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







share|cite|improve this question











$endgroup$

















    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


      1



      $begingroup$


      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 17 hours ago









      Lemniscate

      417211




      417211










      asked 17 hours ago









      BookieBookie

      1227




      1227






















          2 Answers
          2






          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
            17 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
            17 hours ago










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










          • $begingroup$
            I think you're right.
            $endgroup$
            – Stefan Lafon
            11 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
            17 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
            17 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
            17 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
            15 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
            14 hours ago













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3116868%2fconfusion-with-integrating-sinnxsinmx-and-kroenecker-delta%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          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
            17 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
            17 hours ago










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










          • $begingroup$
            I think you're right.
            $endgroup$
            – Stefan Lafon
            11 hours 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
            17 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
            17 hours ago










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










          • $begingroup$
            I think you're right.
            $endgroup$
            – Stefan Lafon
            11 hours 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 17 hours ago









          Stefan LafonStefan Lafon

          2,14518




          2,14518












          • $begingroup$
            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
            17 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
            17 hours ago










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










          • $begingroup$
            I think you're right.
            $endgroup$
            – Stefan Lafon
            11 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
            17 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
            17 hours ago










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










          • $begingroup$
            I think you're right.
            $endgroup$
            – Stefan Lafon
            11 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
          17 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
          17 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
          17 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
          17 hours ago












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




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












          $begingroup$
          I think you're right.
          $endgroup$
          – Stefan Lafon
          11 hours ago




          $begingroup$
          I think you're right.
          $endgroup$
          – Stefan Lafon
          11 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
            17 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
            17 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
            17 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
            15 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
            14 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
            17 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
            17 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
            17 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
            15 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
            14 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 17 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
            17 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
            17 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
            17 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
            15 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
            14 hours ago




















          • $begingroup$
            This seem to not jive with the answer given above. I'm not sure what to think.
            $endgroup$
            – Bookie
            17 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
            17 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
            17 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
            15 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
            14 hours ago


















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




          $begingroup$
          This seem to not jive with the answer given above. I'm not sure what to think.
          $endgroup$
          – Bookie
          17 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
          17 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
          17 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
          17 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
          17 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
          15 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
          15 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
          14 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
          14 hours ago




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3116868%2fconfusion-with-integrating-sinnxsinmx-and-kroenecker-delta%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          How to label and detect the document text images

          Image
          1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C...
          Read more

          Vallis Paradisi

          Image
          Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para...
          Read more

          Tabula Rosettana

          Image
          Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
          Read more