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





      $begingroup$


      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 7 hours ago









      Lemniscate

      417211




      417211










      asked 8 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
            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













          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
            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




















          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

          Fluorita

          Image
          Fluorita Fluorita (blau) amb pirita (daurat). Fórmula química CaF 2 Epònim fluor Localitat tipus Jáchymov Classificació Categoria Halurs Nickel-Strunz 10a ed. 3.AB.25 Nickel-Strunz 9a ed. 3.AB.25 Nickel-Strunz 8a ed. III/A.08 Dana 9.2.1.1 Heys 8.4.7 Propietats Sistema cristal·lí Cúbic Hàbit cristal·lí Sol presentar cubs; menys freqüentment dodecaedres. De vegades hexaoctaedres i tetrahexaedres. Per la combinació d'aquestes formes, les arestes dels cubs són sovint modificades. De vegades els cristalls poden presentar diferències de creixement entre les cares. Sovint s'observen cristalls compostos per sobrecreixement. Pot ser massiva, compacta, terrosa, columnar (poc freqüent), globular en agregats o botroïdal. Estructura cristal·lina a = 5.4626Å Simetria Classe (H-M): m 3 m (4/ m 3 2/ m ) - Hexoctaèdric; Grup espacial: Fm 3 m Color Blanc, groc, verd, violeta, vermell, rosa, blau o negre Mac...
          Read more

          Hulsita

          Image
          Hulsita Hulsita de la localitat tipus Fórmula química Fe 2+ 2 Fe 3+ O 2 BO 3 Epònim Alfred Hulse Brooks Localitat tipus mont Brooks, península de Seward, Nome Borough, Alaska, Estats Units Classificació Categoria borats Nickel-Strunz 10a ed. 6.AB.45 Nickel-Strunz 9a ed. 6.AB.45 Nickel-Strunz 8a ed. V/G.03 Dana 24.2.3.1 Heys 9.6.4 Propietats Sistema cristal·lí monoclínic Estructura cristal·lina a = 10,68Å; b = 3,09Å; c = 5,43Å; β = 94,15° Color negre Macles en [001]; un membre de la macla es troba girat 120° en relació a l'altre Exfoliació bona en {110} Duresa 3 Lluïssor vítria, submetàl·lica Densitat 4,28 g/cm 3 (mesurada); Més informació Estatus IMA mineral heretat (G) i mineral heretat (G) Any d'aprovació 1908 Referències [1] La hulsita és un mineral de la classe dels borats, que pertany al grup de la pinakiolita. Rep el seu nom en honor del nord-americà ...
          Read more

          Península de Txukotka

          Image
          Península de Txukotka Tipus Península Ubicació 66° N, 172° O  /  66°N , 172°O  / 66; -172 Mar mar dels Txuktxis La península de Txukotka a l'extrem oriental de Sibèria La península de Txukotka o península dels Txuktxis (en rus Чуко́тский полуо́стров , Txukotski poluóstrov ) és una península que es troba a l'extrem oriental d'Euràsia. Contingut 1 Descripció 2 Població 3 Vegeu també 4 Bibliografia Descripció L'estret de Bering, entre la península de Txukotka i la península de Seward a Alaska Foto de satèl·lit de la zona amb la península de Txukotka a l'esquerra El seu punt més extrem és el cap Dezhnev a la vora del nucli d'Uelen, un dels poquíssims llocs habitats de la zona. Al nord limita amb el mar dels Txuktxis i amb el mar de Bering al sud i amb l'estret de Bering a l'est. A la costa sud es troben la badia de Sant Lawrence i la badia de Kresta. La penínsu...
          Read more