On the evalution of an infinite sum












10












$begingroup$



I wish to show that
$$sum_{n = 0}^infty (-1)^n left [frac{2n + 1/2}{(2n + 1/2)^2 + 1} + frac{2n + 3/2}{(2n + 3/2)^2 + 1} right] = frac{pi}{sqrt{2}} frac{cosh left (frac{pi}{2} right )}{cosh (pi)}.$$




The reason I wish to find such a sum is as follows.
The question here called for the evaluation (I have added its value) of
$$int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx = frac{pi}{sqrt{2}} frac{cosh left (frac{pi}{2} right )}{cosh (pi)}.$$



As one of the comments, the OP remarked that they would like to see different approaches to the evaluation of the integral so I thought I would try my hand at one that does not rely on contour integration and the residue theorem. My approach was as follows:
begin{align}
int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx &= int_0^1 frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx + int_1^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx\
&= int_0^1 frac{cos (ln x) (x + 1)}{sqrt{x} (1 + x^2)} , dx,
end{align}
after a substitution of $x mapsto 1/x$ has been enforced in the second of the integrals. Now if we enforce a substitution of $x mapsto e^{-x}$ one arrives at
$$int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx = int_0^infty frac{cos x cosh (x/2)}{cosh x} , dx.$$
Writing the hyperbolic functions in terms of exponentials we have
begin{align}
int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx &= int_0^infty frac{cos x (e^{-x/2} + e^{-3x/2})}{1 + e^{-2x}} , dx\
&= text{Re} sum_{n = 0}^infty (-1)^n int_0^infty left [e^{-(2n + 1/2 - i) x} + e^{-(2n + 3/2 - i)x} right ] , dx\
&= text{Re} sum_{n = 0}^infty (-1)^n left [frac{1}{2n + 1/2 - i} + frac{1}{2n + 3/2 - i} right ] tag1\
&= sum_{n = 0}^infty (-1)^n left [frac{2n + 1/2}{(2n + 1/2)^2 + 1} + frac{2n + 3/2}{(2n + 3/2)^2 + 1} right],
end{align}
which brings me to my sum.




Some thoughts on finding this sum



Rewriting the sum $S$ in (1) as follows:
begin{align}
S &= text{Re} cdot frac{1}{4} sum_{n = 0}^infty left [frac{1}{n + 1/8 - i/4} + frac{1}{n + 3/8 - i/4} - frac{1}{n + 5/8 - i/4} - frac{1}{n + 7/8 - i/4} right ]\
&= text{Re} cdot frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 7/8 - i/4} right ) + frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 5/8 - i/4} right )\
& qquad - frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 3/8 - i/4} right ) - frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 1/8 - i/4} right )\
&= frac{1}{4} text{Re} left [psi left (frac{7}{8} - frac{i}{4} right ) + psi left (frac{5}{8} - frac{i}{4} right ) - psi left (frac{3}{8} - frac{i}{4} right ) - psi left (frac{1}{8} - frac{i}{4} right ) right ].
end{align}
Here $psi (z)$ is the digamma function. I was rather hoping to use the reflexion formula for the digamma function, but alas it does not seem to take me any closer to a final real solution.




Final thought



While it would be nice to see how to evaluate this sum, perhaps my approach was not the best so alternative methods to evaluate the integral that avoid this sum and do not rely on contour integration would also be welcome.






integration sequences-and-series definite-integrals improper-integrals closed-form







share|cite|improve this question











$endgroup$

















    10












    $begingroup$



    I wish to show that
    $$sum_{n = 0}^infty (-1)^n left [frac{2n + 1/2}{(2n + 1/2)^2 + 1} + frac{2n + 3/2}{(2n + 3/2)^2 + 1} right] = frac{pi}{sqrt{2}} frac{cosh left (frac{pi}{2} right )}{cosh (pi)}.$$




    The reason I wish to find such a sum is as follows.
    The question here called for the evaluation (I have added its value) of
    $$int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx = frac{pi}{sqrt{2}} frac{cosh left (frac{pi}{2} right )}{cosh (pi)}.$$



    As one of the comments, the OP remarked that they would like to see different approaches to the evaluation of the integral so I thought I would try my hand at one that does not rely on contour integration and the residue theorem. My approach was as follows:
    begin{align}
    int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx &= int_0^1 frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx + int_1^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx\
    &= int_0^1 frac{cos (ln x) (x + 1)}{sqrt{x} (1 + x^2)} , dx,
    end{align}
    after a substitution of $x mapsto 1/x$ has been enforced in the second of the integrals. Now if we enforce a substitution of $x mapsto e^{-x}$ one arrives at
    $$int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx = int_0^infty frac{cos x cosh (x/2)}{cosh x} , dx.$$
    Writing the hyperbolic functions in terms of exponentials we have
    begin{align}
    int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx &= int_0^infty frac{cos x (e^{-x/2} + e^{-3x/2})}{1 + e^{-2x}} , dx\
    &= text{Re} sum_{n = 0}^infty (-1)^n int_0^infty left [e^{-(2n + 1/2 - i) x} + e^{-(2n + 3/2 - i)x} right ] , dx\
    &= text{Re} sum_{n = 0}^infty (-1)^n left [frac{1}{2n + 1/2 - i} + frac{1}{2n + 3/2 - i} right ] tag1\
    &= sum_{n = 0}^infty (-1)^n left [frac{2n + 1/2}{(2n + 1/2)^2 + 1} + frac{2n + 3/2}{(2n + 3/2)^2 + 1} right],
    end{align}
    which brings me to my sum.




    Some thoughts on finding this sum



    Rewriting the sum $S$ in (1) as follows:
    begin{align}
    S &= text{Re} cdot frac{1}{4} sum_{n = 0}^infty left [frac{1}{n + 1/8 - i/4} + frac{1}{n + 3/8 - i/4} - frac{1}{n + 5/8 - i/4} - frac{1}{n + 7/8 - i/4} right ]\
    &= text{Re} cdot frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 7/8 - i/4} right ) + frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 5/8 - i/4} right )\
    & qquad - frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 3/8 - i/4} right ) - frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 1/8 - i/4} right )\
    &= frac{1}{4} text{Re} left [psi left (frac{7}{8} - frac{i}{4} right ) + psi left (frac{5}{8} - frac{i}{4} right ) - psi left (frac{3}{8} - frac{i}{4} right ) - psi left (frac{1}{8} - frac{i}{4} right ) right ].
    end{align}
    Here $psi (z)$ is the digamma function. I was rather hoping to use the reflexion formula for the digamma function, but alas it does not seem to take me any closer to a final real solution.




    Final thought



    While it would be nice to see how to evaluate this sum, perhaps my approach was not the best so alternative methods to evaluate the integral that avoid this sum and do not rely on contour integration would also be welcome.






    integration sequences-and-series definite-integrals improper-integrals closed-form







    share|cite|improve this question











    $endgroup$















      10












      10








      10


      2



      $begingroup$



      I wish to show that
      $$sum_{n = 0}^infty (-1)^n left [frac{2n + 1/2}{(2n + 1/2)^2 + 1} + frac{2n + 3/2}{(2n + 3/2)^2 + 1} right] = frac{pi}{sqrt{2}} frac{cosh left (frac{pi}{2} right )}{cosh (pi)}.$$




      The reason I wish to find such a sum is as follows.
      The question here called for the evaluation (I have added its value) of
      $$int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx = frac{pi}{sqrt{2}} frac{cosh left (frac{pi}{2} right )}{cosh (pi)}.$$



      As one of the comments, the OP remarked that they would like to see different approaches to the evaluation of the integral so I thought I would try my hand at one that does not rely on contour integration and the residue theorem. My approach was as follows:
      begin{align}
      int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx &= int_0^1 frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx + int_1^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx\
      &= int_0^1 frac{cos (ln x) (x + 1)}{sqrt{x} (1 + x^2)} , dx,
      end{align}
      after a substitution of $x mapsto 1/x$ has been enforced in the second of the integrals. Now if we enforce a substitution of $x mapsto e^{-x}$ one arrives at
      $$int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx = int_0^infty frac{cos x cosh (x/2)}{cosh x} , dx.$$
      Writing the hyperbolic functions in terms of exponentials we have
      begin{align}
      int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx &= int_0^infty frac{cos x (e^{-x/2} + e^{-3x/2})}{1 + e^{-2x}} , dx\
      &= text{Re} sum_{n = 0}^infty (-1)^n int_0^infty left [e^{-(2n + 1/2 - i) x} + e^{-(2n + 3/2 - i)x} right ] , dx\
      &= text{Re} sum_{n = 0}^infty (-1)^n left [frac{1}{2n + 1/2 - i} + frac{1}{2n + 3/2 - i} right ] tag1\
      &= sum_{n = 0}^infty (-1)^n left [frac{2n + 1/2}{(2n + 1/2)^2 + 1} + frac{2n + 3/2}{(2n + 3/2)^2 + 1} right],
      end{align}
      which brings me to my sum.




      Some thoughts on finding this sum



      Rewriting the sum $S$ in (1) as follows:
      begin{align}
      S &= text{Re} cdot frac{1}{4} sum_{n = 0}^infty left [frac{1}{n + 1/8 - i/4} + frac{1}{n + 3/8 - i/4} - frac{1}{n + 5/8 - i/4} - frac{1}{n + 7/8 - i/4} right ]\
      &= text{Re} cdot frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 7/8 - i/4} right ) + frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 5/8 - i/4} right )\
      & qquad - frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 3/8 - i/4} right ) - frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 1/8 - i/4} right )\
      &= frac{1}{4} text{Re} left [psi left (frac{7}{8} - frac{i}{4} right ) + psi left (frac{5}{8} - frac{i}{4} right ) - psi left (frac{3}{8} - frac{i}{4} right ) - psi left (frac{1}{8} - frac{i}{4} right ) right ].
      end{align}
      Here $psi (z)$ is the digamma function. I was rather hoping to use the reflexion formula for the digamma function, but alas it does not seem to take me any closer to a final real solution.




      Final thought



      While it would be nice to see how to evaluate this sum, perhaps my approach was not the best so alternative methods to evaluate the integral that avoid this sum and do not rely on contour integration would also be welcome.






      integration sequences-and-series definite-integrals improper-integrals closed-form







      share|cite|improve this question











      $endgroup$





      I wish to show that
      $$sum_{n = 0}^infty (-1)^n left [frac{2n + 1/2}{(2n + 1/2)^2 + 1} + frac{2n + 3/2}{(2n + 3/2)^2 + 1} right] = frac{pi}{sqrt{2}} frac{cosh left (frac{pi}{2} right )}{cosh (pi)}.$$




      The reason I wish to find such a sum is as follows.
      The question here called for the evaluation (I have added its value) of
      $$int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx = frac{pi}{sqrt{2}} frac{cosh left (frac{pi}{2} right )}{cosh (pi)}.$$



      As one of the comments, the OP remarked that they would like to see different approaches to the evaluation of the integral so I thought I would try my hand at one that does not rely on contour integration and the residue theorem. My approach was as follows:
      begin{align}
      int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx &= int_0^1 frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx + int_1^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx\
      &= int_0^1 frac{cos (ln x) (x + 1)}{sqrt{x} (1 + x^2)} , dx,
      end{align}
      after a substitution of $x mapsto 1/x$ has been enforced in the second of the integrals. Now if we enforce a substitution of $x mapsto e^{-x}$ one arrives at
      $$int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx = int_0^infty frac{cos x cosh (x/2)}{cosh x} , dx.$$
      Writing the hyperbolic functions in terms of exponentials we have
      begin{align}
      int_0^infty frac{sqrt{x} cos (ln x)}{x^2 + 1} , dx &= int_0^infty frac{cos x (e^{-x/2} + e^{-3x/2})}{1 + e^{-2x}} , dx\
      &= text{Re} sum_{n = 0}^infty (-1)^n int_0^infty left [e^{-(2n + 1/2 - i) x} + e^{-(2n + 3/2 - i)x} right ] , dx\
      &= text{Re} sum_{n = 0}^infty (-1)^n left [frac{1}{2n + 1/2 - i} + frac{1}{2n + 3/2 - i} right ] tag1\
      &= sum_{n = 0}^infty (-1)^n left [frac{2n + 1/2}{(2n + 1/2)^2 + 1} + frac{2n + 3/2}{(2n + 3/2)^2 + 1} right],
      end{align}
      which brings me to my sum.




      Some thoughts on finding this sum



      Rewriting the sum $S$ in (1) as follows:
      begin{align}
      S &= text{Re} cdot frac{1}{4} sum_{n = 0}^infty left [frac{1}{n + 1/8 - i/4} + frac{1}{n + 3/8 - i/4} - frac{1}{n + 5/8 - i/4} - frac{1}{n + 7/8 - i/4} right ]\
      &= text{Re} cdot frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 7/8 - i/4} right ) + frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 5/8 - i/4} right )\
      & qquad - frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 3/8 - i/4} right ) - frac{1}{4} sum_{n = 0}^infty left (frac{1}{n + 1} - frac{1}{n + 1/8 - i/4} right )\
      &= frac{1}{4} text{Re} left [psi left (frac{7}{8} - frac{i}{4} right ) + psi left (frac{5}{8} - frac{i}{4} right ) - psi left (frac{3}{8} - frac{i}{4} right ) - psi left (frac{1}{8} - frac{i}{4} right ) right ].
      end{align}
      Here $psi (z)$ is the digamma function. I was rather hoping to use the reflexion formula for the digamma function, but alas it does not seem to take me any closer to a final real solution.




      Final thought



      While it would be nice to see how to evaluate this sum, perhaps my approach was not the best so alternative methods to evaluate the integral that avoid this sum and do not rely on contour integration would also be welcome.






      integration sequences-and-series definite-integrals improper-integrals closed-form




      integration sequences-and-series definite-integrals improper-integrals closed-form






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 52 mins ago







      omegadot

















      asked 1 hour ago









      omegadotomegadot

      5,2792728




      5,2792728






















          1 Answer
          1






          active

          oldest

          votes


















          8












          $begingroup$

          $$
          begin{align}newcommand{Re}{operatorname{Re}}
          &sum_{n=0}^infty(-1)^nleft[frac{2n+1/2}{(2n+1/2)^2+1}+frac{2n+3/2}{(2n+3/2)^2+1}right]tag1\
          &=frac12sum_{ninmathbb{Z}}(-1)^nleft[frac{2n+1/2}{(2n+1/2)^2+1}+frac{2n+3/2}{(2n+3/2)^2+1}right]tag2\
          &=frac14sum_{ninmathbb{Z}}(-1)^nleft[frac{n+frac14}{left(n+frac14right)^2+frac14}+frac{n+frac34}{left(n+frac34right)^2+frac14}right]tag3\
          &=frac18sum_{ninmathbb{Z}}(-1)^nleft[frac1{n+frac14-frac i2}+frac1{n+frac14+frac i2}+frac1{n+frac34-frac i2}+frac1{n+frac34+frac i2}right]tag4\
          &=frac18left[fracpi{sinleft(pi!left(frac14-frac i2right)right)}+fracpi{sinleft(pi!left(frac14+frac i2right)right)}+fracpi{sinleft(pi!left(frac34-frac i2right)right)}+fracpi{sinleft(pi!left(frac34+frac i2right)right)}right]tag5\
          &=frac{pisqrt2}8left[
          frac{coshleft(fracpi2right)+isinhleft(fracpi2right)}{cosh(pi)}+
          frac{coshleft(fracpi2right)-isinhleft(fracpi2right)}{cosh(pi)}+
          frac{coshleft(fracpi2right)-isinhleft(fracpi2right)}{cosh(pi)}right.\
          &left.phantom{=frac{pisqrt2}8}+
          frac{coshleft(fracpi2right)+isinhleft(fracpi2right)}{cosh(pi)}right]tag6\
          &=fracpi{sqrt2}frac{cosh(pi/2)}{cosh(pi)}tag7
          end{align}
          $$
          Explanation:
          $(2)$: use symmetry
          $(3)$: pull factor of $frac12$ out front
          $(4)$: partial fractions
          $(5)$: use $(3)$ from this answer
          $(6)$: evaluate the sine of a complex number
          $(7)$: simplify






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Given the form of the final answer, I thought there would be a Mittag-Leffler expansion hiding in there somewhere. Nice work and thanks (+1).
            $endgroup$
            – omegadot
            37 mins 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%2f3088089%2fon-the-evalution-of-an-infinite-sum%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          8












          $begingroup$

          $$
          begin{align}newcommand{Re}{operatorname{Re}}
          &sum_{n=0}^infty(-1)^nleft[frac{2n+1/2}{(2n+1/2)^2+1}+frac{2n+3/2}{(2n+3/2)^2+1}right]tag1\
          &=frac12sum_{ninmathbb{Z}}(-1)^nleft[frac{2n+1/2}{(2n+1/2)^2+1}+frac{2n+3/2}{(2n+3/2)^2+1}right]tag2\
          &=frac14sum_{ninmathbb{Z}}(-1)^nleft[frac{n+frac14}{left(n+frac14right)^2+frac14}+frac{n+frac34}{left(n+frac34right)^2+frac14}right]tag3\
          &=frac18sum_{ninmathbb{Z}}(-1)^nleft[frac1{n+frac14-frac i2}+frac1{n+frac14+frac i2}+frac1{n+frac34-frac i2}+frac1{n+frac34+frac i2}right]tag4\
          &=frac18left[fracpi{sinleft(pi!left(frac14-frac i2right)right)}+fracpi{sinleft(pi!left(frac14+frac i2right)right)}+fracpi{sinleft(pi!left(frac34-frac i2right)right)}+fracpi{sinleft(pi!left(frac34+frac i2right)right)}right]tag5\
          &=frac{pisqrt2}8left[
          frac{coshleft(fracpi2right)+isinhleft(fracpi2right)}{cosh(pi)}+
          frac{coshleft(fracpi2right)-isinhleft(fracpi2right)}{cosh(pi)}+
          frac{coshleft(fracpi2right)-isinhleft(fracpi2right)}{cosh(pi)}right.\
          &left.phantom{=frac{pisqrt2}8}+
          frac{coshleft(fracpi2right)+isinhleft(fracpi2right)}{cosh(pi)}right]tag6\
          &=fracpi{sqrt2}frac{cosh(pi/2)}{cosh(pi)}tag7
          end{align}
          $$
          Explanation:
          $(2)$: use symmetry
          $(3)$: pull factor of $frac12$ out front
          $(4)$: partial fractions
          $(5)$: use $(3)$ from this answer
          $(6)$: evaluate the sine of a complex number
          $(7)$: simplify






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Given the form of the final answer, I thought there would be a Mittag-Leffler expansion hiding in there somewhere. Nice work and thanks (+1).
            $endgroup$
            – omegadot
            37 mins ago
















          8












          $begingroup$

          $$
          begin{align}newcommand{Re}{operatorname{Re}}
          &sum_{n=0}^infty(-1)^nleft[frac{2n+1/2}{(2n+1/2)^2+1}+frac{2n+3/2}{(2n+3/2)^2+1}right]tag1\
          &=frac12sum_{ninmathbb{Z}}(-1)^nleft[frac{2n+1/2}{(2n+1/2)^2+1}+frac{2n+3/2}{(2n+3/2)^2+1}right]tag2\
          &=frac14sum_{ninmathbb{Z}}(-1)^nleft[frac{n+frac14}{left(n+frac14right)^2+frac14}+frac{n+frac34}{left(n+frac34right)^2+frac14}right]tag3\
          &=frac18sum_{ninmathbb{Z}}(-1)^nleft[frac1{n+frac14-frac i2}+frac1{n+frac14+frac i2}+frac1{n+frac34-frac i2}+frac1{n+frac34+frac i2}right]tag4\
          &=frac18left[fracpi{sinleft(pi!left(frac14-frac i2right)right)}+fracpi{sinleft(pi!left(frac14+frac i2right)right)}+fracpi{sinleft(pi!left(frac34-frac i2right)right)}+fracpi{sinleft(pi!left(frac34+frac i2right)right)}right]tag5\
          &=frac{pisqrt2}8left[
          frac{coshleft(fracpi2right)+isinhleft(fracpi2right)}{cosh(pi)}+
          frac{coshleft(fracpi2right)-isinhleft(fracpi2right)}{cosh(pi)}+
          frac{coshleft(fracpi2right)-isinhleft(fracpi2right)}{cosh(pi)}right.\
          &left.phantom{=frac{pisqrt2}8}+
          frac{coshleft(fracpi2right)+isinhleft(fracpi2right)}{cosh(pi)}right]tag6\
          &=fracpi{sqrt2}frac{cosh(pi/2)}{cosh(pi)}tag7
          end{align}
          $$
          Explanation:
          $(2)$: use symmetry
          $(3)$: pull factor of $frac12$ out front
          $(4)$: partial fractions
          $(5)$: use $(3)$ from this answer
          $(6)$: evaluate the sine of a complex number
          $(7)$: simplify






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Given the form of the final answer, I thought there would be a Mittag-Leffler expansion hiding in there somewhere. Nice work and thanks (+1).
            $endgroup$
            – omegadot
            37 mins ago














          8












          8








          8





          $begingroup$

          $$
          begin{align}newcommand{Re}{operatorname{Re}}
          &sum_{n=0}^infty(-1)^nleft[frac{2n+1/2}{(2n+1/2)^2+1}+frac{2n+3/2}{(2n+3/2)^2+1}right]tag1\
          &=frac12sum_{ninmathbb{Z}}(-1)^nleft[frac{2n+1/2}{(2n+1/2)^2+1}+frac{2n+3/2}{(2n+3/2)^2+1}right]tag2\
          &=frac14sum_{ninmathbb{Z}}(-1)^nleft[frac{n+frac14}{left(n+frac14right)^2+frac14}+frac{n+frac34}{left(n+frac34right)^2+frac14}right]tag3\
          &=frac18sum_{ninmathbb{Z}}(-1)^nleft[frac1{n+frac14-frac i2}+frac1{n+frac14+frac i2}+frac1{n+frac34-frac i2}+frac1{n+frac34+frac i2}right]tag4\
          &=frac18left[fracpi{sinleft(pi!left(frac14-frac i2right)right)}+fracpi{sinleft(pi!left(frac14+frac i2right)right)}+fracpi{sinleft(pi!left(frac34-frac i2right)right)}+fracpi{sinleft(pi!left(frac34+frac i2right)right)}right]tag5\
          &=frac{pisqrt2}8left[
          frac{coshleft(fracpi2right)+isinhleft(fracpi2right)}{cosh(pi)}+
          frac{coshleft(fracpi2right)-isinhleft(fracpi2right)}{cosh(pi)}+
          frac{coshleft(fracpi2right)-isinhleft(fracpi2right)}{cosh(pi)}right.\
          &left.phantom{=frac{pisqrt2}8}+
          frac{coshleft(fracpi2right)+isinhleft(fracpi2right)}{cosh(pi)}right]tag6\
          &=fracpi{sqrt2}frac{cosh(pi/2)}{cosh(pi)}tag7
          end{align}
          $$
          Explanation:
          $(2)$: use symmetry
          $(3)$: pull factor of $frac12$ out front
          $(4)$: partial fractions
          $(5)$: use $(3)$ from this answer
          $(6)$: evaluate the sine of a complex number
          $(7)$: simplify






          share|cite|improve this answer











          $endgroup$



          $$
          begin{align}newcommand{Re}{operatorname{Re}}
          &sum_{n=0}^infty(-1)^nleft[frac{2n+1/2}{(2n+1/2)^2+1}+frac{2n+3/2}{(2n+3/2)^2+1}right]tag1\
          &=frac12sum_{ninmathbb{Z}}(-1)^nleft[frac{2n+1/2}{(2n+1/2)^2+1}+frac{2n+3/2}{(2n+3/2)^2+1}right]tag2\
          &=frac14sum_{ninmathbb{Z}}(-1)^nleft[frac{n+frac14}{left(n+frac14right)^2+frac14}+frac{n+frac34}{left(n+frac34right)^2+frac14}right]tag3\
          &=frac18sum_{ninmathbb{Z}}(-1)^nleft[frac1{n+frac14-frac i2}+frac1{n+frac14+frac i2}+frac1{n+frac34-frac i2}+frac1{n+frac34+frac i2}right]tag4\
          &=frac18left[fracpi{sinleft(pi!left(frac14-frac i2right)right)}+fracpi{sinleft(pi!left(frac14+frac i2right)right)}+fracpi{sinleft(pi!left(frac34-frac i2right)right)}+fracpi{sinleft(pi!left(frac34+frac i2right)right)}right]tag5\
          &=frac{pisqrt2}8left[
          frac{coshleft(fracpi2right)+isinhleft(fracpi2right)}{cosh(pi)}+
          frac{coshleft(fracpi2right)-isinhleft(fracpi2right)}{cosh(pi)}+
          frac{coshleft(fracpi2right)-isinhleft(fracpi2right)}{cosh(pi)}right.\
          &left.phantom{=frac{pisqrt2}8}+
          frac{coshleft(fracpi2right)+isinhleft(fracpi2right)}{cosh(pi)}right]tag6\
          &=fracpi{sqrt2}frac{cosh(pi/2)}{cosh(pi)}tag7
          end{align}
          $$
          Explanation:
          $(2)$: use symmetry
          $(3)$: pull factor of $frac12$ out front
          $(4)$: partial fractions
          $(5)$: use $(3)$ from this answer
          $(6)$: evaluate the sine of a complex number
          $(7)$: simplify







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 51 mins ago

























          answered 1 hour ago









          robjohn♦robjohn

          266k27305628




          266k27305628












          • $begingroup$
            Given the form of the final answer, I thought there would be a Mittag-Leffler expansion hiding in there somewhere. Nice work and thanks (+1).
            $endgroup$
            – omegadot
            37 mins ago


















          • $begingroup$
            Given the form of the final answer, I thought there would be a Mittag-Leffler expansion hiding in there somewhere. Nice work and thanks (+1).
            $endgroup$
            – omegadot
            37 mins ago
















          $begingroup$
          Given the form of the final answer, I thought there would be a Mittag-Leffler expansion hiding in there somewhere. Nice work and thanks (+1).
          $endgroup$
          – omegadot
          37 mins ago




          $begingroup$
          Given the form of the final answer, I thought there would be a Mittag-Leffler expansion hiding in there somewhere. Nice work and thanks (+1).
          $endgroup$
          – omegadot
          37 mins 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%2f3088089%2fon-the-evalution-of-an-infinite-sum%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

          Olav Thon

          Image
          Olav Thon Fødd 29. juni 1923 Ål Yrke forretningsdrivande Olav Thon på Commons Olav Thon (fødd 29. juni 1923) er ein norsk, stort sett sjølvlært forretningsmann, fødd i Ål i Hallingdal, som har gjort store pengar på særleg kjøp, bygging og utvikling av fast eigedom, mellom anna kjøpesentre, hotell og restaurantar, i både Noreg og somme andre land. Han vart Ridder av 1. klasse av St. Olavs Orden i 2003, og er som ivrig friluftsmann æresmedlem i Den Norske Turistforening etter å ha ytt store økonomiske bidrag til foreininga sine hytter og ruter i fjellet. Han er kjend for alltid å gå med ei karakteristisk strikkelue. Olav Thon i samtale med Trond Helleland på sentralbanksjefens årstale i februar 2016 I ung alder reiste Olav Thon til Newcastle for å lære seg engelsk, og han var seinare ei tid i Sveits, der han gjorde forretningar i import og eksport. Deretter tok han til å kjøpe faste eigedomar i Noreg med stor økonomisk suksess, verksemda voks snøgt og v...
          Read more

          Waikiki

          Image
          Denne artikkelen kan ha godt av ein språkvask , som reinskar opp målføringa og/eller innfører same språkstilen overalt. Denne artikkelen kan ha godt av ei opprydding for å nå ein høgare standard og/eller for å verta i tråd med standardoppsettet. Sjå korleis du redigerer ei side og stilmanualen for hjelp. Waikiki Land   USA Stat Hawaii Stad Honolulu Wikimedia Commons: Waikiki Utsyn over Waikiki Waikīkī er ein bydel i Honolulu, sør på øya Oahu, som er ei av Hawaii-øyane. Waikīkīstranda er ein av dei meste berømte strendene i verda. Waikīkī strekkjer seg langs stillehavskysten og har ei rekkje velrenommerte hotell. Bydelen er Hawaii-turismens høyborg. På Waikīkī Beach står ei statue av den lokale sportsstjerna Duka Kahanamoku. Kjelder | Denne artikkelen bygger på «Waikiki» frå Wikipedia på bokmål, den 4. november 2018. Denne artikkelen manglar kjelder eller referansar . Hjelp Wikipedia med å finna truverdige kjelder...
          Read more

          Tårekanal

          Image
          Tåreapparatet ( a ) Tårekjertel ( b ) Øvre tårepunkt ( c ) Øvre tårekanalikkel ( d ) Tåresekk ( e ) Nedre tårepunkt ( f ) Nedre tårekanalikkel ( g ) Tårekanal Tårekanalen eller nase-tårekanalen (av latin ductus nasolacrimalis ) er ein ductus som frakter overflødige tårer frå tåresekken til nasehola. Opninga inn mot nedre nasegang er delvis tildekt av ei ufullstendig slimhinnefold kalla Hasners ventil. Overflødige tårer strøymer gjennom tårekanalen og trekker inn i nedre nasegang. Dette er grunnen til at ein kan få rennande nase når ein gret, og at ein somme gonger kan kjenne saltsmaken av tårer i munnen. Både tårekanalen og tåresekken er tekt med fleirlaga sylinderepitel som inneheld slimproduserande begerceller, og er begge omgjevne av bindevev. Dakryostenose (tårekanalstenose) er ei innsnevring av tårekanalen som fører til epifora, ei overflod av tårer. Medfødd dakryostenose kan føra til cystisk utviding av tårekanalen ved at det dannast ein dakryocystocele. Personar m...
          Read more