Repeated integration by parts
For integrals \(\int P(x) e^{ax} \dx\) or \(\int P(x) \sin(bx) \dx\) with \(P\) a polynomial, each integration by parts with \(u\) the polynomial factor lowers its degree by one, so \(\deg P + 1\) integrations by parts reduce the integral to an elementary one.
The tabular method records this: one column lists \(P\) and its successive derivatives, the other lists \(e^{ax}\) (or \(\sin(bx)\)) and its successive antiderivatives, and the integral is the sum of the products along the diagonals with alternating signs starting from \(+\).
For constants \(a,b,c\) and \(\omega\neq 0\), the derivatives of \(t^3+at^2+bt+c\) are \(3t^2+2at+b\), \(6t+2a\), \(6\), \(0\), and successive antiderivatives of \(\sin(\omega t)\) are \(-\frac{\cos(\omega t)}{\omega}\), \(-\frac{\sin(\omega t)}{\omega^2}\), \(\frac{\cos(\omega t)}{\omega^3}\), \(\frac{\sin(\omega t)}{\omega^4}\). Pairing them along the diagonals with the signs \(+,-,+,-\) gives \[ \int (t^3+at^2+bt+c)\sin(\omega t)\,dt =-(t^3+at^2+bt+c)\frac{\cos(\omega t)}{\omega} +(3t^2+2at+b)\frac{\sin(\omega t)}{\omega^2} +(6t+2a)\frac{\cos(\omega t)}{\omega^3} -6\,\frac{\sin(\omega t)}{\omega^4}+C, \] and the process stops when the derivatives reach \(0\).
\[ \int x^2 e^x \dx = x^2e^x - 2xe^x + 2e^x + C. \]
Standard power series
| Series | Valid for |
|---|---|
| \(\frac{1}{1-x}=\sum_{n\geq 0}x^n=1+x+x^2+x^3+\cdots\) | \(x\in(-1,1)\) |
| \(e^x=\sum_{n\geq 0}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\) | \(x\in\RR\) |
| \(\cos x=\sum_{n\geq 0}(-1)^n\frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots\) | \(x\in\RR\) |
| \(\sin x=\sum_{n\geq 0}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\) | \(x\in\RR\) |
| \(\ln(1+x)=\sum_{n\geq 1}(-1)^{n+1}\frac{x^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\) | \(x\in(-1,1]\) |
| \(\arctan x=\sum_{n\geq 0}(-1)^n\frac{x^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdots\) | \(x\in[-1,1]\) |
Since \(\cos\) is even and \(\sin\) is odd, their series contain only even and only odd powers respectively; substituting into the exponential series gives, for example, \(e^{17x}=\sum_{n\geq 0}\frac{17^nx^n}{n!}\).
For \(\abs x<1\), \(\sum_{n\geq0} x^n = \frac{1}{1-x}\). Differentiating term by term gives \(\sum_{n\geq1} n x^{n-1} = \frac{1}{(1-x)^2}\), and multiplying by \(x\) gives \(\sum_{n\geq1} n x^n = \frac{x}{(1-x)^2}\). Repeating the operation \(x\frac{d}{dx}\) gives \(\sum_n n^kx^n\) for each \(k\geq 1\).