Hello David,
I don't know if I understood well your question.
Did you asked why, if you write
> display2d:false;
> assume(x>0);
> f(d,x):=integrate(z**(d-1)*exp(-z),z,0,x);
> taylor(f(d,x),d,1,1);
you get this answer
(%o5) at('integrate(z^(d-1)*%e^-z,z,0,x),d = 1) +
(at('integrate(z^(d-1)*%e^-z*log(z),z,0,x),
d = 1))*(d-1)
and not
(%o5) 1-%e^-x +
(at('integrate(z^(d-1)*%e^-z*log(z),z,0,x),
d = 1))*(d-1)
isn't?
I tried your code, and I had the same result.
I think it's quite strange, but I'm not an expert
of symbolic integration, so I cannot help you.
Best regards,
Giovanni Gherdovich
p4.vert.ukl.yahoo.com uncompressed/chunked Wed Aug 30 11:13:38 GMT 2006
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