三角函数积分一例

2022-08-07 16:26 作者:艾琳娜的糖果屋 0人读过 | 我要投稿

一个老问题，还是挺复杂的，计算 $%0A%5Cint_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%7B%5Cfrac%7Bx%5E2%5Cln%20%5Cleft(%202%5Ccos%20x%20%5Cright)%7D%7B%5Cleft(%20x%5E2%2B%5Cln%20%5E2%5Cleft(%202%5Ccos%20x%20%5Cright)%20%5Cright)%20%5E2%7Ddx%7D%0A$

首先注意到 $%0A%5Cint_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%7B%5Cfrac%7Bx%5E2%5Cln%20%5Cleft(%202%5Ccos%20x%20%5Cright)%7D%7B%5Cleft(%20x%5E2%2B%5Cln%20%5Cleft(%202%5Ccos%20x%20%5Cright)%20%5Cright)%20%5E2%7Ddx%7D%3D%5Cmathrm%7BIm%7D%5Cfrac%7B1%7D%7B2%7D%5Cint_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%7B%5Cfrac%7Bx%7D%7B%5Cln%20%5E2%5Cleft(%201%2Be%5E%7B-2ix%7D%20%5Cright)%7Ddx%7D%3D%5Cfrac%7B%5Cmathrm%7BIm%7D%7D%7B8%7D%5Cint_0%5E%7B%5Cpi%7D%7B%5Cfrac%7Bx%7D%7B%5Cln%20%5E2%5Cleft(%201%2Be%5E%7B-ix%7D%20%5Cright)%7Ddx%7D%3D-%5Cfrac%7B%5Cmathrm%7BIm%7D%7D%7B8%7D%5Cint_L%7B%5Cfrac%7B%5Cln%20z%7D%7Bz%5Cln%20%5E2%5Cleft(%201%2B%5Cfrac%7B1%7D%7Bz%7D%20%5Cright)%7Ddz%7D%0A%0A$

这里的路径为 $%0AL%3AC_%7B%5Cdelta%7D%5Crightarrow%201-%5Cdelta%20%5Crightarrow%201%5Crightarrow%20C_%7BR%3D1%7D%5Crightarrow%20C_%7B%5Cepsilon%7D%5Crightarrow%201-%5Cepsilon%20%5Crightarrow%201-%5Cdelta%20%5Crightarrow%20C_%7B%5Cdelta%7D%0A%0A$

根据柯西积分定理有 $%0A%5Cint_C%7B%5Cfrac%7B%5Cln%20z%7D%7Bz%5Cln%20%5E2%5Cleft(%201%2B%5Cfrac%7B1%7D%7Bz%7D%20%5Cright)%7Ddz%7D%2B%5Cint_0%5E1%7B%5Cfrac%7B%5Cln%20x%7D%7Bx%5Cln%20%5E2%5Cleft(%201%2B%5Cfrac%7B1%7D%7Bx%7D%20%5Cright)%7Ddx%7D%2B%5Cint_1%5E0%7B%5Cfrac%7B%5Cln%20x%2Bi%5Cpi%7D%7B-x%5Cleft(%20%5Cln%20%5Cleft(%201-x%20%5Cright)%20-%5Cln%20x-i%5Cpi%20%5Cright)%20%5E2%7D-dx%7D%3D0%0A%0A%0A$

化解一下就可以得到 $%0A%5Cmathrm%7BIm%7D%5Cint_C%7B%5Cfrac%7B%5Cln%20z%7D%7Bz%5Cln%20%5E2%5Cleft(%201%2B%5Cfrac%7B1%7D%7Bz%7D%20%5Cright)%7Ddz%7D%3D%5Cmathrm%7BIm%7D%5Cint_0%5E1%7B%5Cfrac%7B%5Cln%20x%2Bi%5Cpi%7D%7Bx%5Cleft(%20%5Cln%20%5Cleft(%20%5Cfrac%7B1-x%7D%7Bx%7D%20%5Cright)%20-i%5Cpi%20%5Cright)%20%5E2%7Ddx%7D%3D%5Cmathrm%7BIm%7D%5Cint_0%5E1%7B%5Cfrac%7B%5Cleft(%20%5Cln%20x%2Bi%5Cpi%20%5Cright)%20%5Cleft(%20%5Cln%20%5Cleft(%20%5Cfrac%7B1-x%7D%7Bx%7D%20%5Cright)%20%2Bi%5Cpi%20%5Cright)%20%5E2%7D%7Bx%5Cleft(%20%5Cln%20%5E2%5Cleft(%20%5Cfrac%7B1-x%7D%7Bx%7D%20%5Cright)%20%2B%5Cpi%20%5E2%20%5Cright)%20%5E2%7Ddx%7D%3D%5Cint_0%5E1%7B%5Cfrac%7B2%5Cpi%20%5Cln%20x%5Cln%20%5Cleft(%20%5Cfrac%7B1-x%7D%7Bx%7D%20%5Cright)%20%2B%5Cpi%20%5Cleft(%20%5Cln%20%5E2%5Cleft(%20%5Cfrac%7B1-x%7D%7Bx%7D%20%5Cright)%20-%5Cpi%20%5E2%20%5Cright)%7D%7Bx%5Cleft(%20%5Cln%20%5E2%5Cleft(%20%5Cfrac%7B1-x%7D%7Bx%7D%20%5Cright)%20%2B%5Cpi%20%5E2%20%5Cright)%20%5E2%7Ddx%7D%5Cxrightarrow%7B%5Cfrac%7B1-x%7D%7Bx%7D%3Du%7D%0A%5C%5C%0A%3D-2%5Cpi%20%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cln%20%5Cleft(%201%2Bu%20%5Cright)%20%5Cln%20u%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%20%5E2%7Ddx%7D-%5Cpi%20%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20%5Cln%20%5E2u-%5Cpi%20%5E2%20%5Cright)%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%20%5E2%7Ddx%7D%0A%0A$ $%0A$

首先计算后面那个积分 $%0A%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20%5Cln%20%5E2u-%5Cpi%20%5E2%20%5Cright)%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%20%5E2%7Ddx%7D%3D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20%5Cln%20%5E2u-%5Cpi%20%5E2%20%5Cright)%7D%7Bu%5Cleft(%201%2Bu%20%5Cright)%20%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%20%5E2%7Ddx%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cln%20%5E2u-%5Cpi%20%5E2%7D%7Bu%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%20%5E2%7Ddu%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bx%5E2-%5Cpi%20%5E2%7D%7B%5Cleft(%20x%5E2%2B%5Cpi%20%5E2%20%5Cright)%20%5E2%7Ddx%7D%3D0%0A%0A$

其次对于前一部分可以先分部积分处理一下 $%0A%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cln%20%5Cleft(%201%2Bu%20%5Cright)%20%5Cln%20u%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%20%5E2%7Ddx%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bu%5Cln%20%5Cleft(%201%2Bu%20%5Cright)%7D%7B%5Cleft(%201%2Bu%20%5Cright)%7Dd%5Cleft(%20-%5Cfrac%7B1%7D%7B%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7D%20%5Cright)%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B1%7D%7B%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7Dd%5Cleft(%20%5Cfrac%7Bu%5Cln%20%5Cleft(%201%2Bu%20%5Cright)%7D%7B%5Cleft(%201%2Bu%20%5Cright)%7D%20%5Cright)%7D%0A%0A%0A%0A%0A$

$%3D%5Cfrac%7B1%7D%7B2%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20%5Cln%20%5Cleft(%201%2Bu%20%5Cright)%20%2B%5Cfrac%7Bu%7D%7B1%2Bu%7D%20%5Cright)%20%5Cleft(%201%2Bu%20%5Cright)%20-u%5Cln%20%5Cleft(%201%2Bu%20%5Cright)%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5E2%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7Ddu%7D%0A%0A%3D%5Cfrac%7B1%7D%7B2%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cln%20%5Cleft(%201%2Bu%20%5Cright)%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5E2%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7Ddu%7D%2B%5Cfrac%7B1%7D%7B2%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bu%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5E2%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7Ddu%7D%5C%2C%5C%2C%20%0A$

再利用一个常用结论 $%0A%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B1%7D%7B%5Cleft(%20a%2Bu%20%5Cright)%20%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7Ddu%7D%3D%5Cfrac%7B1%7D%7B%5Cln%20a%7D-%5Cfrac%7B1%7D%7Ba-1%7D%5Cleft(%20a%3E0%20%5Cright)%20%0A%0A$ 可以方便的计算出后面的积分 $%0A%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bu%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5E2%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7Ddu%7D%3D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B1%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7Ddu%7D-%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B1%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5E2%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7Ddu%7D$

$%3D%5Clim_%7Ba%5Crightarrow%201%7D%20%5Cleft%5C%7B%20%5Cfrac%7B1%7D%7B%5Cln%20a%7D-%5Cfrac%7B1%7D%7Ba-1%7D%2B%5Cfrac%7Bd%7D%7Bda%7D%5Cleft(%20%5Cfrac%7B1%7D%7B%5Cln%20a%7D-%5Cfrac%7B1%7D%7Ba-1%7D%20%5Cright)%20%5Cright%5C%7D%20%3D%5Cfrac%7B1%7D%7B2%7D-%5Cfrac%7B1%7D%7B12%7D%3D%5Cfrac%7B5%7D%7B12%7D%0A%0A$

对于前一部分，设置参数 $a$ ， $%0AJ%5Cleft(%20a%20%5Cright)%20%3D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cln%20%5Cleft(%201%2Bau%20%5Cright)%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5E2%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7Ddu%7D%5C%2C%5C%2CJ%5Cleft(%200%20%5Cright)%20%3D0%0A%0A$

$J'%5Cleft(%20a%20%5Cright)%20%3D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bu%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5E2%5Cleft(%201%2Bau%20%5Cright)%20%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7Ddu%7D%0A%0A$ ，部分分式拆分一下有

$%0A%3D-%5Cfrac%7Ba%7D%7B%5Cleft(%201-a%20%5Cright)%20%5E2%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bdu%7D%7B%5Cleft(%201%2Bau%20%5Cright)%20%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7D%7D%2B%5Cfrac%7B1%7D%7B%5Cleft(%201-a%20%5Cright)%20%5E2%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bdu%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7D%7D-%5Cfrac%7B1%7D%7B1-a%7D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bdu%7D%7B%5Cleft(%201%2Bu%20%5Cright)%20%5E2%5Cleft(%20%5Cln%20%5E2u%2B%5Cpi%20%5E2%20%5Cright)%7D%7D%0A%0A$

$%0A%3D%5Cfrac%7B1%7D%7B%5Cleft(%201-a%20%5Cright)%20%5E2%5Cln%20a%7D%2B%5Cfrac%7Ba%7D%7B%5Cleft(%201-a%20%5Cright)%20%5E3%7D%2B%5Cfrac%7B1%7D%7B2%5Cleft(%201-a%20%5Cright)%20%5E2%7D-%5Cfrac%7B1%7D%7B12%5Cleft(%201-a%20%5Cright)%7D%5C%2C%5C%2C%20%0A$

于是可以得到 $%0AJ%3DJ%5Cleft(%201%20%5Cright)%20%3D%5Cint_0%5E1%7B%5Cleft(%20%5Cfrac%7B1%7D%7B%5Cleft(%201-a%20%5Cright)%20%5E2%5Cln%20a%7D%2B%5Cfrac%7Ba%7D%7B%5Cleft(%201-a%20%5Cright)%20%5E3%7D%2B%5Cfrac%7B1%7D%7B2%5Cleft(%201-a%20%5Cright)%20%5E2%7D-%5Cfrac%7B1%7D%7B12%5Cleft(%201-a%20%5Cright)%7D%20%5Cright)%20da%7D%0A%0A$

$%0A%3D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cleft(%20-%5Cfrac%7B1%7D%7B%5Cleft(%201-e%5E%7B-x%7D%20%5Cright)%20%5E2x%7D%2B%5Cfrac%7Be%5E%7B-x%7D%7D%7B%5Cleft(%201-e%5E%7B-x%7D%20%5Cright)%20%5E3%7D%2B%5Cfrac%7B1%7D%7B2%5Cleft(%201-e%5E%7B-x%7D%20%5Cright)%20%5E2%7D-%5Cfrac%7B1%7D%7B12%5Cleft(%201-e%5E%7B-x%7D%20%5Cright)%7D%20%5Cright)%20e%5E%7B-x%7Ddx%7D%0A%0A$

为了求得结果，依然是设置参数 $%0AI%5Cleft(%20s%20%5Cright)%20%3D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cleft(%20-%5Cfrac%7B1%7D%7B%5Cleft(%201-e%5E%7B-x%7D%20%5Cright)%20%5E2x%7D%2B%5Cfrac%7Be%5E%7B-x%7D%7D%7B%5Cleft(%201-e%5E%7B-x%7D%20%5Cright)%20%5E3%7D%2B%5Cfrac%7B1%7D%7B2%5Cleft(%201-e%5E%7B-x%7D%20%5Cright)%20%5E2%7D-%5Cfrac%7B1%7D%7B12%5Cleft(%201-e%5E%7B-x%7D%20%5Cright)%7D%20%5Cright)%20e%5E%7B-x%7Dx%5Esdx%7D%0A%0A$

$%0A%3D%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cleft(%20-%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7Bne%5E%7B-nx%7Dx%5E%7Bs-1%7D%7D%2B%5Cfrac%7B1%7D%7B2%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7Bn%5Cleft(%20n%2B1%20%5Cright)%20e%5E%7B-%5Cleft(%20n%2B1%20%5Cright)%20x%7Dx%5Es%7D%2B%5Cfrac%7B1%7D%7B2%7D%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7Bne%5E%7B-nx%7Dx%5Es%7D-%5Cfrac%7B1%7D%7B12%7D%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7Be%5E%7B-%5Cleft(%20n%2B1%20%5Cright)%20x%7Dx%5Es%7D%20%5Cright)%20dx%7D%0A%0A$

$%0A%3D-%5CGamma%20%5Cleft(%20s%20%5Cright)%20%5Czeta%20%5Cleft(%20s-1%20%5Cright)%20%2B%5Cfrac%7B1%7D%7B2%7D%5CGamma%20%5Cleft(%20s%2B1%20%5Cright)%20%5Cleft(%20%5Czeta%20%5Cleft(%20s-1%20%5Cright)%20-%5Czeta%20%5Cleft(%20s%20%5Cright)%20%5Cright)%20%2B%5Cfrac%7B1%7D%7B2%7D%5CGamma%20%5Cleft(%20s%2B1%20%5Cright)%20%5Czeta%20%5Cleft(%20s%20%5Cright)%20-%5Cfrac%7B1%7D%7B12%7D%5CGamma%20%5Cleft(%20s%2B1%20%5Cright)%20%5Czeta%20%5Cleft(%201%2Bs%20%5Cright)%20%0A%0A$

$%0A%3D%5Czeta%20%5Cleft(%20s-1%20%5Cright)%20%5CGamma%20%5Cleft(%20s%20%5Cright)%20%5Cleft(%20%5Cfrac%7Bs%7D%7B2%7D-1%20%5Cright)%20-%5Cfrac%7B1%7D%7B12%7D%5CGamma%20%5Cleft(%20s%2B1%20%5Cright)%20%5Czeta%20%5Cleft(%201%2Bs%20%5Cright)%20%0A%0A$

再利用以下泰勒展开和洛朗展开 $%5Czeta%20%5Cleft(%20s-1%20%5Cright)%20%3D-%5Cfrac%7B1%7D%7B12%7D%2B%5Cleft(%20%5Cfrac%7B1%7D%7B12%7D-%5Cln%20A%20%5Cright)%20s%2BO%5Cleft(%20s%20%5Cright)%20%5C%2C%5C%2C%5CGamma%20%5Cleft(%20s%20%5Cright)%20%3D%5Cfrac%7B1%7D%7Bs%7D-%5Cgamma%20%2BO%5Cleft(%20s%20%5Cright)%20%5C%2C%5C%2C%5Czeta%20%5Cleft(%201%2Bs%20%5Cright)%20%3D%5Cfrac%7B1%7D%7Bs%7D%2B%5Cgamma%20%2BO%5Cleft(%20s%20%5Cright)%20%5C%2C%5C%2C%5Cleft(%20s%5Crightarrow%200%20%5Cright)%20%0A%0A$

（A是 $%0AGlaisher-Kinkelin%5C%2C%5C%2CConstant%0A$ ）不难得到

$%0A%5Czeta%20%5Cleft(%20s-1%20%5Cright)%20%5CGamma%20%5Cleft(%20s%20%5Cright)%20%5Cleft(%20%5Cfrac%7Bs%7D%7B2%7D-1%20%5Cright)%20-%5Cfrac%7B1%7D%7B12%7D%5CGamma%20%5Cleft(%20s%2B1%20%5Cright)%20%5Czeta%20%5Cleft(%201%2Bs%20%5Cright)%20%3D%5Cleft(%20%5Cln%20A-%5Cfrac%7B%5Cgamma%7D%7B12%7D-%5Cfrac%7B1%7D%7B8%7D%20%5Cright)%20%2BO%5Cleft(%20s%20%5Cright)%20%0A$

令 $s%5Crightarrow%200$ 我们就得到了 $%0A%5Cint_0%5E%7B%5Cinfty%7D%7B%5Cleft(%20-%5Cfrac%7B1%7D%7B%5Cleft(%201-e%5E%7B-x%7D%20%5Cright)%20%5E2x%7D%2B%5Cfrac%7Be%5E%7B-x%7D%7D%7B%5Cleft(%201-e%5E%7B-x%7D%20%5Cright)%20%5E3%7D%2B%5Cfrac%7B1%7D%7B2%5Cleft(%201-e%5E%7B-x%7D%20%5Cright)%20%5E2%7D-%5Cfrac%7B1%7D%7B12%5Cleft(%201-e%5E%7B-x%7D%20%5Cright)%7D%20%5Cright)%20e%5E%7B-x%7Ddx%7D%3D%5Cln%20A-%5Cfrac%7B%5Cgamma%7D%7B12%7D-%5Cfrac%7B1%7D%7B8%7D%5C%2C%5C%2C%20%20%20%0A%0A$

整理一下上述结果就有

$%0A%5Cint_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%7B%5Cfrac%7Bx%5E2%5Cln%20%5Cleft(%202%5Ccos%20x%20%5Cright)%7D%7B%5Cleft(%20x%5E2%2B%5Cln%20%5E2%5Cleft(%202%5Ccos%20x%20%5Cright)%20%5Cright)%20%5E2%7Ddx%7D%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%5Cleft%5C%7B%20%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20%5Cln%20A-%5Cfrac%7B%5Cgamma%7D%7B12%7D-%5Cfrac%7B1%7D%7B8%7D%20%5Cright)%20%2B%5Cfrac%7B5%7D%7B24%7D%20%5Cright%5C%7D%20%3D%5Cfrac%7B%5Cpi%20%5Cln%20A%7D%7B8%7D-%5Cfrac%7B%5Cpi%20%5Cgamma%7D%7B96%7D%2B%5Cfrac%7B7%5Cpi%7D%7B192%7D%0A%0A$

这个计算也是用到很多技巧了，分部积分、倒代换、参数积分求导、围道积分、渐近展开都给用到了，还是很有代表性的。

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