一个无穷级数

2023-01-15 12:49 作者:艾琳娜的糖果屋 0人读过 | 我要投稿

在推导一个二重级数的时候得到了一个副产物 $%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5En%5Ccosh%20%5Cleft(%20n%5Cpi%20%5Cright)%7D%7Bn%5E2%5Csinh%20%5E2%5Cleft(%20n%5Cpi%20%5Cright)%7D%7D%3D%5Cfrac%7B%5Cpi%20%5E2%7D%7B45%7D-%5Cfrac%7BG%7D%7B3%7D%0A%0A$ ，形式比较和谐，下面来推导一下。

先考虑级数 $%0A$

$%5Csum_%7B%5Cleft(%20n%2Cm%20%5Cright)%20%5Cne%20%5Cleft(%200%2C0%20%5Cright)%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5E%7Bn%2Bm%7D%7D%7B%5Cleft(%20n%5E2%2Bm%5E2%20%5Cright)%20%5Es%7D%7D%3D%0A%5Cfrac%7B1%7D%7B%5CGamma%20%5Cleft(%20s%20%5Cright)%7D%5Cint_0%5E%7B%5Cinfty%7D%7Bt%5E%7Bs-1%7D%5Csum_%7B%5Cleft(%20n%2Cm%20%5Cright)%20%5Cne%20%5Cleft(%200%2C0%20%5Cright)%7D%7B%5Cleft(%20-1%20%5Cright)%20%5E%7Bn%2Bm%7De%5E%7B-%5Cleft(%20n%5E2%2Bm%5E2%20%5Cright)%20t%7D%7Ddt%7D%0A%0A$

$%0A%3D%5Cfrac%7B1%7D%7B%5CGamma%20%5Cleft(%20s%20%5Cright)%7D%5Cint_0%5E%7B%5Cinfty%7D%7Bt%5E%7Bs-1%7D%5Cleft(%20%5Cvartheta%20_%7B4%7D%5E%7B2%7D%5Cleft(%20e%5E%7B-t%7D%20%5Cright)%20-1%20%5Cright)%20dt%7D%5C%20%5C%20%5C%0A%5Cvartheta%20_4%5Cleft(%20q%20%5Cright)%20%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20-1%20%5Cright)%20%5Enq%5E%7Bn%5E2%7D%7D%0A%0A%0A%0A$

同时利用雅可比nd函数的傅里叶展开 $%0A%5Cmathrm%7Bnd%7D%5Cleft(%202Kv%20%5Cright)%20%3D%5Cfrac%7B%5Cpi%7D%7B2Kk%5Cprime%7D%2B%5Cfrac%7B2%5Cpi%7D%7BKk%5Cprime%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5Enq%5En%7D%7B1%2Bq%5E%7B2n%7D%7D%5Ccos%20%5Cleft(%202n%5Cpi%20v%20%5Cright)%7D$

令 $v%3D0$ 得到

$%0A%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5Enq%5En%7D%7B1%2Bq%5E%7B2n%7D%7D%7D%3D%5Cfrac%7B1%7D%7B4%7D%5Cleft(%20%5Cvartheta%20_%7B4%7D%5E%7B2%7D%5Cleft(%20q%20%5Cright)%20-1%20%5Cright)%20%0A$ ，继续带入积分，再次展开整理

$%0A%5Cfrac%7B1%7D%7B%5CGamma%20%5Cleft(%20s%20%5Cright)%7D%5Cint_0%5E%7B%5Cinfty%7D%7Bt%5E%7Bs-1%7D%5Cleft(%20%5Cvartheta%20_%7B4%7D%5E%7B2%7D%5Cleft(%20e%5E%7B-t%7D%20%5Cright)%20-1%20%5Cright)%20dt%7D%3D%5Cfrac%7B4%7D%7B%5CGamma%20%5Cleft(%20s%20%5Cright)%7D%5Cint_0%5E%7B%5Cinfty%7D%7Bt%5E%7Bs-1%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5Enq%5En%7D%7B1%2Bq%5E%7B2n%7D%7D%7Ddt%7D%0A$

$%0A%3D%5Cfrac%7B4%7D%7B%5CGamma%20%5Cleft(%20s%20%5Cright)%7D%5Cint_0%5E%7B%5Cinfty%7D%7Bt%5E%7Bs-1%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20-1%20%5Cright)%20%5Enq%5En%5Csum_%7Bk%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20-1%20%5Cright)%20%5Ekq%5E%7B2nk%7D%7D%7Ddt%7D%3D%5Cfrac%7B4%7D%7B%5CGamma%20%5Cleft(%20s%20%5Cright)%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20-1%20%5Cright)%20%5En%5Csum_%7Bk%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20-1%20%5Cright)%20%5Ek%7D%7D%5Cint_0%5E%7B%5Cinfty%7D%7Bt%5E%7Bs-1%7De%5E%7B-n%5Cleft(%201%2B2k%20%5Cright)%20t%7Ddt%7D%0A$

$%0A%3D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20-1%20%5Cright)%20%5En%5Cfrac%7B1%7D%7Bn%5Es%7D%5Csum_%7Bk%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20-1%20%5Cright)%20%5Ek%5Cfrac%7B1%7D%7B%5Cleft(%202k%2B1%20%5Cright)%20%5Es%7D%7D%7D%3D-4%5Ceta%20%5Cleft(%20s%20%5Cright)%20%5Cbeta%20%5Cleft(%20s%20%5Cright)%20%0A%0A$

于是得到：

$%0A%5Ceta%20%5Cleft(%20s%20%5Cright)%20%3D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5E%7Bn-1%7D%7D%7Bn%5Es%7D%7D%3D%5Cleft(%201-2%5E%7B1-s%7D%20%5Cright)%20%5Czeta%20%5Cleft(%20s%20%5Cright)%20%0A%0A$ ； $%0A%5Cbeta%20%5Cleft(%20s%20%5Cright)%20%3D%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5En%7D%7B%5Cleft(%202n%2B1%20%5Cright)%20%5Es%7D%7D%0A%0A$

另一方面当 $%0A%5Cmathrm%7BRe%7Ds%3E1%0A%0A$ 时

$%0A%5Csum_%7B%5Cleft(%20n%2Cm%20%5Cright)%20%5Cne%20%5Cleft(%200%2C0%20%5Cright)%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5E%7Bn%2Bm%7D%7D%7B%5Cleft(%20n%5E2%2Bm%5E2%20%5Cright)%20%5Es%7D%7D%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20-1%20%5Cright)%20%5En%5Csum_%7Bm%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5Em%7D%7B%5Cleft(%20n%5E2%2Bm%5E2%20%5Cright)%20%5Es%7D%7D%7D%0A$

$%0A%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20-1%20%5Cright)%20%5En%5Cleft(%202%5Csum_%7Bm%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5Em%7D%7B%5Cleft(%20n%5E2%2Bm%5E2%20%5Cright)%20%5Es%7D%2B%5Cfrac%7B1%7D%7Bn%5E%7B2s%7D%7D%7D%20%5Cright)%7D%3D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Csum_%7Bm%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B4%5Cleft(%20-1%20%5Cright)%20%5E%7Bm%2Bn%7D%7D%7B%5Cleft(%20n%5E2%2Bm%5E2%20%5Cright)%20%5Es%7D%7D%7D%2B2%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5En%7D%7Bn%5E%7B2s%7D%7D%7D%2B2%5Csum_%7Bm%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5Em%7D%7Bm%5E%7B2s%7D%7D%7D%0A$

所以有

$%0A%7B%20%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Csum_%7Bm%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5E%7Bm%2Bn%7D%7D%7B%5Cleft(%20n%5E2%2Bm%5E2%20%5Cright)%20%5Es%7D%7D%7D%3D%5Ceta%20%5Cleft(%202s%20%5Cright)%20-%5Ceta%20%5Cleft(%20s%20%5Cright)%20%5Cbeta%20%5Cleft(%20s%20%5Cright)%20%7D%0A%0A$ 现在考虑 $s%3D2$

以及围道积分 $%0A%5Coint%7B%5Cfrac%7B%5Cpi%20%5Ccsc%20%5Cleft(%20%5Cpi%20z%20%5Cright)%7D%7Bz%5E2%2Bx%5E2%7Ddz%7D%0A%0A$ ，路径为一无穷大的正方形，不难得到

$%0A%5Csum_%7Bm%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5Em%7D%7Bm%5E2%2Bx%5E2%7D%7D%3D%5Cfrac%7B%5Cpi%7D%7Bx%5Csinh%20%5Cleft(%20%5Cpi%20x%20%5Cright)%7D%5Cleft(%20x%3E0%20%5Cright)%20%0A%0A$

两边求导后化解有

$%0A%5Csum_%7Bm%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5Em%7D%7B%5Cleft(%20m%5E2%2Bx%5E2%20%5Cright)%20%5E2%7D%7D%3D%5Cfrac%7B%5Cpi%20%5Cleft(%20%5Cpi%20x%5Ccosh%20%5Cleft(%20%5Cpi%20x%20%5Cright)%20%2B%5Csinh%20%5Cleft(%20%5Cpi%20x%20%5Cright)%20%5Cright)%7D%7B4x%5E3%5Csinh%20%5E2%5Cleft(%20%5Cpi%20x%20%5Cright)%7D-%5Cfrac%7B1%7D%7B2x%5E4%7D%0A$

带入二重级数

$%0A%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Csum_%7Bm%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5E%7Bm%2Bn%7D%7D%7B%5Cleft(%20n%5E2%2Bm%5E2%20%5Cright)%20%5E2%7D%7D%7D%3D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20-1%20%5Cright)%20%5En%5Cleft(%20%5Cfrac%7B%5Cpi%20%5Cleft(%20%5Cpi%20n%5Ccosh%20%5Cleft(%20%5Cpi%20n%20%5Cright)%20%2B%5Csinh%20%5Cleft(%20%5Cpi%20n%20%5Cright)%20%5Cright)%7D%7B4n%5E3%5Csinh%20%5E2%5Cleft(%20%5Cpi%20n%20%5Cright)%7D-%5Cfrac%7B1%7D%7B2n%5E4%7D%20%5Cright)%7D%0A$

$%0A%3D%5Cfrac%7B%5Cpi%20%5E2%7D%7B4%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5En%5Ccosh%20%5Cleft(%20n%5Cpi%20%5Cright)%7D%7Bn%5E2%5Csinh%20%5E2%5Cleft(%20n%5Cpi%20%5Cright)%7D%7D%2B%5Cfrac%7B%5Cpi%7D%7B4%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5En%7D%7Bn%5E3%5Csinh%20%5Cleft(%20n%5Cpi%20%5Cright)%7D%7D%2B%5Cfrac%7B1%7D%7B2%7D%5Ceta%20%5Cleft(%204%20%5Cright)%20%3D%5Ceta%20%5Cleft(%204%20%5Cright)%20-%5Ceta%20%5Cleft(%202%20%5Cright)%20%5Cbeta%20%5Cleft(%202%20%5Cright)%20%0A%0A$

化解得到

$%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5En%5Ccosh%20%5Cleft(%20n%5Cpi%20%5Cright)%7D%7Bn%5E2%5Csinh%20%5E2%5Cleft(%20n%5Cpi%20%5Cright)%7D%7D%3D%5Cfrac%7B4%7D%7B%5Cpi%20%5E2%7D%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%5Ceta%20%5Cleft(%204%20%5Cright)%20-%5Ceta%20%5Cleft(%202%20%5Cright)%20%5Cbeta%20%5Cleft(%202%20%5Cright)%20-%5Cfrac%7B%5Cpi%7D%7B4%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5En%7D%7Bn%5E3%5Csinh%20%5Cleft(%20n%5Cpi%20%5Cright)%7D%7D%20%5Cright)%20%0A$

下面计算 $%0A%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5En%7D%7Bn%5E3%5Csinh%20%5Cleft(%20n%5Cpi%20%5Cright)%7D%7D%0A$

再次利用上述分式展开有

$%0A%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5En%7D%7Bn%5E3%5Csinh%20%5Cleft(%20n%5Cpi%20%5Cright)%7D%7D%3D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5En%7D%7Bn%5E3%7D%5Cleft(%20%5Cfrac%7B1%7D%7Bn%5Cpi%7D%2B%5Cfrac%7B2%7D%7B%5Cpi%7D%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bn%5Cleft(%20-1%20%5Cright)%20%5Ek%7D%7Bk%5E2%2Bn%5E2%7D%7D%20%5Cright)%7D%3D-%5Cfrac%7B%5Ceta%20%5Cleft(%204%20%5Cright)%7D%7B%5Cpi%7D%2B%5Cfrac%7B2%7D%7B%5Cpi%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5E%7Bn%2Bk%7D%7D%7Bn%5E2%5Cleft(%20k%5E2%2Bn%5E2%20%5Cright)%7D%7D%7D%0A$

$%0A%3D-%5Cfrac%7B%5Ceta%20%5Cleft(%204%20%5Cright)%7D%7B%5Cpi%7D%2B%5Cfrac%7B2%7D%7B%5Cpi%7D%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5Ek%7D%7Bk%5E2%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20-1%20%5Cright)%20%5En%5Cleft(%20%5Cfrac%7B1%7D%7Bn%5E2%7D-%5Cfrac%7B1%7D%7Bn%5E2%2Bk%5E2%7D%20%5Cright)%7D%7D$

$%3D-%5Cfrac%7B%5Ceta%20%5Cleft(%204%20%5Cright)%7D%7B%5Cpi%7D%2B%5Cfrac%7B2%7D%7B%5Cpi%7D%5Cleft(%20%5Ceta%20%5E2%5Cleft(%202%20%5Cright)%20-%5Csum_%7Bk%3D1%7D%5E%7B%5Cinfty%7D%7B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5E%7Bn%2Bk%7D%7D%7Bk%5E2%5Cleft(%20n%5E2%2Bk%5E2%20%5Cright)%7D%7D%7D%20%5Cright)%20%0A%0A%0A%3D-%5Cfrac%7B%5Ceta%20%5Cleft(%204%20%5Cright)%7D%7B%5Cpi%7D%2B%5Cfrac%7B%5Ceta%20%5E2%5Cleft(%202%20%5Cright)%7D%7B%5Cpi%7D%0A$

$%0A%3D-%5Cfrac%7B%5Cpi%20%5E3%7D%7B360%7D%0A$

于是最终我们得到

$%0A%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20-1%20%5Cright)%20%5En%5Ccosh%20%5Cleft(%20n%5Cpi%20%5Cright)%7D%7Bn%5E2%5Csinh%20%5E2%5Cleft(%20n%5Cpi%20%5Cright)%7D%7D%3D%5Cfrac%7B4%7D%7B%5Cpi%20%5E2%7D%5Cleft(%20%5Cfrac%7B7%7D%7B16%7D%5Cfrac%7B%5Cpi%20%5E4%7D%7B90%7D-%5Cfrac%7B%5Cpi%20%5E2%5Cbeta%20%5Cleft(%202%20%5Cright)%7D%7B12%7D%2B%5Cfrac%7B%5Cpi%20%5E4%7D%7B4%5Ctimes%20360%7D%20%5Cright)%20%3D%5Cfrac%7B%5Cpi%20%5E2%7D%7B45%7D-%5Cfrac%7BG%7D%7B3%7D%0A$