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













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      edited 52 mins ago







      omegadot

















      asked 1 hour ago









      omegadotomegadot

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          1 Answer
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          $begingroup$

          $$
          begin{align}newcommand{Re}{operatorname{Re}}
          &sum_{n=0}^infty(-1)^nleft[frac{2n+1/2}{(2n+1/2)^2+1}+frac{2n+3/2}{(2n+3/2)^2+1}right]tag1\
          &=frac12sum_{ninmathbb{Z}}(-1)^nleft[frac{2n+1/2}{(2n+1/2)^2+1}+frac{2n+3/2}{(2n+3/2)^2+1}right]tag2\
          &=frac14sum_{ninmathbb{Z}}(-1)^nleft[frac{n+frac14}{left(n+frac14right)^2+frac14}+frac{n+frac34}{left(n+frac34right)^2+frac14}right]tag3\
          &=frac18sum_{ninmathbb{Z}}(-1)^nleft[frac1{n+frac14-frac i2}+frac1{n+frac14+frac i2}+frac1{n+frac34-frac i2}+frac1{n+frac34+frac i2}right]tag4\
          &=frac18left[fracpi{sinleft(pi!left(frac14-frac i2right)right)}+fracpi{sinleft(pi!left(frac14+frac i2right)right)}+fracpi{sinleft(pi!left(frac34-frac i2right)right)}+fracpi{sinleft(pi!left(frac34+frac i2right)right)}right]tag5\
          &=frac{pisqrt2}8left[
          frac{coshleft(fracpi2right)+isinhleft(fracpi2right)}{cosh(pi)}+
          frac{coshleft(fracpi2right)-isinhleft(fracpi2right)}{cosh(pi)}+
          frac{coshleft(fracpi2right)-isinhleft(fracpi2right)}{cosh(pi)}right.\
          &left.phantom{=frac{pisqrt2}8}+
          frac{coshleft(fracpi2right)+isinhleft(fracpi2right)}{cosh(pi)}right]tag6\
          &=fracpi{sqrt2}frac{cosh(pi/2)}{cosh(pi)}tag7
          end{align}
          $$
          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











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          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


















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