Re: series expansion : a question

In article <33119411.1186861057132.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx> tommy1729 <tommy1729@xxxxxxxxx> writes:
considering series expansion ..

a power series is taylor
an nth derivate series is taylor

You are doing things the wrong way. A Taylor series exists for a function
that is analytical in a circle around the origin. In that case the Taylor
series is convergent within that circle. When a Taylor series does exist
we can get the n-th derivative of it by taking the n-th derivative of the
terms, constructing a new Taylor series. And we can integrate the function
by integrating the terms (where we have to insert suitable constant terms).
A function for which the n-th derivative does not exist at x = 0 does not
have a Taylor series expansion at all. (An example is x^3.|x|, which has
first, second, third and fourth derivatives, but for which the fifth
derivative at x = 0 does not exist, so it is not analytical in x = 0 and
so does not have a Taylor series expansion.)

what is the analogue for an integral ??
wich series can be expressed in the nth integral ??

See above. Although I do not understand the second question.
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