While studying Fourier analysis last semester, I saw an interesting identity: ∑ n = 1 ∞ 1 n 2 − α 2 = 1 2 α 2 − π 2 α tan ⁡ π α whenever α ∈ C ∖ Z , which I learned two proofs using Fourier series and residue calculus.More explicitly, we can deduce the theorem using Fourier series of f ( θ ) = e i ( π − θ ) α on [ 0 , 2 π ] or contour integral of the function g ( z ) = π ( z 2 − α 2 ) tan ⁡ π z along large circles.But these techniques, as long as I know, wasn't fully developed at Euler's time.So what was Euler's method to prove this identity? Is there any proof at elementary level?

Question

While studying Fourier analysis last semester, I saw an interesting identity:      ∑          n      =      1              ∞            1                  n        2            −              α        2              =      1          2              α        2              −      π          2      α      tan      ⁡      π      α      whenever   α  ∈      C    ∖      Z  , which I learned two proofs using Fourier series and residue calculus.More explicitly, we can deduce the theorem using Fourier series of   f  (  θ  )  =      e          i      (      π      −      θ      )      α       on   [  0  ,  2  π  ] or contour integral of the function   g  (  z  )  =      π          (              z        2            −              α        2            )      tan      ⁡      π      z       along large circles.But these techniques, as long as I know, wasn&#039;t fully developed at Euler&#039;s time.So what was Euler&#039;s method to prove this identity? Is there any proof at elementary level?

Jazmin Cameron · Accepted Answer

Euler was the first to give a representation of the sine function as an infinite product:  (  ∗  )    sin  ⁡  (  π  α  )  =  π  α      ∏          n      =      1              ∞        (                    n        2            −              α        2                    n      2        )  ,which was formally proved by Weierstrass about 100 years later.Now taking &quot;  ln&quot; on by sides of (*) gives  ln  ⁡  (  sin  ⁡  (  π  α  )  )  =  ln  ⁡  (  π  α  )  +      ∑          n      =      1              ∞        ln  ⁡  (                    n        2            −              α        2                    n      2        )  ,and after taking derivatives on both sides we arrive at      ∑          n      =      1              ∞            1                  n        2            −              α        2              =      1          2              α        2              −      π          2      α      tan      ⁡      π      α        .

Jaxson Mack · Accepted Answer

detailed solution path  π  cot  ⁡  π  z  =      1    z    +      ∑          n      ≥      1            (          1              z        −        n              +          1              z        +        n              )    =      lim          k      →      ∞            ∑          n      =      −      k        k        1          z      −      n      hence                              ∑                      n            =            1                    ∞                          1                                    n              2                        −                          α              2                                                          =                  lim                      k            →            ∞                                    ∑                      n            =            1                    k                          1                                    n              2                        −                          α              2                                                                        =        −                  1                      2            α                                    lim                      k            →            ∞                                    ∑                      n            =            1                    k                          1                      α            −            n                          +                  1                      α            +            n                                                            =        −                  1                      2            α                                    lim                      k            →            ∞                                    (          −                      1            α                    +                      ∑                          n              =              −              k                        k                                1                          α              −              n                                )                                                  =        −                  1                      2            α                                    (          π          cot          ⁡          π          α          −                      1            α                    )                                                  =                              1            −            π            α            cot            ⁡            π            α                                2                          α              2                                                                        =                  1                      2                          α              2                                      −                  π                      2            α            tan            ⁡            π            α

While studying Fourier analysis last semester, I saw an interesting identity: sim_(n=1)^infty(1)/(n^2−alpha^2)=1/(2alpha^2)−pi/(2alpha tan pi alpha) whenever alpha in CC∖ZZ, which I learned two proofs using Fourier series and residue calculus.

Answered question

Answer & Explanation

New Questions in Integral Calculus