[積分] 瑕積分
(1)
∞ x^(p-1) 1 x^(p-1)
∫-------- dx =∫--------- dx = B[p,1-p]
0 (1+x) 0 (1-x)^p
∞ x^(-p) 1 x^(-p)
= ∫-------- dx =∫ ----------- dx = B[1-p,p]
0 (1+x) 0 (1-x)^(1-p)
= Γ[p]Γ[1-p]
π
= -------- = πCsc[pπ] (0<p<1)
Sin[pπ]
(2)
∞ x^a 1
∫--------- dx = ----------- B[a+1,m-(a+1)]
0 (x+b)^m b^(m-(a+1))
(3)
1 1
Γ[x] = ∫(ln---)^(x-1)dt (x>0)
0 t
(4)
∞ Cos[ax]
∫--------- dx = (π/2)e^(-｜a｜)
0 1+x^2
好難喔！不知道怎麼推導出來的，請各位大大幫忙解解看^^
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