# Math Insight

### 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, $$\int_0^{T} {dx\over 1-x}\;dx= T+{T^2\over 2}+{T^3\over 3}+{T^4\over 4}+\ldots$$