Find the limit of $int_{0}^{pi }frac{sin nx}{nx} mathrm dx $.












1












$begingroup$


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







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$endgroup$












  • $begingroup$
    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


















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 '18 at 19:35







Normal

















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




















  • $begingroup$
    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












2 Answers
2






active

oldest

votes


















1












$begingroup$

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 $ntoinfty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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





















3












$begingroup$

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 $$lim _{nto infty}int_{0}^{pi }frac{sin nx}{nx} mathrm dx=0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $ntoinfty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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 $ntoinfty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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 $ntoinfty$.






share|cite|improve this answer









$endgroup$



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 $ntoinfty$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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













3












$begingroup$

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 $$lim _{nto infty}int_{0}^{pi }frac{sin nx}{nx} mathrm dx=0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















3












$begingroup$

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 $$lim _{nto infty}int_{0}^{pi }frac{sin nx}{nx} mathrm dx=0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















3












3








3





$begingroup$

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 $$lim _{nto infty}int_{0}^{pi }frac{sin nx}{nx} mathrm dx=0$$






share|cite|improve this answer









$endgroup$



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 $$lim _{nto infty}int_{0}^{pi }frac{sin nx}{nx} mathrm dx=0$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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