# Expanding in a Taylor series

## Homework Statement

If $\int_{0}^{1} f(x) g(x) \ dx$ converges, and assuming $g(x)$ can be expanded in a Taylor series at $x=0$ that converges to $g(x)$ for $|x| < 1$ (and perhaps for $x= -1$ as well), will it always be true that $\int_{0}^{1} f(x) g(x) \ dx = \int_{0}^{1} f(x) \sum_{n=0}^{\infty} a_{n} x^{n} \ dx$?

Will the fact that that the series doesn't converge for $x=1$ ever be an issue?

A couple of examples are $$\int_{0}^{1} \frac{f(x)}{1-x} \ dx = \int_{0}^{1} f(x) \sum_{n=0}^{\infty} x^{n} \ dx$$ and $$\int_{0}^{1} f(x) \ln(1-x) \ dx = -\int_{0}^{1} f(x) \sum_{n=1}^{\infty} \frac{x^{n}}{n} \ dx.$$

## The Attempt at a Solution

I want to say that it will always be true since it's just a single point. But I don't know if that's sufficient justification.

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## Answers and Replies

mfb
Mentor
Is the sum really inside the integral? If yes: the integral (by definition) does not care about the function value at 1 at all.
If it is outside, it gets more interesting.

Stephen Tashi
Science Advisor
What kind of integration has been defined in this analysis course? What definition of a definite integral is used? Are you studying measure theory?