Putting Taylor Series to Work

Some integrals simply have no elementary antiderivative — \(\int \frac{\sin x}{x}\,dx\) and \(\int e^{-x^2}\,dx\) are the classics. But replace the integrand with its Maclaurin series and you can integrate term by term. From \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\), divide by \(x\): \[\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots,\] then integrate each power: \[\int \frac{\sin x}{x}\,dx = C + x - \frac{x^3}{3\cdot 3!} + \frac{x^5}{5\cdot 5!} - \frac{x^7}{7\cdot 7!} + \cdots,\] valid for all \(x\). Don't drop the constant of integration \(C\) — pin it down with a known value, like the integral being 0 at \(x = 0\). The same substitution trick cracks limits: it replaces four rounds of L'Hopital's rule in the worked example below with one line of algebra.

For any real \(k\) and \(|x| < 1\), \[(1+x)^k = \sum_{n=0}^{\infty} \binom{k}{n} x^n, \qquad \binom{k}{n} = \frac{k(k-1)(k-2)\cdots(k-n+1)}{n!}.\] This is nothing more than the Maclaurin series of \(f(x) = (1+x)^k\): since \(f^{(n)}(0) = k(k-1)\cdots(k-n+1)\), the Taylor coefficient formula collapses exactly to the generalized binomial coefficient. When \(k\) is a nonnegative integer, \(\binom{k}{n} = 0\) for \(n > k\) and the series terminates — recovering the ordinary binomial theorem. For negative or fractional \(k\) it's a genuine infinite series, and it only converges for \(|x| < 1\). When \(k\) is negative or fractional, expand the product \(k(k-1)\cdots(k-n+1)\) one factor at a time — guessing the sign pattern is the classic way to slip up.