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简单搞个积

2023-03-26 20:52 作者:艾琳娜的糖果屋  | 我要投稿

我们来计算两个超几何函数的积分%5Cint_0%5E%7B%5Cinfty%7D%7B_2F_1%5Cleft(%20%5Cfrac%7B3%7D%7B4%7D%2C%5Cfrac%7B5%7D%7B6%7D%3B1%3B-x%20%5Cright)%20%5E2%5Cmathrm%7Bd%7Dx%7D%20%5C%5C%0A%5Cint_0%5E%7B%5Cinfty%7D%7B_3F_2%5Cleft(%20%5Cfrac%7B5%7D%7B8%7D%2C%5Cfrac%7B5%7D%7B8%7D%2C%5Cfrac%7B9%7D%7B8%7D%3B%5Cfrac%7B1%7D%7B2%7D%2C%5Cfrac%7B13%7D%7B8%7D%3B-x%20%5Cright)%20%5E2%5Cfrac%7B%5Cmathrm%7Bd%7Dx%7D%7B%5Csqrt%7Bx%7D%7D%7D%0A%0A%0A%0A%0A

先看第一个,因为是无穷区间上的广义积分而且还是平方的形式,可以考虑积分变换里面的卷积定理,这里我们采用梅林变换,因为超几何函数的梅林变换是很容易得到的。

%0A%5Cint_0%5E%7B%5Cinfty%7D%7B%7Bx%5E%7Bs-1%7D%7D_2F_1%5Cleft(%20a%2Cb%3Bc%3B-x%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%3D%5Cfrac%7B%5CGamma%20%5Cleft(%20c%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20a%20%5Cright)%20%5CGamma%20%5Cleft(%20b%20%5Cright)%7D%5Cfrac%7B%5CGamma%20%5Cleft(%20s%20%5Cright)%20%5CGamma%20%5Cleft(%20a-s%20%5Cright)%20%5CGamma%20%5Cleft(%20b-s%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20c-s%20%5Cright)%7D%5C%2C%5C%2C%5Cleft(%200%3C%5Cmathrm%7BRe%7Ds%3C%5Cmin%20%5Cleft%5C%7B%20%5Cmathrm%7BRe%7Da%2C%5Cmathrm%7BRe%7Db%20%5Cright%5C%7D%20%5Cright)%20%0A%0A

这梅林变换利用2F1的积分表达或者拉马努金主定理就能得到。再利用梅林变换的卷积定理有

%0A%5Cint_0%5E%7B%5Cinfty%7D%7B%7Bx%5E%7Bs-1%7D%7D_2F_1%5Cleft(%20a%2Cb%3Bc%3B-x%20%5Cright)%20%5E2%5Cmathrm%7Bd%7Dx%7D%5C%5C%3D%5Cfrac%7B1%7D%7B2%5Cpi%20i%7D%5Cint_%7B%5Ceta%20-i%5Cinfty%7D%5E%7B%5Ceta%20%2Bi%5Cinfty%7D%7B%5Cleft(%20%5Cfrac%7B%5CGamma%20%5Cleft(%20c%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20a%20%5Cright)%20%5CGamma%20%5Cleft(%20b%20%5Cright)%7D%20%5Cright)%20%5E2%5Cfrac%7B%5CGamma%20%5Cleft(%20%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20a-%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20b-%5Cmu%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20c-%5Cmu%20%5Cright)%7D%5Cfrac%7B%5CGamma%20%5Cleft(%20s-%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20a-s%2B%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20b-s%2B%5Cmu%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20c-s%2B%5Cmu%20%5Cright)%7D%5Cmathrm%7Bd%7D%5Cmu%7D%0A%0A%0A

令s=c得到

%0A%5Cint_0%5E%7B%5Cinfty%7D%7B%7Bx%5E%7Bc-1%7D%7D_2F_1%5Cleft(%20a%2Cb%3Bc%3B-x%20%5Cright)%20%5E2%5Cmathrm%7Bd%7Dx%7D%3D%5Cleft(%20%5Cfrac%7B%5CGamma%20%5Cleft(%20c%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20a%20%5Cright)%20%5CGamma%20%5Cleft(%20b%20%5Cright)%7D%20%5Cright)%20%5E2%5Cfrac%7B1%7D%7B2%5Cpi%20i%7D%5Cint_%7B%5Ceta%20-i%5Cinfty%7D%5E%7B%5Ceta%20%2Bi%5Cinfty%7D%7B%5CGamma%20%5Cleft(%20a-c%2B%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20b-c%2B%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20a-%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20b-%5Cmu%20%5Cright)%20%5Cmathrm%7Bd%7D%5Cmu%7D%0A%5C%5C%0A%3D%5Cleft(%20%5Cfrac%7B%5CGamma%20%5Cleft(%20c%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20a%20%5Cright)%20%5CGamma%20%5Cleft(%20b%20%5Cright)%7D%20%5Cright)%20%5E2%5Cfrac%7B%5CGamma%20%5Cleft(%20a-c%2Ba%20%5Cright)%20%5CGamma%20%5Cleft(%20a-c%2Bb%20%5Cright)%20%5CGamma%20%5Cleft(%20b-c%2Ba%20%5Cright)%20%5CGamma%20%5Cleft(%20b-c%2Bb%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20a-c%2Bb-c%2Ba%2Bb%20%5Cright)%7D%0A%5C%5C%0A%3D%5Cleft(%20%5Cfrac%7B%5CGamma%20%5Cleft(%20c%20%5Cright)%20%5CGamma%20%5Cleft(%20a%2Bb-c%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20a%20%5Cright)%20%5CGamma%20%5Cleft(%20b%20%5Cright)%7D%20%5Cright)%20%5E2%5Cfrac%7B%5CGamma%20%5Cleft(%202a-c%20%5Cright)%20%5CGamma%20%5Cleft(%202b-c%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%202a%2B2b-2c%20%5Cright)%7D%0A%0A%0A%5Cleft(%20a-b%5Cnotin%20Z%20%5Cright)%20%0A%0A

带入数据得到第一题的结果

%0A%5Cint_0%5E%7B%5Cinfty%7D%7B_2F_1%5Cleft(%20%5Cfrac%7B3%7D%7B4%7D%2C%5Cfrac%7B5%7D%7B6%7D%3B1%3B-x%20%5Cright)%20%5E2%5Cmathrm%7Bd%7Dx%7D%3D%5Cfrac%7B3%5Csqrt%5B4%5D%7B3%7D%5Cleft(%20%5Csqrt%7B3%7D-1%20%5Cright)%20%5CGamma%20%5E3%5Cleft(%20%5Cfrac%7B1%7D%7B3%7D%20%5Cright)%7D%7B2%5E%7B%5Cfrac%7B5%7D%7B6%7D%7D%5Cpi%20%5E2%7D%0A%0A

再来看3F2的这个积分,难度提升了不少,首先由拉马努金主定理给出梅林变换

%0A%5Cint_0%5E%7B%5Cinfty%7D%7B%7Bx%5E%7Bs-1%7D%7D_3F_2%5Cleft(%20%5Cfrac%7B5%7D%7B8%7D%2C%5Cfrac%7B5%7D%7B8%7D%2C%5Cfrac%7B9%7D%7B8%7D%3B%5Cfrac%7B1%7D%7B2%7D%2C%5Cfrac%7B13%7D%7B8%7D%3B-x%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%3D%5Cfrac%7B%5CGamma%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B13%7D%7B8%7D%20%5Cright)%7D%7B%5CGamma%20%5E2%5Cleft(%20%5Cfrac%7B5%7D%7B8%7D%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B9%7D%7B8%7D%20%5Cright)%7D%5Cfrac%7B%5CGamma%20%5Cleft(%20s%20%5Cright)%20%5CGamma%20%5E2%5Cleft(%20%5Cfrac%7B5%7D%7B8%7D-s%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B9%7D%7B8%7D-s%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D-s%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B13%7D%7B8%7D-s%20%5Cright)%7D%5Cleft(%200%3C%5Cmathrm%7BRe%7Ds%3C%5Cfrac%7B5%7D%7B8%7D%20%5Cright)%20%0A

那么利用卷积定理我们有

%0A%5Cint_0%5E%7B%5Cinfty%7D%7B_3F_2%5Cleft(%20%5Cfrac%7B5%7D%7B8%7D%2C%5Cfrac%7B5%7D%7B8%7D%2C%5Cfrac%7B9%7D%7B8%7D%3B%5Cfrac%7B1%7D%7B2%7D%2C%5Cfrac%7B13%7D%7B8%7D%3B-x%20%5Cright)%20%5E2%5Cfrac%7B%5Cmathrm%7Bd%7Dx%7D%7B%5Csqrt%7Bx%7D%7D%7D%5C%5C%0A%5Cleft(%20%5Cfrac%7B%5CGamma%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B13%7D%7B8%7D%20%5Cright)%7D%7B%5CGamma%20%5E2%5Cleft(%20%5Cfrac%7B5%7D%7B8%7D%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B9%7D%7B8%7D%20%5Cright)%7D%20%5Cright)%20%5E2%5Cfrac%7B1%7D%7B2%5Cpi%20i%7D%5Cint_%7B%5Cgamma%20-i%5Cinfty%7D%5E%7B%5Cgamma%20%2Bi%5Cinfty%7D%7B%5Cfrac%7B%5CGamma%20%5Cleft(%20%5Cmu%20%5Cright)%20%5CGamma%20%5E2%5Cleft(%20%5Cfrac%7B5%7D%7B8%7D-%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B9%7D%7B8%7D-%5Cmu%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D-%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B13%7D%7B8%7D-%5Cmu%20%5Cright)%7D%5Cfrac%7B%5CGamma%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D-%5Cmu%20%5Cright)%20%5CGamma%20%5E2%5Cleft(%20%5Cmu%20%2B%5Cfrac%7B1%7D%7B8%7D%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B5%7D%7B8%7D%2B%5Cmu%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cmu%20%2B%5Cfrac%7B9%7D%7B8%7D%20%5Cright)%7D%5Cmathrm%7Bd%7D%5Cmu%7D%5Cleft(%200%3C%5Cgamma%20%3C%5Cfrac%7B4%7D%7B8%7D%20%5Cright)%20%0A%5C%5C%0A%3D%5Cfrac%7B25%7D%7B2%5Csqrt%7B2%7D%5CGamma%20%5E2%5Cleft(%20%5Cfrac%7B1%7D%7B4%7D%20%5Cright)%7D%5Cfrac%7B32%5Csqrt%7B2%7D%7D%7Bi%7D%5Cint_%7B%5Cgamma%20-i%5Cinfty%7D%5E%7B%5Cgamma%20%2Bi%5Cinfty%7D%7B%5Cfrac%7B%5CGamma%20%5Cleft(%20%5Cfrac%7B5%7D%7B4%7D-2%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B1%7D%7B4%7D%2B2%5Cmu%20%5Cright)%7D%7B%5Cleft(%205-8%5Cmu%20%5Cright)%20%5Cleft(%201%2B8%5Cmu%20%5Cright)%7D%5Cmathrm%7Bd%7D%5Cmu%7D%0A%0A%0A%0A%0A

该问题的关键是如何计算这无穷高直线上的积分,也许会想到像处理barnes积分一样,但是这个地方不好计算,但是注意到γ的选取是任意的,通过待定系数法很容易找到最合适的γ=1/4

于是积分可写为

%0A%5Cfrac%7B25%7D%7B2%5Csqrt%7B2%7D%5CGamma%20%5E2%5Cleft(%20%5Cfrac%7B1%7D%7B4%7D%20%5Cright)%7D%5Cfrac%7B32%5Csqrt%7B2%7D%7D%7Bi%7D%5Cint_%7B%5Cfrac%7B1%7D%7B4%7D-i%5Cinfty%7D%5E%7B%5Cfrac%7B1%7D%7B4%7D%2Bi%5Cinfty%7D%7B%5Cfrac%7B%5CGamma%20%5Cleft(%20%5Cfrac%7B5%7D%7B4%7D-2%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B1%7D%7B4%7D%2B2%5Cmu%20%5Cright)%7D%7B%5Cleft(%205-8%5Cmu%20%5Cright)%20%5Cleft(%201%2B8%5Cmu%20%5Cright)%7D%5Cmathrm%7Bd%7D%5Cmu%7D%3D%0A%5Cfrac%7B200%7D%7B%5CGamma%20%5E2%5Cleft(%20%5Cfrac%7B1%7D%7B4%7D%20%5Cright)%7D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5CGamma%20%5Cleft(%20%5Cfrac%7B3%7D%7B4%7D-iy%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B3%7D%7B4%7D%2Biy%20%5Cright)%7D%7B9%2B16y%5E2%7D%5Cmathrm%7Bd%7Dy%7D%0A

%0A%3D%5Cfrac%7B25%5CGamma%20%5Cleft(%20%5Cfrac%7B3%7D%7B2%7D%20%5Cright)%7D%7B2%5CGamma%20%5E2%5Cleft(%20%5Cfrac%7B1%7D%7B4%7D%20%5Cright)%7D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7B%5Cleft(%20%5Cfrac%7Bx%7D%7B1-x%7D%20%5Cright)%20%5E%7Biy%7D%7D%7B%5Cfrac%7B9%7D%7B16%7D%2By%5E2%7D%5Cmathrm%7Bd%7Dy%5Cint_0%5E1%7Bx%5E%7B-%5Cfrac%7B1%7D%7B4%7D%7D%5Cleft(%201-x%20%5Cright)%20%5E%7B-%5Cfrac%7B1%7D%7B4%7D%7D%5Cmathrm%7Bd%7Dx%7D%7D%3D%5Cfrac%7B25%5CGamma%20%5Cleft(%20%5Cfrac%7B3%7D%7B2%7D%20%5Cright)%7D%7B2%5CGamma%20%5E2%5Cleft(%20%5Cfrac%7B1%7D%7B4%7D%20%5Cright)%7D%5Cfrac%7B4%5Cpi%7D%7B3%7D%5Cint_0%5E1%7Bx%5E%7B-%5Cfrac%7B1%7D%7B4%7D%7D%5Cleft(%201-x%20%5Cright)%20%5E%7B-%5Cfrac%7B1%7D%7B4%7D%7De%5E%7B-%5Cfrac%7B3%7D%7B4%7D%5Cleft%7C%20%5Cln%20%5Cfrac%7Bx%7D%7B1-x%7D%20%5Cright%7C%7D%5Cmathrm%7Bd%7Dx%7D%0A%0A

%0A%3D%5Cfrac%7B25%5Cpi%20%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B3%5CGamma%20%5E2%5Cleft(%20%5Cfrac%7B1%7D%7B4%7D%20%5Cright)%7D%5Cleft(%20%5Cint_0%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7Bx%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5Cleft(%201-x%20%5Cright)%20%5E%7B-1%7D%5Cmathrm%7Bd%7Dx%7D%2B%5Cint_%7B%5Cfrac%7B1%7D%7B2%7D%7D%5E1%7Bx%5E%7B-1%7D%5Cleft(%201-x%20%5Cright)%20%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5Cmathrm%7Bd%7Dx%7D%20%5Cright)%20%3D%5Cfrac%7B50%5Cpi%20%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B3%5CGamma%20%5E2%5Cleft(%20%5Cfrac%7B1%7D%7B4%7D%20%5Cright)%7D%5Cint_0%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%7B%5Cfrac%7B%5Csqrt%7Bx%7D%7D%7B1-x%7D%5Cmathrm%7Bd%7Dx%7D%0A%0A

%0A%3D%5Cfrac%7B50%5Cpi%20%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B3%5CGamma%20%5E2%5Cleft(%20%5Cfrac%7B1%7D%7B4%7D%20%5Cright)%7D%5Cleft(%202%5Cln%20%5Cleft(%201%2B%5Csqrt%7B2%7D%20%5Cright)%20-%5Csqrt%7B2%7D%20%5Cright)%20%0A%0A

注:

%0A%5Cfrac%7B%5CGamma%20%5Cleft(%20p%20%5Cright)%20%5CGamma%20%5Cleft(%20q%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20p%2Bq%20%5Cright)%7D%3DB%5Cleft(%20p%2Cq%20%5Cright)%20%3D%5Cint_0%5E1%7Bx%5E%7Bp-1%7D%5Cleft(%201-x%20%5Cright)%20%5E%7Bq-1%7D%5Cmathrm%7Bd%7Dx%7D%5Cleft(%20%5Cmathrm%7BRe%7Dp%2Cq%3E0%20%5Cright)%20%5C%5C%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7Be%5E%7Bi%5Cleft%7C%20a%20%5Cright%7Cz%7D%7D%7Bz%5E2%2Bb%5E2%7D%5Cmathrm%7Bd%7Dz%7D%3D%5Cfrac%7B%5Cpi%7D%7B%5Cleft%7C%20b%20%5Cright%7C%7De%5E%7B-%5Cleft%7C%20ba%20%5Cright%7C%7D%0A%0A

梅林变换的卷积定理:

%0A%5Cint_0%5E%7B%5Cinfty%7D%7Bx%5E%7Bs-1%7Df_1%5Cleft(%20x%20%5Cright)%20f_2%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%3D%5Cfrac%7B1%7D%7B2%5Cpi%20i%7D%5Cint_%7B%5Cgamma%20-i%5Cinfty%7D%5E%7B%5Cgamma%20%2Bi%5Cinfty%7D%7BF_1%5Cleft(%20%5Cmu%20%5Cright)%20F_2%5Cleft(%20s-%5Cmu%20%5Cright)%20%5Cmathrm%7Bd%7D%5Cmu%7D%0A%0A

对于第二个积分能否直接和barnes积分一样使用留数计算呢?也就是直接计算积分

%0A%5Cfrac%7B1%7D%7B2%5Cpi%20i%7D%5Cint_%7B%5Cgamma%20-i%5Cinfty%7D%5E%7B%5Cgamma%20%2Bi%5Cinfty%7D%7B%5Cfrac%7B%5CGamma%20%5E2%5Cleft(%20%5Cfrac%7B5%7D%7B8%7D-%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B9%7D%7B8%7D-%5Cmu%20%5Cright)%20%5CGamma%20%5E2%5Cleft(%20%5Cmu%20%2B%5Cfrac%7B1%7D%7B8%7D%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cfrac%7B5%7D%7B8%7D%2B%5Cmu%20%5Cright)%7D%7B%5CGamma%20%5Cleft(%20%5Cfrac%7B13%7D%7B8%7D-%5Cmu%20%5Cright)%20%5CGamma%20%5Cleft(%20%5Cmu%20%2B%5Cfrac%7B9%7D%7B8%7D%20%5Cright)%7D%5Cmathrm%7Bd%7D%5Cmu%7D%0A%0A而存在的问题就是需要计算一个4F3以及二阶极点的留数,好像并不容易。



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简单搞个积的评论 (共 条)

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