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簡單搞個積

2023-03-26 20:52 作者:艾琳娜的糖果屋  | 我要投稿

我們來計算兩個超幾何函數(shù)的積分%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

先看第一個,因為是無窮區(qū)間上的廣義積分而且還是平方的形式,可以考慮積分變換里面的卷積定理,這里我們采用梅林變換,因為超幾何函數(shù)的梅林變換是很容易得到的。

%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

這梅林變換利用2F1的積分表達或者拉馬努金主定理就能得到。再利用梅林變換的卷積定理有

%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

帶入數(shù)據(jù)得到第一題的結(jié)果

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再來看3F2的這個積分,難度提升了不少,首先由拉馬努金主定理給出梅林變換

%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

那么利用卷積定理我們有

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該問題的關(guān)鍵是如何計算這無窮高直線上的積分,也許會想到像處理barnes積分一樣,但是這個地方不好計算,但是注意到γ的選取是任意的,通過待定系數(shù)法很容易找到最合適的γ=1/4

于是積分可寫為

%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

梅林變換的卷積定理:

%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

對于第二個積分能否直接和barnes積分一樣使用留數(shù)計算呢?也就是直接計算積分

%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而存在的問題就是需要計算一個4F3以及二階極點的留數(shù),好像并不容易。



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簡單搞個積的評論 (共 條)

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