### Integrating Taylor polynomials: first example

Thinking simultaneously about the difficulty (or impossibility) of
‘direct’ symbolic integration of complicated expressions, by contrast to
the ease of integration of *polynomials*, we might hope to get
some mileage out of *integrating Taylor polynomials*.

As a promising example: on one hand, it's not too hard to compute that $$\int_0^{T} {dx\over 1-x}\;dx=[-\log (1-x)]_0^T=-\log(1-T)$$ On the other hand, if we write out $${1\over 1-x}=1+x+x^2+x^3+x^4+\ldots$$ then we could obtain $$\int_0^{T} (1+x+x^2+x^3+x^4+\ldots)\;dx=[x+{x^2\over 2}+{x^3\over 3}+\ldots]_0^T=$$ $$ = T+{T^2\over 2}+{T^3\over 3}+{T^4\over 4}+\ldots$$ Putting these two together (and changing the variable back to ‘$x$’) gives $$-\log(1-x)= x+{x^2\over 2}+{x^3\over 3}+{x^4\over 4}+\ldots$$

(For the moment let's not worry about what happens to the error term for the Taylor polynomial).

This little computation has several useful interpretations. First, we obtained a Taylor polynomial for $-\log(1-T)$ from that of a geometric series, without going to the trouble of recomputing derivatives. Second, from a different perspective, we have an expression for the integral $$\int_0^{T} {dx\over 1-x}\;dx$$ without necessarily mentioning the logarithm: that is, with some suitable interpretation of the trailing dots, $$\int_0^{T} {dx\over 1-x}\;dx= T+{T^2\over 2}+{T^3\over 3}+{T^4\over 4}+\ldots$$

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