Find the limit of $int_{0}^{pi }frac{sin nx}{nx} mathrm dx $.
$begingroup$
If $a>0,$ show that $lim limits_{n to infty} int_{ a}^{pi} frac{sin nx}{nx} mathrm dx =0,$ what happens if $a=0.$
The first part is easy since the integrand uniformly converges to zero on $[a, pi]$ and for each $n, ; f_n(x)= frac {sin nx}{nx}$ is continuous on $[a,pi].$
But for $a=0$ I don't understand how to proceed. I was thinking of using bounded convergence theorem but that required uniform boundedness of the integrand .
Please give me a hint. Thanks in advance.
real-analysis
asked Nov 23 '18 at 19:28
NormalNormal
647
$endgroup$
add a comment |
$begingroup$
If $a>0,$ show that $lim limits_{n to infty} int_{ a}^{pi} frac{sin nx}{nx} mathrm dx =0,$ what happens if $a=0.$
The first part is easy since the integrand uniformly converges to zero on $[a, pi]$ and for each $n, ; f_n(x)= frac {sin nx}{nx}$ is continuous on $[a,pi].$
But for $a=0$ I don't understand how to proceed. I was thinking of using bounded convergence theorem but that required uniform boundedness of the integrand .
Please give me a hint. Thanks in advance.
real-analysis
asked Nov 23 '18 at 19:28
NormalNormal
647
$endgroup$
$begingroup$
is this what you aimed to write? Please check.
$endgroup$
– user376343
Nov 23 '18 at 19:34
1
$begingroup$
Where is $a$ in the integral?
$endgroup$
– Will M.
Nov 23 '18 at 19:34
$begingroup$
No, sorry, I fixed it
$endgroup$
– Normal
Nov 23 '18 at 19:38
$begingroup$
The integrand is continuous and bounded, so what happens when you integrate on $[0,a]$ with $ato0$?
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 19:45
$begingroup$
The integrand is continuous on [0,pi] considering $ f_n(0) $=1 and hence it is bounded on [0,pi]. Sorry I am a beginner in real analysis ..I could not understand what you said .
$endgroup$
– Normal
Nov 23 '18 at 20:17
add a comment |
$begingroup$
If $a>0,$ show that $lim limits_{n to infty} int_{ a}^{pi} frac{sin nx}{nx} mathrm dx =0,$ what happens if $a=0.$
The first part is easy since the integrand uniformly converges to zero on $[a, pi]$ and for each $n, ; f_n(x)= frac {sin nx}{nx}$ is continuous on $[a,pi].$
But for $a=0$ I don't understand how to proceed. I was thinking of using bounded convergence theorem but that required uniform boundedness of the integrand .
Please give me a hint. Thanks in advance.
real-analysis
asked Nov 23 '18 at 19:28
NormalNormal
647
$endgroup$
If $a>0,$ show that $lim limits_{n to infty} int_{ a}^{pi} frac{sin nx}{nx} mathrm dx =0,$ what happens if $a=0.$
The first part is easy since the integrand uniformly converges to zero on $[a, pi]$ and for each $n, ; f_n(x)= frac {sin nx}{nx}$ is continuous on $[a,pi].$
But for $a=0$ I don't understand how to proceed. I was thinking of using bounded convergence theorem but that required uniform boundedness of the integrand .
Please give me a hint. Thanks in advance.
real-analysis
real-analysis
asked Nov 23 '18 at 19:28
NormalNormal
647
asked Nov 23 '18 at 19:28
NormalNormal
647
edited Nov 23 '18 at 19:35
Normal
asked Nov 23 '18 at 19:28
NormalNormal
647
asked Nov 23 '18 at 19:28
NormalNormal
647
asked Nov 23 '18 at 19:28
NormalNormal
647
647
$begingroup$
is this what you aimed to write? Please check.
$endgroup$
– user376343
Nov 23 '18 at 19:34
1
$begingroup$
Where is $a$ in the integral?
$endgroup$
– Will M.
Nov 23 '18 at 19:34
$begingroup$
No, sorry, I fixed it
$endgroup$
– Normal
Nov 23 '18 at 19:38
$begingroup$
The integrand is continuous and bounded, so what happens when you integrate on $[0,a]$ with $ato0$?
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 19:45
$begingroup$
The integrand is continuous on [0,pi] considering $ f_n(0) $=1 and hence it is bounded on [0,pi]. Sorry I am a beginner in real analysis ..I could not understand what you said .
$endgroup$
– Normal
Nov 23 '18 at 20:17
add a comment |
$begingroup$
is this what you aimed to write? Please check.
$endgroup$
– user376343
Nov 23 '18 at 19:34
1
$begingroup$
Where is $a$ in the integral?
$endgroup$
– Will M.
Nov 23 '18 at 19:34
$begingroup$
No, sorry, I fixed it
$endgroup$
– Normal
Nov 23 '18 at 19:38
$begingroup$
The integrand is continuous and bounded, so what happens when you integrate on $[0,a]$ with $ato0$?
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 19:45
$begingroup$
The integrand is continuous on [0,pi] considering $ f_n(0) $=1 and hence it is bounded on [0,pi]. Sorry I am a beginner in real analysis ..I could not understand what you said .
$endgroup$
– Normal
Nov 23 '18 at 20:17
$begingroup$
is this what you aimed to write? Please check.
$endgroup$
– user376343
Nov 23 '18 at 19:34
$begingroup$
is this what you aimed to write? Please check.
$endgroup$
– user376343
Nov 23 '18 at 19:34
1
1
$begingroup$
Where is $a$ in the integral?
$endgroup$
– Will M.
Nov 23 '18 at 19:34
$begingroup$
Where is $a$ in the integral?
$endgroup$
– Will M.
Nov 23 '18 at 19:34
$begingroup$
No, sorry, I fixed it
$endgroup$
– Normal
Nov 23 '18 at 19:38
$begingroup$
No, sorry, I fixed it
$endgroup$
– Normal
Nov 23 '18 at 19:38
$begingroup$
The integrand is continuous and bounded, so what happens when you integrate on $[0,a]$ with $ato0$?
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 19:45
$begingroup$
The integrand is continuous and bounded, so what happens when you integrate on $[0,a]$ with $ato0$?
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 19:45
$begingroup$
The integrand is continuous on [0,pi] considering $ f_n(0) $=1 and hence it is bounded on [0,pi]. Sorry I am a beginner in real analysis ..I could not understand what you said .
$endgroup$
– Normal
Nov 23 '18 at 20:17
$begingroup$
The integrand is continuous on [0,pi] considering $ f_n(0) $=1 and hence it is bounded on [0,pi]. Sorry I am a beginner in real analysis ..I could not understand what you said .
$endgroup$
– Normal
Nov 23 '18 at 20:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For all $x$, $|f_n(x)|le1$, hence
$$left|int_0^{a}f_n(x)dxright|le (a-0) sup_{[0,a]}|f_n|le a$$
Therefore
$$left|int_0^pi f_n(x)dxright|=left|int_0^a f_n(x)dx+int_a^pi f_n(x)dxright|le a+left|int_a^pi f_n(x)dxright|$$
Now, for all $epsilon>0$, let $a=epsilon$, and the remaining integral term tends to $0$ as $nto0$ (as $a$ is fixed as soon as $epsilon$ is fixed), so there is a $N$ such that for all $n>N$, $left|int_a^pi f_n(x)dxright|<epsilon$.
Hence, for all $epsilon>0$, there is a $N$ such that for all $n>N$, $left|int_0^pi f_n(x)dxright|<2epsilon$.
Hence $int_0^pi f_n(x)dxto0$ as $ntoinfty$.
answered Nov 23 '18 at 21:15
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
$endgroup$
$begingroup$
Thank you ..I have a question....first the given sequences of function are not defined at 0 but since the limit exists at zero and the limiting value is 1 , I can define $ f_n(0) $=1, so did you write $ | f_n(x)| le 1$ with the same thought I have or something else ..If there is an other explanation please tell me ....so then $ f_n(x) $ is uniformly bounded and I can use bounded convergence theorem to conclude that the limiting value is zero..can I do that ?
$endgroup$
– Normal
Nov 23 '18 at 22:19
$begingroup$
@Normal Yes, you can define $f_n(0)=1$ because the limit exists. It's a removable singularity, and the function thus defined is $C^{infty}$. See also en.wikipedia.org/wiki/Sinc_function
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:21
$begingroup$
Thank you... please tell me it will be ok if I use bounded convergence theorem by using the uniform boundedness of $ f_n(x) $
$endgroup$
– Normal
Nov 23 '18 at 22:38
$begingroup$
@Normal Yes, it also works. But you're not really a beginner then :)
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:41
$begingroup$
One more question what if I choose an $ epsilon $ different from a .
$endgroup$
– Normal
Nov 23 '18 at 23:05
|
show 2 more comments
$begingroup$
We have that $$int_{0}^{pi }frac{sin nx}{nx} mathrm dx=int_{0}^{npi }frac{sin nx}{nx} mathrm {1over n}dnx={1over n }int_{0}^{npi} {sin xover x}dx$$also we know that $$int_0^{infty}{sin xover x}dx={pi over 2}$$according to Integration of Sinc function therefore $$lim _{nto infty}int_{0}^{pi }frac{sin nx}{nx} mathrm dx=0$$
answered Nov 23 '18 at 19:43
Mostafa AyazMostafa Ayaz
14.7k3938
$endgroup$
$begingroup$
Thank you .So $ lim_{nto infty} int_{0}^{pi} frac{sin nx}{nx} $=$( lim_{nto infty} {1 over n})(lim_{nto infty} int_{0}^{ npi}{sin x over x}) dx $ =0×(pi/2)=0. Is it what you did?
$endgroup$
– Normal
Nov 23 '18 at 20:05
$begingroup$
You're welcome. Also could be explained that way. Yes...
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 20:10
$begingroup$
Sorry for disturbing you again ,I have a doubt am I allowed to write what I wrote above..I mean I know lim ($ x_n y_n $ ) = lim ($ x_n $)lim ( $ y_n $) when the both limit exists ..but am I allowed to do the same thing in here.
$endgroup$
– Normal
Nov 23 '18 at 21:18
$begingroup$
No mind at all :) Notice that i concluded that $$0le lim int_{0}^{pi }frac{sin nx}{nx} mathrm dx= lim {1over n }int_{0}^{npi} {sin xover x}dxlelim {1over n }int_{0}^{infty} {sin xover x}dx=lim {pi over 2n}=0$$ and used squeeze theorem
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 21:21
$begingroup$
Thanks for answering though I didn't understand the line where you wrote lim $ {1 over n} int_{0}^{npi} { sin x over x} dx le lim {1 over n} int_{0}^{infty} {sin x over x} dx $ but it's alright the function is uniformly bounded defining $ f_n(0) =1 $ so by bounded convergence theorem I can conclude the answer is zero .
$endgroup$
– Normal
Nov 23 '18 at 22:31
|
show 1 more comment
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
For all $x$, $|f_n(x)|le1$, hence
$$left|int_0^{a}f_n(x)dxright|le (a-0) sup_{[0,a]}|f_n|le a$$
Therefore
$$left|int_0^pi f_n(x)dxright|=left|int_0^a f_n(x)dx+int_a^pi f_n(x)dxright|le a+left|int_a^pi f_n(x)dxright|$$
Now, for all $epsilon>0$, let $a=epsilon$, and the remaining integral term tends to $0$ as $nto0$ (as $a$ is fixed as soon as $epsilon$ is fixed), so there is a $N$ such that for all $n>N$, $left|int_a^pi f_n(x)dxright|<epsilon$.
Hence, for all $epsilon>0$, there is a $N$ such that for all $n>N$, $left|int_0^pi f_n(x)dxright|<2epsilon$.
Hence $int_0^pi f_n(x)dxto0$ as $ntoinfty$.
answered Nov 23 '18 at 21:15
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
$endgroup$
$begingroup$
Thank you ..I have a question....first the given sequences of function are not defined at 0 but since the limit exists at zero and the limiting value is 1 , I can define $ f_n(0) $=1, so did you write $ | f_n(x)| le 1$ with the same thought I have or something else ..If there is an other explanation please tell me ....so then $ f_n(x) $ is uniformly bounded and I can use bounded convergence theorem to conclude that the limiting value is zero..can I do that ?
$endgroup$
– Normal
Nov 23 '18 at 22:19
$begingroup$
@Normal Yes, you can define $f_n(0)=1$ because the limit exists. It's a removable singularity, and the function thus defined is $C^{infty}$. See also en.wikipedia.org/wiki/Sinc_function
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:21
$begingroup$
Thank you... please tell me it will be ok if I use bounded convergence theorem by using the uniform boundedness of $ f_n(x) $
$endgroup$
– Normal
Nov 23 '18 at 22:38
$begingroup$
@Normal Yes, it also works. But you're not really a beginner then :)
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:41
$begingroup$
One more question what if I choose an $ epsilon $ different from a .
$endgroup$
– Normal
Nov 23 '18 at 23:05
|
show 2 more comments
$begingroup$
For all $x$, $|f_n(x)|le1$, hence
$$left|int_0^{a}f_n(x)dxright|le (a-0) sup_{[0,a]}|f_n|le a$$
Therefore
$$left|int_0^pi f_n(x)dxright|=left|int_0^a f_n(x)dx+int_a^pi f_n(x)dxright|le a+left|int_a^pi f_n(x)dxright|$$
Now, for all $epsilon>0$, let $a=epsilon$, and the remaining integral term tends to $0$ as $nto0$ (as $a$ is fixed as soon as $epsilon$ is fixed), so there is a $N$ such that for all $n>N$, $left|int_a^pi f_n(x)dxright|<epsilon$.
Hence, for all $epsilon>0$, there is a $N$ such that for all $n>N$, $left|int_0^pi f_n(x)dxright|<2epsilon$.
Hence $int_0^pi f_n(x)dxto0$ as $ntoinfty$.
answered Nov 23 '18 at 21:15
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
$endgroup$
$begingroup$
Thank you ..I have a question....first the given sequences of function are not defined at 0 but since the limit exists at zero and the limiting value is 1 , I can define $ f_n(0) $=1, so did you write $ | f_n(x)| le 1$ with the same thought I have or something else ..If there is an other explanation please tell me ....so then $ f_n(x) $ is uniformly bounded and I can use bounded convergence theorem to conclude that the limiting value is zero..can I do that ?
$endgroup$
– Normal
Nov 23 '18 at 22:19
$begingroup$
@Normal Yes, you can define $f_n(0)=1$ because the limit exists. It's a removable singularity, and the function thus defined is $C^{infty}$. See also en.wikipedia.org/wiki/Sinc_function
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:21
$begingroup$
Thank you... please tell me it will be ok if I use bounded convergence theorem by using the uniform boundedness of $ f_n(x) $
$endgroup$
– Normal
Nov 23 '18 at 22:38
$begingroup$
@Normal Yes, it also works. But you're not really a beginner then :)
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:41
$begingroup$
One more question what if I choose an $ epsilon $ different from a .
$endgroup$
– Normal
Nov 23 '18 at 23:05
|
show 2 more comments
$begingroup$
For all $x$, $|f_n(x)|le1$, hence
$$left|int_0^{a}f_n(x)dxright|le (a-0) sup_{[0,a]}|f_n|le a$$
Therefore
$$left|int_0^pi f_n(x)dxright|=left|int_0^a f_n(x)dx+int_a^pi f_n(x)dxright|le a+left|int_a^pi f_n(x)dxright|$$
Now, for all $epsilon>0$, let $a=epsilon$, and the remaining integral term tends to $0$ as $nto0$ (as $a$ is fixed as soon as $epsilon$ is fixed), so there is a $N$ such that for all $n>N$, $left|int_a^pi f_n(x)dxright|<epsilon$.
Hence, for all $epsilon>0$, there is a $N$ such that for all $n>N$, $left|int_0^pi f_n(x)dxright|<2epsilon$.
Hence $int_0^pi f_n(x)dxto0$ as $ntoinfty$.
answered Nov 23 '18 at 21:15
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
$endgroup$
For all $x$, $|f_n(x)|le1$, hence
$$left|int_0^{a}f_n(x)dxright|le (a-0) sup_{[0,a]}|f_n|le a$$
Therefore
$$left|int_0^pi f_n(x)dxright|=left|int_0^a f_n(x)dx+int_a^pi f_n(x)dxright|le a+left|int_a^pi f_n(x)dxright|$$
Now, for all $epsilon>0$, let $a=epsilon$, and the remaining integral term tends to $0$ as $nto0$ (as $a$ is fixed as soon as $epsilon$ is fixed), so there is a $N$ such that for all $n>N$, $left|int_a^pi f_n(x)dxright|<epsilon$.
Hence, for all $epsilon>0$, there is a $N$ such that for all $n>N$, $left|int_0^pi f_n(x)dxright|<2epsilon$.
Hence $int_0^pi f_n(x)dxto0$ as $ntoinfty$.
answered Nov 23 '18 at 21:15
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
answered Nov 23 '18 at 21:15
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
answered Nov 23 '18 at 21:15
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
answered Nov 23 '18 at 21:15
Jean-Claude ArbautJean-Claude Arbaut
14.7k63464
14.7k63464
$begingroup$
Thank you ..I have a question....first the given sequences of function are not defined at 0 but since the limit exists at zero and the limiting value is 1 , I can define $ f_n(0) $=1, so did you write $ | f_n(x)| le 1$ with the same thought I have or something else ..If there is an other explanation please tell me ....so then $ f_n(x) $ is uniformly bounded and I can use bounded convergence theorem to conclude that the limiting value is zero..can I do that ?
$endgroup$
– Normal
Nov 23 '18 at 22:19
$begingroup$
@Normal Yes, you can define $f_n(0)=1$ because the limit exists. It's a removable singularity, and the function thus defined is $C^{infty}$. See also en.wikipedia.org/wiki/Sinc_function
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:21
$begingroup$
Thank you... please tell me it will be ok if I use bounded convergence theorem by using the uniform boundedness of $ f_n(x) $
$endgroup$
– Normal
Nov 23 '18 at 22:38
$begingroup$
@Normal Yes, it also works. But you're not really a beginner then :)
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:41
$begingroup$
One more question what if I choose an $ epsilon $ different from a .
$endgroup$
– Normal
Nov 23 '18 at 23:05
|
show 2 more comments
$begingroup$
Thank you ..I have a question....first the given sequences of function are not defined at 0 but since the limit exists at zero and the limiting value is 1 , I can define $ f_n(0) $=1, so did you write $ | f_n(x)| le 1$ with the same thought I have or something else ..If there is an other explanation please tell me ....so then $ f_n(x) $ is uniformly bounded and I can use bounded convergence theorem to conclude that the limiting value is zero..can I do that ?
$endgroup$
– Normal
Nov 23 '18 at 22:19
$begingroup$
@Normal Yes, you can define $f_n(0)=1$ because the limit exists. It's a removable singularity, and the function thus defined is $C^{infty}$. See also en.wikipedia.org/wiki/Sinc_function
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:21
$begingroup$
Thank you... please tell me it will be ok if I use bounded convergence theorem by using the uniform boundedness of $ f_n(x) $
$endgroup$
– Normal
Nov 23 '18 at 22:38
$begingroup$
@Normal Yes, it also works. But you're not really a beginner then :)
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:41
$begingroup$
One more question what if I choose an $ epsilon $ different from a .
$endgroup$
– Normal
Nov 23 '18 at 23:05
$begingroup$
Thank you ..I have a question....first the given sequences of function are not defined at 0 but since the limit exists at zero and the limiting value is 1 , I can define $ f_n(0) $=1, so did you write $ | f_n(x)| le 1$ with the same thought I have or something else ..If there is an other explanation please tell me ....so then $ f_n(x) $ is uniformly bounded and I can use bounded convergence theorem to conclude that the limiting value is zero..can I do that ?
$endgroup$
– Normal
Nov 23 '18 at 22:19
$begingroup$
Thank you ..I have a question....first the given sequences of function are not defined at 0 but since the limit exists at zero and the limiting value is 1 , I can define $ f_n(0) $=1, so did you write $ | f_n(x)| le 1$ with the same thought I have or something else ..If there is an other explanation please tell me ....so then $ f_n(x) $ is uniformly bounded and I can use bounded convergence theorem to conclude that the limiting value is zero..can I do that ?
$endgroup$
– Normal
Nov 23 '18 at 22:19
$begingroup$
@Normal Yes, you can define $f_n(0)=1$ because the limit exists. It's a removable singularity, and the function thus defined is $C^{infty}$. See also en.wikipedia.org/wiki/Sinc_function
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:21
$begingroup$
@Normal Yes, you can define $f_n(0)=1$ because the limit exists. It's a removable singularity, and the function thus defined is $C^{infty}$. See also en.wikipedia.org/wiki/Sinc_function
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:21
$begingroup$
Thank you... please tell me it will be ok if I use bounded convergence theorem by using the uniform boundedness of $ f_n(x) $
$endgroup$
– Normal
Nov 23 '18 at 22:38
$begingroup$
Thank you... please tell me it will be ok if I use bounded convergence theorem by using the uniform boundedness of $ f_n(x) $
$endgroup$
– Normal
Nov 23 '18 at 22:38
$begingroup$
@Normal Yes, it also works. But you're not really a beginner then :)
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:41
$begingroup$
@Normal Yes, it also works. But you're not really a beginner then :)
$endgroup$
– Jean-Claude Arbaut
Nov 23 '18 at 22:41
$begingroup$
One more question what if I choose an $ epsilon $ different from a .
$endgroup$
– Normal
Nov 23 '18 at 23:05
$begingroup$
One more question what if I choose an $ epsilon $ different from a .
$endgroup$
– Normal
Nov 23 '18 at 23:05
|
show 2 more comments
$begingroup$
We have that $$int_{0}^{pi }frac{sin nx}{nx} mathrm dx=int_{0}^{npi }frac{sin nx}{nx} mathrm {1over n}dnx={1over n }int_{0}^{npi} {sin xover x}dx$$also we know that $$int_0^{infty}{sin xover x}dx={pi over 2}$$according to Integration of Sinc function therefore $$lim _{nto infty}int_{0}^{pi }frac{sin nx}{nx} mathrm dx=0$$
answered Nov 23 '18 at 19:43
Mostafa AyazMostafa Ayaz
14.7k3938
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Thank you .So $ lim_{nto infty} int_{0}^{pi} frac{sin nx}{nx} $=$( lim_{nto infty} {1 over n})(lim_{nto infty} int_{0}^{ npi}{sin x over x}) dx $ =0×(pi/2)=0. Is it what you did?
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– Normal
Nov 23 '18 at 20:05
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You're welcome. Also could be explained that way. Yes...
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– Mostafa Ayaz
Nov 23 '18 at 20:10
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Sorry for disturbing you again ,I have a doubt am I allowed to write what I wrote above..I mean I know lim ($ x_n y_n $ ) = lim ($ x_n $)lim ( $ y_n $) when the both limit exists ..but am I allowed to do the same thing in here.
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– Normal
Nov 23 '18 at 21:18
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No mind at all :) Notice that i concluded that $$0le lim int_{0}^{pi }frac{sin nx}{nx} mathrm dx= lim {1over n }int_{0}^{npi} {sin xover x}dxlelim {1over n }int_{0}^{infty} {sin xover x}dx=lim {pi over 2n}=0$$ and used squeeze theorem
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– Mostafa Ayaz
Nov 23 '18 at 21:21
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Thanks for answering though I didn't understand the line where you wrote lim $ {1 over n} int_{0}^{npi} { sin x over x} dx le lim {1 over n} int_{0}^{infty} {sin x over x} dx $ but it's alright the function is uniformly bounded defining $ f_n(0) =1 $ so by bounded convergence theorem I can conclude the answer is zero .
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– Normal
Nov 23 '18 at 22:31
|
show 1 more comment
$begingroup$
We have that $$int_{0}^{pi }frac{sin nx}{nx} mathrm dx=int_{0}^{npi }frac{sin nx}{nx} mathrm {1over n}dnx={1over n }int_{0}^{npi} {sin xover x}dx$$also we know that $$int_0^{infty}{sin xover x}dx={pi over 2}$$according to Integration of Sinc function therefore $$lim _{nto infty}int_{0}^{pi }frac{sin nx}{nx} mathrm dx=0$$
answered Nov 23 '18 at 19:43
Mostafa AyazMostafa Ayaz
14.7k3938
$endgroup$
$begingroup$
Thank you .So $ lim_{nto infty} int_{0}^{pi} frac{sin nx}{nx} $=$( lim_{nto infty} {1 over n})(lim_{nto infty} int_{0}^{ npi}{sin x over x}) dx $ =0×(pi/2)=0. Is it what you did?
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– Normal
Nov 23 '18 at 20:05
$begingroup$
You're welcome. Also could be explained that way. Yes...
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 20:10
$begingroup$
Sorry for disturbing you again ,I have a doubt am I allowed to write what I wrote above..I mean I know lim ($ x_n y_n $ ) = lim ($ x_n $)lim ( $ y_n $) when the both limit exists ..but am I allowed to do the same thing in here.
$endgroup$
– Normal
Nov 23 '18 at 21:18
$begingroup$
No mind at all :) Notice that i concluded that $$0le lim int_{0}^{pi }frac{sin nx}{nx} mathrm dx= lim {1over n }int_{0}^{npi} {sin xover x}dxlelim {1over n }int_{0}^{infty} {sin xover x}dx=lim {pi over 2n}=0$$ and used squeeze theorem
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– Mostafa Ayaz
Nov 23 '18 at 21:21
$begingroup$
Thanks for answering though I didn't understand the line where you wrote lim $ {1 over n} int_{0}^{npi} { sin x over x} dx le lim {1 over n} int_{0}^{infty} {sin x over x} dx $ but it's alright the function is uniformly bounded defining $ f_n(0) =1 $ so by bounded convergence theorem I can conclude the answer is zero .
$endgroup$
– Normal
Nov 23 '18 at 22:31
|
show 1 more comment
$begingroup$
We have that $$int_{0}^{pi }frac{sin nx}{nx} mathrm dx=int_{0}^{npi }frac{sin nx}{nx} mathrm {1over n}dnx={1over n }int_{0}^{npi} {sin xover x}dx$$also we know that $$int_0^{infty}{sin xover x}dx={pi over 2}$$according to Integration of Sinc function therefore $$lim _{nto infty}int_{0}^{pi }frac{sin nx}{nx} mathrm dx=0$$
answered Nov 23 '18 at 19:43
Mostafa AyazMostafa Ayaz
14.7k3938
$endgroup$
We have that $$int_{0}^{pi }frac{sin nx}{nx} mathrm dx=int_{0}^{npi }frac{sin nx}{nx} mathrm {1over n}dnx={1over n }int_{0}^{npi} {sin xover x}dx$$also we know that $$int_0^{infty}{sin xover x}dx={pi over 2}$$according to Integration of Sinc function therefore $$lim _{nto infty}int_{0}^{pi }frac{sin nx}{nx} mathrm dx=0$$
answered Nov 23 '18 at 19:43
Mostafa AyazMostafa Ayaz
14.7k3938
answered Nov 23 '18 at 19:43
Mostafa AyazMostafa Ayaz
14.7k3938
answered Nov 23 '18 at 19:43
Mostafa AyazMostafa Ayaz
14.7k3938
answered Nov 23 '18 at 19:43
Mostafa AyazMostafa Ayaz
14.7k3938
14.7k3938
$begingroup$
Thank you .So $ lim_{nto infty} int_{0}^{pi} frac{sin nx}{nx} $=$( lim_{nto infty} {1 over n})(lim_{nto infty} int_{0}^{ npi}{sin x over x}) dx $ =0×(pi/2)=0. Is it what you did?
$endgroup$
– Normal
Nov 23 '18 at 20:05
$begingroup$
You're welcome. Also could be explained that way. Yes...
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 20:10
$begingroup$
Sorry for disturbing you again ,I have a doubt am I allowed to write what I wrote above..I mean I know lim ($ x_n y_n $ ) = lim ($ x_n $)lim ( $ y_n $) when the both limit exists ..but am I allowed to do the same thing in here.
$endgroup$
– Normal
Nov 23 '18 at 21:18
$begingroup$
No mind at all :) Notice that i concluded that $$0le lim int_{0}^{pi }frac{sin nx}{nx} mathrm dx= lim {1over n }int_{0}^{npi} {sin xover x}dxlelim {1over n }int_{0}^{infty} {sin xover x}dx=lim {pi over 2n}=0$$ and used squeeze theorem
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 21:21
$begingroup$
Thanks for answering though I didn't understand the line where you wrote lim $ {1 over n} int_{0}^{npi} { sin x over x} dx le lim {1 over n} int_{0}^{infty} {sin x over x} dx $ but it's alright the function is uniformly bounded defining $ f_n(0) =1 $ so by bounded convergence theorem I can conclude the answer is zero .
$endgroup$
– Normal
Nov 23 '18 at 22:31
|
show 1 more comment
$begingroup$
Thank you .So $ lim_{nto infty} int_{0}^{pi} frac{sin nx}{nx} $=$( lim_{nto infty} {1 over n})(lim_{nto infty} int_{0}^{ npi}{sin x over x}) dx $ =0×(pi/2)=0. Is it what you did?
$endgroup$
– Normal
Nov 23 '18 at 20:05
$begingroup$
You're welcome. Also could be explained that way. Yes...
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 20:10
$begingroup$
Sorry for disturbing you again ,I have a doubt am I allowed to write what I wrote above..I mean I know lim ($ x_n y_n $ ) = lim ($ x_n $)lim ( $ y_n $) when the both limit exists ..but am I allowed to do the same thing in here.
$endgroup$
– Normal
Nov 23 '18 at 21:18
$begingroup$
No mind at all :) Notice that i concluded that $$0le lim int_{0}^{pi }frac{sin nx}{nx} mathrm dx= lim {1over n }int_{0}^{npi} {sin xover x}dxlelim {1over n }int_{0}^{infty} {sin xover x}dx=lim {pi over 2n}=0$$ and used squeeze theorem
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 21:21
$begingroup$
Thanks for answering though I didn't understand the line where you wrote lim $ {1 over n} int_{0}^{npi} { sin x over x} dx le lim {1 over n} int_{0}^{infty} {sin x over x} dx $ but it's alright the function is uniformly bounded defining $ f_n(0) =1 $ so by bounded convergence theorem I can conclude the answer is zero .
$endgroup$
– Normal
Nov 23 '18 at 22:31
$begingroup$
Thank you .So $ lim_{nto infty} int_{0}^{pi} frac{sin nx}{nx} $=$( lim_{nto infty} {1 over n})(lim_{nto infty} int_{0}^{ npi}{sin x over x}) dx $ =0×(pi/2)=0. Is it what you did?
$endgroup$
– Normal
Nov 23 '18 at 20:05
$begingroup$
Thank you .So $ lim_{nto infty} int_{0}^{pi} frac{sin nx}{nx} $=$( lim_{nto infty} {1 over n})(lim_{nto infty} int_{0}^{ npi}{sin x over x}) dx $ =0×(pi/2)=0. Is it what you did?
$endgroup$
– Normal
Nov 23 '18 at 20:05
$begingroup$
You're welcome. Also could be explained that way. Yes...
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 20:10
$begingroup$
You're welcome. Also could be explained that way. Yes...
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 20:10
$begingroup$
Sorry for disturbing you again ,I have a doubt am I allowed to write what I wrote above..I mean I know lim ($ x_n y_n $ ) = lim ($ x_n $)lim ( $ y_n $) when the both limit exists ..but am I allowed to do the same thing in here.
$endgroup$
– Normal
Nov 23 '18 at 21:18
$begingroup$
Sorry for disturbing you again ,I have a doubt am I allowed to write what I wrote above..I mean I know lim ($ x_n y_n $ ) = lim ($ x_n $)lim ( $ y_n $) when the both limit exists ..but am I allowed to do the same thing in here.
$endgroup$
– Normal
Nov 23 '18 at 21:18
$begingroup$
No mind at all :) Notice that i concluded that $$0le lim int_{0}^{pi }frac{sin nx}{nx} mathrm dx= lim {1over n }int_{0}^{npi} {sin xover x}dxlelim {1over n }int_{0}^{infty} {sin xover x}dx=lim {pi over 2n}=0$$ and used squeeze theorem
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 21:21
$begingroup$
No mind at all :) Notice that i concluded that $$0le lim int_{0}^{pi }frac{sin nx}{nx} mathrm dx= lim {1over n }int_{0}^{npi} {sin xover x}dxlelim {1over n }int_{0}^{infty} {sin xover x}dx=lim {pi over 2n}=0$$ and used squeeze theorem
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 21:21
$begingroup$
Thanks for answering though I didn't understand the line where you wrote lim $ {1 over n} int_{0}^{npi} { sin x over x} dx le lim {1 over n} int_{0}^{infty} {sin x over x} dx $ but it's alright the function is uniformly bounded defining $ f_n(0) =1 $ so by bounded convergence theorem I can conclude the answer is zero .
$endgroup$
– Normal
Nov 23 '18 at 22:31
$begingroup$
Thanks for answering though I didn't understand the line where you wrote lim $ {1 over n} int_{0}^{npi} { sin x over x} dx le lim {1 over n} int_{0}^{infty} {sin x over x} dx $ but it's alright the function is uniformly bounded defining $ f_n(0) =1 $ so by bounded convergence theorem I can conclude the answer is zero .
$endgroup$
– Normal
Nov 23 '18 at 22:31
|
show 1 more comment
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is this what you aimed to write? Please check.
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– user376343
Nov 23 '18 at 19:34
1
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Where is $a$ in the integral?
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– Will M.
Nov 23 '18 at 19:34
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No, sorry, I fixed it
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– Normal
Nov 23 '18 at 19:38
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The integrand is continuous and bounded, so what happens when you integrate on $[0,a]$ with $ato0$?
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– Jean-Claude Arbaut
Nov 23 '18 at 19:45
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The integrand is continuous on [0,pi] considering $ f_n(0) $=1 and hence it is bounded on [0,pi]. Sorry I am a beginner in real analysis ..I could not understand what you said .
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– Normal
Nov 23 '18 at 20:17