如果我整合以下内容,我会得到答案:
Integrate[(nor*x^ap*(1 - x)^bp*(1 - cp*x)*Exp[-t*al]), {x, 0.001, 1},
Assumptions -> {-1 < nor < 1, 0 < ap < 3, 0 < bp < 5, 0 < cp < 5,
0 < t < 2, 0 < al < 1}]
,答案是
E^(-al t) nor ((1. Gamma[1 + ap] Gamma[1 + bp])/Gamma[2 + ap + bp] - (
1. cp Gamma[2 + ap] Gamma[1 + bp])/Gamma[3 + ap + bp] +
0.001^ap (-((0.001 Hypergeometric2F1[1 + ap, -bp, 2 + ap, 0.001])/(
1. + ap)) + (
1.*10^-6 cp Hypergeometric2F1[2 + ap, -bp, 3 + ap, 0.001])/(
2. + ap)))
但如果我想这样做,
Integrate[(nor*x^ap*(1 - x)^bp*(1 - cp*x)*Exp[-t*al*(1 - x)]), {x,
0.001, 1},
Assumptions -> {-1 < nor < 1, 0 < ap < 3, 0 < bp < 5, 0 < cp < 5,
0 < t < 2, 0.5 < al < 1}]
它没有给出最终表达。它给出了:
Integrate[(nor*x^ap*(1 - x)^bp*(1 - cp*x)*Exp[-t*al*(1 - x)]), {x,
0.001, 1},
Assumptions -> {-1 < nor < 1, 0 < ap < 3, 0 < bp < 5, 0 < cp < 5,
0 < t < 2, 0.5 < al < 1}]
与积分相同。 你能帮我弄清楚如何用指数中的附加(1-x)得到第二个积分的答案吗?