditions on $a$ and $b$? When you say "prove it by splitting the integral in integrals of length $2pi$", the length of $[a,b]$ may be less than $2pi$, in which case you may have to consider some cases. Also, could you clarify what you mean in the sentence I quoted?
– user23793
Nov 28 '18 at 2:34
@user23793 Usually the length of the interval is large, a lot greater than $2pi.$ There are no other conditions on $a$ and $b$. We just have to know that both sides are treated as functions of $b$, while $a$ is considered to be constant in terms of $a$. I also edited the question about this clarification.
– Sachpazis Stelios
Nov 28 '18 at 2:39
The reason I ask is that there might be a singularity at $0$ if $0in [a,b]$. It seems that there is some missing information.
– user23793
Nov 28 '18 at 2:44
I suspect you do a Taylor expansion for this problem.
– user23793
Nov 28 '18 at 2:09
I suspect you do a Taylor expansion for this problem.
– user23793
Nov 28 '18 at 2:09
@user23793 And what do you do after that actually?
– Sachpazis Stelios
Nov 28 '18 at 2:11
@user23793 And what do you do after that actually?
– Sachpazis Stelios
Nov 28 '18 at 2:11
What are the conditions on $a$ and $b$? When you say "prove it by splitting the integral in integrals of length $2pi$", the length of $[a,b]$ may be less than $2pi$, in which case you may have to consider some cases. Also, could you clarify what you mean in the sentence I quoted?
– user23793
Nov 28 '18 at 2:34
What are the conditions on $a$ and $b$? When you say "prove it by splitting the integral in integrals of length $2pi$", the length of $[a,b]$ may be less than $2pi$, in which case you may have to consider some cases. Also, could you clarify what you mean in the sentence I quoted?
– user23793
Nov 28 '18 at 2:34
@user23793 Usually the length of the interval is large, a lot greater than $2pi.$ There are no other conditions on $a$ and $b$. We just have to know that both sides are treated as functions of $b$, while $a$ is considered to be constant in terms of $a$. I also edited the question about this clarification.
– Sachpazis Stelios
Nov 28 '18 at 2:39
@user23793 Usually the length of the interval is large, a lot greater than $2pi.$ There are no other conditions on $a$ and $b$. We just have to know that both sides are treated as functions of $b$, while $a$ is considered to be constant in terms of $a$. I also edited the question about this clarification.
– Sachpazis Stelios
Nov 28 '18 at 2:39
The reason I ask is that there might be a singularity at $0$ if $0in [a,b]$. It seems that there is some missing information.
– user23793
Nov 28 '18 at 2:44
The reason I ask is that there might be a singularity at $0$ if $0in [a,b]$. It seems that there is some missing information.
– user23793
Nov 28 '18 at 2:44
add a comment |
1 Answer
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We assume that $ageq 4pi$ and $b>a$.
One way to prove the inequality is to partition the range $[a,b]$ into the sets $I_n = [ 2 pi n , 2pi (n+1)] $. In particular, we have that
$$int_{a}^{b}frac{|cos (x)|}{x}dx = sum_{n=n_0}^{n_1} int_{I_n cap [a,b]}
frac{|cos (x)|}{x}dx$$
with $n_0 = lfloor a/2pi rfloor$ and $n_1 = lceil b/2pi rceil$.
Now, we have that
$$int_{a}^{b}frac{|cos (x)|}{x}dx leq sum_{n=n_0}^{n_1} int_{I_n}
frac{|cos (x)|}{x}dx leq sum_{n=n_0}^{n_1} int_{I_n}
frac{|cos (x)|}{2pi n}dx = frac{2}{pi} sum_{n=n_0}^{n_1}frac{1}{2pi n} leq frac{2}{pi} int_{lfloor a/2pi rfloor -1}^{lceil b/2pi rceil} frac{dn}{2pi n} \= frac{2}{pi} logleft(frac{b}aright) + O(1),.$$
answered Nov 28 '18 at 5:11
Fabian
19.4k3674
Ahh, I knew that it had to be something simple. That time I was stuck. Anyway, thanks for the answer!
– Sachpazis Stelios
Nov 29 '18 at 1:32
Just a comment. You don't have the $2pi$ in the denominators after using the integral equality. However, the rest remains valid, because the $2pi$'s will cancel in the logarithm at the end.
– Sachpazis Stelios
Nov 29 '18 at 1:39
add a comment |
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1 Answer
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1 Answer
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votes
We assume that $ageq 4pi$ and $b>a$.
One way to prove the inequality is to partition the range $[a,b]$ into the sets $I_n = [ 2 pi n , 2pi (n+1)] $. In particular, we have that
$$int_{a}^{b}frac{|cos (x)|}{x}dx = sum_{n=n_0}^{n_1} int_{I_n cap [a,b]}
frac{|cos (x)|}{x}dx$$
with $n_0 = lfloor a/2pi rfloor$ and $n_1 = lceil b/2pi rceil$.
Now, we have that
$$int_{a}^{b}frac{|cos (x)|}{x}dx leq sum_{n=n_0}^{n_1} int_{I_n}
frac{|cos (x)|}{x}dx leq sum_{n=n_0}^{n_1} int_{I_n}
frac{|cos (x)|}{2pi n}dx = frac{2}{pi} sum_{n=n_0}^{n_1}frac{1}{2pi n} leq frac{2}{pi} int_{lfloor a/2pi rfloor -1}^{lceil b/2pi rceil} frac{dn}{2pi n} \= frac{2}{pi} logleft(frac{b}aright) + O(1),.$$
answered Nov 28 '18 at 5:11
Fabian
19.4k3674
Ahh, I knew that it had to be something simple. That time I was stuck. Anyway, thanks for the answer!
– Sachpazis Stelios
Nov 29 '18 at 1:32
Just a comment. You don't have the $2pi$ in the denominators after using the integral equality. However, the rest remains valid, because the $2pi$'s will cancel in the logarithm at the end.
– Sachpazis Stelios
Nov 29 '18 at 1:39
add a comment |
We assume that $ageq 4pi$ and $b>a$.
One way to prove the inequality is to partition the range $[a,b]$ into the sets $I_n = [ 2 pi n , 2pi (n+1)] $. In particular, we have that
$$int_{a}^{b}frac{|cos (x)|}{x}dx = sum_{n=n_0}^{n_1} int_{I_n cap [a,b]}
frac{|cos (x)|}{x}dx$$
with $n_0 = lfloor a/2pi rfloor$ and $n_1 = lceil b/2pi rceil$.
Now, we have that
$$int_{a}^{b}frac{|cos (x)|}{x}dx leq sum_{n=n_0}^{n_1} int_{I_n}
frac{|cos (x)|}{x}dx leq sum_{n=n_0}^{n_1} int_{I_n}
frac{|cos (x)|}{2pi n}dx = frac{2}{pi} sum_{n=n_0}^{n_1}frac{1}{2pi n} leq frac{2}{pi} int_{lfloor a/2pi rfloor -1}^{lceil b/2pi rceil} frac{dn}{2pi n} \= frac{2}{pi} logleft(frac{b}aright) + O(1),.$$
answered Nov 28 '18 at 5:11
Fabian
19.4k3674
Ahh, I knew that it had to be something simple. That time I was stuck. Anyway, thanks for the answer!
– Sachpazis Stelios
Nov 29 '18 at 1:32
Just a comment. You don't have the $2pi$ in the denominators after using the integral equality. However, the rest remains valid, because the $2pi$'s will cancel in the logarithm at the end.
– Sachpazis Stelios
Nov 29 '18 at 1:39
add a comment |
We assume that $ageq 4pi$ and $b>a$.
One way to prove the inequality is to partition the range $[a,b]$ into the sets $I_n = [ 2 pi n , 2pi (n+1)] $. In particular, we have that
$$int_{a}^{b}frac{|cos (x)|}{x}dx = sum_{n=n_0}^{n_1} int_{I_n cap [a,b]}
frac{|cos (x)|}{x}dx$$
with $n_0 = lfloor a/2pi rfloor$ and $n_1 = lceil b/2pi rceil$.
Now, we have that
$$int_{a}^{b}frac{|cos (x)|}{x}dx leq sum_{n=n_0}^{n_1} int_{I_n}
frac{|cos (x)|}{x}dx leq sum_{n=n_0}^{n_1} int_{I_n}
frac{|cos (x)|}{2pi n}dx = frac{2}{pi} sum_{n=n_0}^{n_1}frac{1}{2pi n} leq frac{2}{pi} int_{lfloor a/2pi rfloor -1}^{lceil b/2pi rceil} frac{dn}{2pi n} \= frac{2}{pi} logleft(frac{b}aright) + O(1),.$$
answered Nov 28 '18 at 5:11
Fabian
19.4k3674
We assume that $ageq 4pi$ and $b>a$.
One way to prove the inequality is to partition the range $[a,b]$ into the sets $I_n = [ 2 pi n , 2pi (n+1)] $. In particular, we have that
$$int_{a}^{b}frac{|cos (x)|}{x}dx = sum_{n=n_0}^{n_1} int_{I_n cap [a,b]}
frac{|cos (x)|}{x}dx$$
with $n_0 = lfloor a/2pi rfloor$ and $n_1 = lceil b/2pi rceil$.
Now, we have that
$$int_{a}^{b}frac{|cos (x)|}{x}dx leq sum_{n=n_0}^{n_1} int_{I_n}
frac{|cos (x)|}{x}dx leq sum_{n=n_0}^{n_1} int_{I_n}
frac{|cos (x)|}{2pi n}dx = frac{2}{pi} sum_{n=n_0}^{n_1}frac{1}{2pi n} leq frac{2}{pi} int_{lfloor a/2pi rfloor -1}^{lceil b/2pi rceil} frac{dn}{2pi n} \= frac{2}{pi} logleft(frac{b}aright) + O(1),.$$
answered Nov 28 '18 at 5:11
Fabian
19.4k3674
answered Nov 28 '18 at 5:11
Fabian
19.4k3674
answered Nov 28 '18 at 5:11
Fabian
19.4k3674
answered Nov 28 '18 at 5:11
Fabian
19.4k3674
19.4k3674
Ahh, I knew that it had to be something simple. That time I was stuck. Anyway, thanks for the answer!
– Sachpazis Stelios
Nov 29 '18 at 1:32
Just a comment. You don't have the $2pi$ in the denominators after using the integral equality. However, the rest remains valid, because the $2pi$'s will cancel in the logarithm at the end.
– Sachpazis Stelios
Nov 29 '18 at 1:39
add a comment |
Ahh, I knew that it had to be something simple. That time I was stuck. Anyway, thanks for the answer!
– Sachpazis Stelios
Nov 29 '18 at 1:32
Just a comment. You don't have the $2pi$ in the denominators after using the integral equality. However, the rest remains valid, because the $2pi$'s will cancel in the logarithm at the end.
– Sachpazis Stelios
Nov 29 '18 at 1:39
Ahh, I knew that it had to be something simple. That time I was stuck. Anyway, thanks for the answer!
– Sachpazis Stelios
Nov 29 '18 at 1:32
Ahh, I knew that it had to be something simple. That time I was stuck. Anyway, thanks for the answer!
– Sachpazis Stelios
Nov 29 '18 at 1:32
Just a comment. You don't have the $2pi$ in the denominators after using the integral equality. However, the rest remains valid, because the $2pi$'s will cancel in the logarithm at the end.
– Sachpazis Stelios
Nov 29 '18 at 1:39
Just a comment. You don't have the $2pi$ in the denominators after using the integral equality. However, the rest remains valid, because the $2pi$'s will cancel in the logarithm at the end.
– Sachpazis Stelios
Nov 29 '18 at 1:39
add a comment |
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I suspect you do a Taylor expansion for this problem.
– user23793
Nov 28 '18 at 2:09
@user23793 And what do you do after that actually?
– Sachpazis Stelios
Nov 28 '18 at 2:11
What are the conditions on $a$ and $b$? When you say "prove it by splitting the integral in integrals of length $2pi$", the length of $[a,b]$ may be less than $2pi$, in which case you may have to consider some cases. Also, could you clarify what you mean in the sentence I quoted?
– user23793
Nov 28 '18 at 2:34
@user23793 Usually the length of the interval is large, a lot greater than $2pi.$ There are no other conditions on $a$ and $b$. We just have to know that both sides are treated as functions of $b$, while $a$ is considered to be constant in terms of $a$. I also edited the question about this clarification.
– Sachpazis Stelios
Nov 28 '18 at 2:39
The reason I ask is that there might be a singularity at $0$ if $0in [a,b]$. It seems that there is some missing information.
– user23793
Nov 28 '18 at 2:44