[微積]求一積分值
∫
∞ 1
∫ -------------------------- dv
-∞ (π^2)(1+v^2)[1+(2u-v)^2]
此為求2個Cauchy distribution平均分配
X_1 X_2 iid f(x)=(π^-1)(1+x^2)^-1
U=(X_1+X_2)/2 V=X_1 去做變數變換
想請問這個部分的積分該如何做呢?
有試過分布積分不過越變越複雜...
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