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发表于 2008-4-9 01:22:36
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来自 安徽合肥
我举的例子太简单了,下午在验证一个很复杂的积分公式,总是出现一大串的表达讨论式子,我把其中一表达式写出来:
Integrate[1/(x^2+y^2+z^2)^(5/2),{x,a,b}],你再验证一下,是否是一大串讨论式子,一下午没有算出一个出来,其实积分出来的,真郁闷,难道是我的mathematica6.0出问题了?
附上我的算的结果,看看到底怎么了
In[3]:=Integrate[1/(x^2 + y^2 + z^2)^(5/2), {x, a, b}]
Out[3]:=If[(-Im Re[a] + Im[a] Re)/(Im[a] - Im) >=
0 && (Re[(a - Sqrt[-y^2 - z^2])/(a - b)] <= 0 ||
Re[(a - Sqrt[-y^2 - z^2])/(a - b)] >= 1 ||
Im[(a - Sqrt[-y^2 - z^2])/(a - b)] !=
0) && (Re[(a + Sqrt[-y^2 - z^2])/(a - b)] <= 0 ||
Re[(a + Sqrt[-y^2 - z^2])/(a - b)] >= 1 ||
Im[(a + Sqrt[-y^2 - z^2])/(a - b)] != 0), -(
a (2 a^2 + 3 (y^2 + z^2)))/(
3 (y^2 + z^2)^2 (a^2 + y^2 + z^2)^(3/2)) + (
b (2 b^2 + 3 (y^2 + z^2)))/(
3 (y^2 + z^2)^2 (b^2 + y^2 + z^2)^(3/2)),
Integrate[1/(x^2 + y^2 + z^2)^(5/2), {x, a, b},
Assumptions -> ! ((-Im Re[a] + Im[a] Re)/(Im[a] - Im) >=
0 && (Re[(a - Sqrt[-y^2 - z^2])/(a - b)] <= 0 ||
Re[(a - Sqrt[-y^2 - z^2])/(a - b)] >= 1 ||
Im[(a - Sqrt[-y^2 - z^2])/(a - b)] !=
0) && (Re[(a + Sqrt[-y^2 - z^2])/(a - b)] <= 0 ||
Re[(a + Sqrt[-y^2 - z^2])/(a - b)] >= 1 ||
Im[(a + Sqrt[-y^2 - z^2])/(a - b)] != 0))]] |
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