Confusion with integrating sin(nx)sin(mx) and Kroenecker delta












3












$begingroup$


The specific integral I'm working with is the following:
$$ int_0^asin(npi y/a)sin(n'pi y/a) $$
This is supposed to come out to $0$ in the case that $ n neq n' $ and $frac{1}{2}a$ in the case that $n= n'$. I can obtain this result sometimes, but the method I'm using currently is giving me a value of $0$ all the time. I'm applying a product-to-sum formula and then integrating.
Only resources I've managed to find on this equate the following expression (which is my final result before applying limits of integration) to the Kroenecker delta:
$$ frac{sin((n-m)pi)}{(n-m)pi} - frac{sin((n+m)pi)}{(n+m)pi} $$



Basically asserting that this evaluates to $1$ when $ n = m$ and to $0$ when $ n neq m $. I've been staring at this for minutes now and I feel like I'm going insane. It seems obvious to me that if I set $n =m$ the whole thing evaluates to $0$ regardless. We get $sin(0)$ in the first term and we get $sin(2npi)$ in the second term, which also is $0$ because $n$ is an integer. What am I missing here?






integration







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    3












    $begingroup$


    The specific integral I'm working with is the following:
    $$ int_0^asin(npi y/a)sin(n'pi y/a) $$
    This is supposed to come out to $0$ in the case that $ n neq n' $ and $frac{1}{2}a$ in the case that $n= n'$. I can obtain this result sometimes, but the method I'm using currently is giving me a value of $0$ all the time. I'm applying a product-to-sum formula and then integrating.
    Only resources I've managed to find on this equate the following expression (which is my final result before applying limits of integration) to the Kroenecker delta:
    $$ frac{sin((n-m)pi)}{(n-m)pi} - frac{sin((n+m)pi)}{(n+m)pi} $$



    Basically asserting that this evaluates to $1$ when $ n = m$ and to $0$ when $ n neq m $. I've been staring at this for minutes now and I feel like I'm going insane. It seems obvious to me that if I set $n =m$ the whole thing evaluates to $0$ regardless. We get $sin(0)$ in the first term and we get $sin(2npi)$ in the second term, which also is $0$ because $n$ is an integer. What am I missing here?






    integration







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      The specific integral I'm working with is the following:
      $$ int_0^asin(npi y/a)sin(n'pi y/a) $$
      This is supposed to come out to $0$ in the case that $ n neq n' $ and $frac{1}{2}a$ in the case that $n= n'$. I can obtain this result sometimes, but the method I'm using currently is giving me a value of $0$ all the time. I'm applying a product-to-sum formula and then integrating.
      Only resources I've managed to find on this equate the following expression (which is my final result before applying limits of integration) to the Kroenecker delta:
      $$ frac{sin((n-m)pi)}{(n-m)pi} - frac{sin((n+m)pi)}{(n+m)pi} $$



      Basically asserting that this evaluates to $1$ when $ n = m$ and to $0$ when $ n neq m $. I've been staring at this for minutes now and I feel like I'm going insane. It seems obvious to me that if I set $n =m$ the whole thing evaluates to $0$ regardless. We get $sin(0)$ in the first term and we get $sin(2npi)$ in the second term, which also is $0$ because $n$ is an integer. What am I missing here?






      integration







      share|cite|improve this question











      $endgroup$




      The specific integral I'm working with is the following:
      $$ int_0^asin(npi y/a)sin(n'pi y/a) $$
      This is supposed to come out to $0$ in the case that $ n neq n' $ and $frac{1}{2}a$ in the case that $n= n'$. I can obtain this result sometimes, but the method I'm using currently is giving me a value of $0$ all the time. I'm applying a product-to-sum formula and then integrating.
      Only resources I've managed to find on this equate the following expression (which is my final result before applying limits of integration) to the Kroenecker delta:
      $$ frac{sin((n-m)pi)}{(n-m)pi} - frac{sin((n+m)pi)}{(n+m)pi} $$



      Basically asserting that this evaluates to $1$ when $ n = m$ and to $0$ when $ n neq m $. I've been staring at this for minutes now and I feel like I'm going insane. It seems obvious to me that if I set $n =m$ the whole thing evaluates to $0$ regardless. We get $sin(0)$ in the first term and we get $sin(2npi)$ in the second term, which also is $0$ because $n$ is an integer. What am I missing here?






      integration




      integration






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      edited 7 hours ago









      Lemniscate

      417211




      417211










      asked 8 hours ago









      BookieBookie

      1227




      1227






















          2 Answers
          2






          active

          oldest

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          5












          $begingroup$

          Look at the denominator of the first term. It also is equal to $0$ when $m=n$. So you get $frac 0 0$.
          In fact, the first term is to be interpreted as the limit when $mrightarrow n$, and it converges to $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This makes sense to me. To clarify further, I would take something like L'Hopital's and choose to evaluate the indeterminate expression as a limit? Then I would receive 1 as desired. This is valid?
            $endgroup$
            – Bookie
            7 hours ago










          • $begingroup$
            Yes, that's one way to think about it. But as @GReyes said, you don't have to think of it as a limit. His answer shows a way to evaluate at $m=n$.
            $endgroup$
            – Stefan Lafon
            7 hours ago










          • $begingroup$
            Using limits here creates confusion. This integral can be (trivially) computed in all cases.
            $endgroup$
            – GReyes
            5 hours ago










          • $begingroup$
            I think you're right.
            $endgroup$
            – Stefan Lafon
            1 hour ago



















          2












          $begingroup$

          The problem is that when $n=n'$, when you transform into a sum, you get
          $$
          frac{1}{2}left[cosleft(frac{(n-n')pi y}{a}right)-cosleft(frac{(n+n')pi y}{a}right)right]
          $$
          When $n=n'$ the first cosine is just $=1$ and integrates as $y$, NOT as the corresponding sine.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This seem to not jive with the answer given above. I'm not sure what to think.
            $endgroup$
            – Bookie
            7 hours ago










          • $begingroup$
            You do not need any kind of limit procedure here. When $n=n'$ your integrand has a first piece equal to $1/2$ and integrates to $y/2$. Please let me know what is not clear.
            $endgroup$
            – GReyes
            7 hours ago










          • $begingroup$
            The formula to transform a product of sines into a sum of cosines is general. What does not make sense is to integrate $cos 0$ as $sin 0$.
            $endgroup$
            – GReyes
            7 hours ago










          • $begingroup$
            @Bookie Forget about the limit argument. You are only trying to prove this for integer values of $n$ and $n'$ so you can't take any "limits as $n$ tends to something." When $n = n'$ you are just integrating $sin^2(npi y/a)$ which you learned how to do in Calculus 2 - it doesn't need clever arguments about limits.
            $endgroup$
            – alephzero
            5 hours ago










          • $begingroup$
            The basic formulas here are $cos(p+q) = cos p cos q - sin p sin q$ and $cos(p-q) = cos p cos q + sin p sin q$. So $sin p sin q = (cos(p-q) - cos(p+q))/2$. When $p = q$, you have $sin^2 p = (1 - cos 2p)/2$. That's all there is to it.
            $endgroup$
            – alephzero
            4 hours ago













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          2 Answers
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          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          Look at the denominator of the first term. It also is equal to $0$ when $m=n$. So you get $frac 0 0$.
          In fact, the first term is to be interpreted as the limit when $mrightarrow n$, and it converges to $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This makes sense to me. To clarify further, I would take something like L'Hopital's and choose to evaluate the indeterminate expression as a limit? Then I would receive 1 as desired. This is valid?
            $endgroup$
            – Bookie
            7 hours ago










          • $begingroup$
            Yes, that's one way to think about it. But as @GReyes said, you don't have to think of it as a limit. His answer shows a way to evaluate at $m=n$.
            $endgroup$
            – Stefan Lafon
            7 hours ago










          • $begingroup$
            Using limits here creates confusion. This integral can be (trivially) computed in all cases.
            $endgroup$
            – GReyes
            5 hours ago










          • $begingroup$
            I think you're right.
            $endgroup$
            – Stefan Lafon
            1 hour ago
















          5












          $begingroup$

          Look at the denominator of the first term. It also is equal to $0$ when $m=n$. So you get $frac 0 0$.
          In fact, the first term is to be interpreted as the limit when $mrightarrow n$, and it converges to $1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This makes sense to me. To clarify further, I would take something like L'Hopital's and choose to evaluate the indeterminate expression as a limit? Then I would receive 1 as desired. This is valid?
            $endgroup$
            – Bookie
            7 hours ago










          • $begingroup$
            Yes, that's one way to think about it. But as @GReyes said, you don't have to think of it as a limit. His answer shows a way to evaluate at $m=n$.
            $endgroup$
            – Stefan Lafon
            7 hours ago










          • $begingroup$
            Using limits here creates confusion. This integral can be (trivially) computed in all cases.
            $endgroup$
            – GReyes
            5 hours ago










          • $begingroup$
            I think you're right.
            $endgroup$
            – Stefan Lafon
            1 hour ago














          5












          5








          5





          $begingroup$

          Look at the denominator of the first term. It also is equal to $0$ when $m=n$. So you get $frac 0 0$.
          In fact, the first term is to be interpreted as the limit when $mrightarrow n$, and it converges to $1$.






          share|cite|improve this answer









          $endgroup$



          Look at the denominator of the first term. It also is equal to $0$ when $m=n$. So you get $frac 0 0$.
          In fact, the first term is to be interpreted as the limit when $mrightarrow n$, and it converges to $1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 7 hours ago









          Stefan LafonStefan Lafon

          2,13018




          2,13018












          • $begingroup$
            This makes sense to me. To clarify further, I would take something like L'Hopital's and choose to evaluate the indeterminate expression as a limit? Then I would receive 1 as desired. This is valid?
            $endgroup$
            – Bookie
            7 hours ago










          • $begingroup$
            Yes, that's one way to think about it. But as @GReyes said, you don't have to think of it as a limit. His answer shows a way to evaluate at $m=n$.
            $endgroup$
            – Stefan Lafon
            7 hours ago










          • $begingroup$
            Using limits here creates confusion. This integral can be (trivially) computed in all cases.
            $endgroup$
            – GReyes
            5 hours ago










          • $begingroup$
            I think you're right.
            $endgroup$
            – Stefan Lafon
            1 hour ago


















          • $begingroup$
            This makes sense to me. To clarify further, I would take something like L'Hopital's and choose to evaluate the indeterminate expression as a limit? Then I would receive 1 as desired. This is valid?
            $endgroup$
            – Bookie
            7 hours ago










          • $begingroup$
            Yes, that's one way to think about it. But as @GReyes said, you don't have to think of it as a limit. His answer shows a way to evaluate at $m=n$.
            $endgroup$
            – Stefan Lafon
            7 hours ago










          • $begingroup$
            Using limits here creates confusion. This integral can be (trivially) computed in all cases.
            $endgroup$
            – GReyes
            5 hours ago










          • $begingroup$
            I think you're right.
            $endgroup$
            – Stefan Lafon
            1 hour ago
















          $begingroup$
          This makes sense to me. To clarify further, I would take something like L'Hopital's and choose to evaluate the indeterminate expression as a limit? Then I would receive 1 as desired. This is valid?
          $endgroup$
          – Bookie
          7 hours ago




          $begingroup$
          This makes sense to me. To clarify further, I would take something like L'Hopital's and choose to evaluate the indeterminate expression as a limit? Then I would receive 1 as desired. This is valid?
          $endgroup$
          – Bookie
          7 hours ago












          $begingroup$
          Yes, that's one way to think about it. But as @GReyes said, you don't have to think of it as a limit. His answer shows a way to evaluate at $m=n$.
          $endgroup$
          – Stefan Lafon
          7 hours ago




          $begingroup$
          Yes, that's one way to think about it. But as @GReyes said, you don't have to think of it as a limit. His answer shows a way to evaluate at $m=n$.
          $endgroup$
          – Stefan Lafon
          7 hours ago












          $begingroup$
          Using limits here creates confusion. This integral can be (trivially) computed in all cases.
          $endgroup$
          – GReyes
          5 hours ago




          $begingroup$
          Using limits here creates confusion. This integral can be (trivially) computed in all cases.
          $endgroup$
          – GReyes
          5 hours ago












          $begingroup$
          I think you're right.
          $endgroup$
          – Stefan Lafon
          1 hour ago




          $begingroup$
          I think you're right.
          $endgroup$
          – Stefan Lafon
          1 hour ago











          2












          $begingroup$

          The problem is that when $n=n'$, when you transform into a sum, you get
          $$
          frac{1}{2}left[cosleft(frac{(n-n')pi y}{a}right)-cosleft(frac{(n+n')pi y}{a}right)right]
          $$
          When $n=n'$ the first cosine is just $=1$ and integrates as $y$, NOT as the corresponding sine.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This seem to not jive with the answer given above. I'm not sure what to think.
            $endgroup$
            – Bookie
            7 hours ago










          • $begingroup$
            You do not need any kind of limit procedure here. When $n=n'$ your integrand has a first piece equal to $1/2$ and integrates to $y/2$. Please let me know what is not clear.
            $endgroup$
            – GReyes
            7 hours ago










          • $begingroup$
            The formula to transform a product of sines into a sum of cosines is general. What does not make sense is to integrate $cos 0$ as $sin 0$.
            $endgroup$
            – GReyes
            7 hours ago










          • $begingroup$
            @Bookie Forget about the limit argument. You are only trying to prove this for integer values of $n$ and $n'$ so you can't take any "limits as $n$ tends to something." When $n = n'$ you are just integrating $sin^2(npi y/a)$ which you learned how to do in Calculus 2 - it doesn't need clever arguments about limits.
            $endgroup$
            – alephzero
            5 hours ago










          • $begingroup$
            The basic formulas here are $cos(p+q) = cos p cos q - sin p sin q$ and $cos(p-q) = cos p cos q + sin p sin q$. So $sin p sin q = (cos(p-q) - cos(p+q))/2$. When $p = q$, you have $sin^2 p = (1 - cos 2p)/2$. That's all there is to it.
            $endgroup$
            – alephzero
            4 hours ago


















          2












          $begingroup$

          The problem is that when $n=n'$, when you transform into a sum, you get
          $$
          frac{1}{2}left[cosleft(frac{(n-n')pi y}{a}right)-cosleft(frac{(n+n')pi y}{a}right)right]
          $$
          When $n=n'$ the first cosine is just $=1$ and integrates as $y$, NOT as the corresponding sine.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This seem to not jive with the answer given above. I'm not sure what to think.
            $endgroup$
            – Bookie
            7 hours ago










          • $begingroup$
            You do not need any kind of limit procedure here. When $n=n'$ your integrand has a first piece equal to $1/2$ and integrates to $y/2$. Please let me know what is not clear.
            $endgroup$
            – GReyes
            7 hours ago










          • $begingroup$
            The formula to transform a product of sines into a sum of cosines is general. What does not make sense is to integrate $cos 0$ as $sin 0$.
            $endgroup$
            – GReyes
            7 hours ago










          • $begingroup$
            @Bookie Forget about the limit argument. You are only trying to prove this for integer values of $n$ and $n'$ so you can't take any "limits as $n$ tends to something." When $n = n'$ you are just integrating $sin^2(npi y/a)$ which you learned how to do in Calculus 2 - it doesn't need clever arguments about limits.
            $endgroup$
            – alephzero
            5 hours ago










          • $begingroup$
            The basic formulas here are $cos(p+q) = cos p cos q - sin p sin q$ and $cos(p-q) = cos p cos q + sin p sin q$. So $sin p sin q = (cos(p-q) - cos(p+q))/2$. When $p = q$, you have $sin^2 p = (1 - cos 2p)/2$. That's all there is to it.
            $endgroup$
            – alephzero
            4 hours ago
















          2












          2








          2





          $begingroup$

          The problem is that when $n=n'$, when you transform into a sum, you get
          $$
          frac{1}{2}left[cosleft(frac{(n-n')pi y}{a}right)-cosleft(frac{(n+n')pi y}{a}right)right]
          $$
          When $n=n'$ the first cosine is just $=1$ and integrates as $y$, NOT as the corresponding sine.






          share|cite|improve this answer









          $endgroup$



          The problem is that when $n=n'$, when you transform into a sum, you get
          $$
          frac{1}{2}left[cosleft(frac{(n-n')pi y}{a}right)-cosleft(frac{(n+n')pi y}{a}right)right]
          $$
          When $n=n'$ the first cosine is just $=1$ and integrates as $y$, NOT as the corresponding sine.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 7 hours ago









          GReyesGReyes

          1,41015




          1,41015












          • $begingroup$
            This seem to not jive with the answer given above. I'm not sure what to think.
            $endgroup$
            – Bookie
            7 hours ago










          • $begingroup$
            You do not need any kind of limit procedure here. When $n=n'$ your integrand has a first piece equal to $1/2$ and integrates to $y/2$. Please let me know what is not clear.
            $endgroup$
            – GReyes
            7 hours ago










          • $begingroup$
            The formula to transform a product of sines into a sum of cosines is general. What does not make sense is to integrate $cos 0$ as $sin 0$.
            $endgroup$
            – GReyes
            7 hours ago










          • $begingroup$
            @Bookie Forget about the limit argument. You are only trying to prove this for integer values of $n$ and $n'$ so you can't take any "limits as $n$ tends to something." When $n = n'$ you are just integrating $sin^2(npi y/a)$ which you learned how to do in Calculus 2 - it doesn't need clever arguments about limits.
            $endgroup$
            – alephzero
            5 hours ago










          • $begingroup$
            The basic formulas here are $cos(p+q) = cos p cos q - sin p sin q$ and $cos(p-q) = cos p cos q + sin p sin q$. So $sin p sin q = (cos(p-q) - cos(p+q))/2$. When $p = q$, you have $sin^2 p = (1 - cos 2p)/2$. That's all there is to it.
            $endgroup$
            – alephzero
            4 hours ago




















          • $begingroup$
            This seem to not jive with the answer given above. I'm not sure what to think.
            $endgroup$
            – Bookie
            7 hours ago










          • $begingroup$
            You do not need any kind of limit procedure here. When $n=n'$ your integrand has a first piece equal to $1/2$ and integrates to $y/2$. Please let me know what is not clear.
            $endgroup$
            – GReyes
            7 hours ago










          • $begingroup$
            The formula to transform a product of sines into a sum of cosines is general. What does not make sense is to integrate $cos 0$ as $sin 0$.
            $endgroup$
            – GReyes
            7 hours ago










          • $begingroup$
            @Bookie Forget about the limit argument. You are only trying to prove this for integer values of $n$ and $n'$ so you can't take any "limits as $n$ tends to something." When $n = n'$ you are just integrating $sin^2(npi y/a)$ which you learned how to do in Calculus 2 - it doesn't need clever arguments about limits.
            $endgroup$
            – alephzero
            5 hours ago










          • $begingroup$
            The basic formulas here are $cos(p+q) = cos p cos q - sin p sin q$ and $cos(p-q) = cos p cos q + sin p sin q$. So $sin p sin q = (cos(p-q) - cos(p+q))/2$. When $p = q$, you have $sin^2 p = (1 - cos 2p)/2$. That's all there is to it.
            $endgroup$
            – alephzero
            4 hours ago


















          $begingroup$
          This seem to not jive with the answer given above. I'm not sure what to think.
          $endgroup$
          – Bookie
          7 hours ago




          $begingroup$
          This seem to not jive with the answer given above. I'm not sure what to think.
          $endgroup$
          – Bookie
          7 hours ago












          $begingroup$
          You do not need any kind of limit procedure here. When $n=n'$ your integrand has a first piece equal to $1/2$ and integrates to $y/2$. Please let me know what is not clear.
          $endgroup$
          – GReyes
          7 hours ago




          $begingroup$
          You do not need any kind of limit procedure here. When $n=n'$ your integrand has a first piece equal to $1/2$ and integrates to $y/2$. Please let me know what is not clear.
          $endgroup$
          – GReyes
          7 hours ago












          $begingroup$
          The formula to transform a product of sines into a sum of cosines is general. What does not make sense is to integrate $cos 0$ as $sin 0$.
          $endgroup$
          – GReyes
          7 hours ago




          $begingroup$
          The formula to transform a product of sines into a sum of cosines is general. What does not make sense is to integrate $cos 0$ as $sin 0$.
          $endgroup$
          – GReyes
          7 hours ago












          $begingroup$
          @Bookie Forget about the limit argument. You are only trying to prove this for integer values of $n$ and $n'$ so you can't take any "limits as $n$ tends to something." When $n = n'$ you are just integrating $sin^2(npi y/a)$ which you learned how to do in Calculus 2 - it doesn't need clever arguments about limits.
          $endgroup$
          – alephzero
          5 hours ago




          $begingroup$
          @Bookie Forget about the limit argument. You are only trying to prove this for integer values of $n$ and $n'$ so you can't take any "limits as $n$ tends to something." When $n = n'$ you are just integrating $sin^2(npi y/a)$ which you learned how to do in Calculus 2 - it doesn't need clever arguments about limits.
          $endgroup$
          – alephzero
          5 hours ago












          $begingroup$
          The basic formulas here are $cos(p+q) = cos p cos q - sin p sin q$ and $cos(p-q) = cos p cos q + sin p sin q$. So $sin p sin q = (cos(p-q) - cos(p+q))/2$. When $p = q$, you have $sin^2 p = (1 - cos 2p)/2$. That's all there is to it.
          $endgroup$
          – alephzero
          4 hours ago






          $begingroup$
          The basic formulas here are $cos(p+q) = cos p cos q - sin p sin q$ and $cos(p-q) = cos p cos q + sin p sin q$. So $sin p sin q = (cos(p-q) - cos(p+q))/2$. When $p = q$, you have $sin^2 p = (1 - cos 2p)/2$. That's all there is to it.
          $endgroup$
          – alephzero
          4 hours ago




















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