I have the following very simple code. A power series in a-parameter, function f[x], is integrated with respect to x, and the result is expanded as a power series around a=0.

The result that "Mathematica" gives is wrong (it gives only the first term ignoring the other terms).


f[x_] := f0[x] + a f1[x] + a^2 f2[x] + a^3 f3[x]
Integrate[ f[x], x]
Series[%, {a, 0, 2}]
Out[1]= \[Integral](f0[x] + a f1[x] + a^2 f2[x] + a^3 f3[x]) \[DifferentialD]x
Out[2]= \[Integral]f0[x] \[DifferentialD]x
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    – rhermans
    Oct 23, 2014 at 8:26
  • $\begingroup$ The integration does not even do anything. Why even worry about what Series does? What is Series supposed to be when given an un-evaluated integral anyway? Is this even defined somewhere? Series work on functions. $\endgroup$
    – Nasser
    Oct 23, 2014 at 8:37
  • 2
    $\begingroup$ Strange, it works if you force Distributivity Series[Distribute[Integrate[f[x], x]], {a, 0, 2}]. I disagree with @Nasser, what it should do is well defined, and should be the same as in Series[Sum[Subscript[c, i] a^i, {i, 0, 3}], {a, 0, 2}] . Or am I thinking this wrong? Bug? $\endgroup$
    – rhermans
    Oct 23, 2014 at 8:56
  • $\begingroup$ @rhermans I only studies Series expansion for functions at school. Teacher never mentioned series expansion for integrals. So I do not know. You could be right. I'll ask the teacher tomorrow when I go to school. $\endgroup$
    – Nasser
    Oct 23, 2014 at 9:41
  • $\begingroup$ @KDH, It seems that no one answers, the link explains what to do. $\endgroup$
    – rhermans
    Oct 24, 2014 at 9:26

1 Answer 1


Response from WRI:

Thank you for bringing this issue to our attention.

I reproduced the behavior you describe and it appears to be an incorrect result. I have sent a note about this issue to our development team.

We are always interested in improving Mathematica and I want to thank you again for taking the time to contact us about this issue.

So it's reasonable to consider it as a bug.


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