33 Antidifferentiation
33.1 Substitution
Theorem 33.1 (Substitution) Let \(g\) be a continuously differentiable function on \([a,b]\) and \(f\) be continuous on the range of \(g\), with \(F\) an antiderivative of \(f\). Then \[\int_{[a,b]}f(g(x))g^\prime(x) = F\circ g\Bigg|_{[a,b]}\]
Proof. Note \(g\) is continuous as it is differentiable. As compositions and products of continuous functions are continuous, \(f(g(x))g^\prime(x)\) is a continuous function, and hence integrable. Thus by the fundamental theorem of calculus we can evaluate its integral by finding an antiderivative. The chain rule readily confirms \(F\circ g\) is such a function as \[(F(g(x)))^\prime = F^\prime(g(x))g^\prime(x)=f(g(x))g^\prime(x)\] Thus \(\int_{[a,b]}f(g(x))g^\prime(x) = F(g(b))-F(g(a))\).
This justifies the familiar use of u-substitution in Calculus
Example 33.1 To integrate \((2x+1)^5\) on the interval \([a,b]\), note that we may write \[(2x+1)^5=f(g(x))\] for \(f(x)=x^5\) and \(g(x)=2x+1\). Then \(2(2x+1)^5\) is the derivative of \(\frac{1}{6}(2x+1)^6\), so \[\int_{[a,b]}2(2x+1)^5 = \frac{1}{6}(2x+1)^6\Bigg|_{[a,b]}\] By the linearity of the integral \(\int_{[a,b]}2(2x+1)^5=2\int_{[a,b]}(2x+1)^5\) and solving for this yields \[\int_{[a,b]}(2x+1)^5 = \frac{(2x+1)^6}{12}\Bigg|_{[a,b]}=\frac{(2b+1)^6-(2a+1)^5}{12}\]
Exercise 33.1
Corollary 33.1 (Inverse Substitution) Subbing \(x=f(t)\) instead of \(u=g(x)\) (like trig sub)
Examples
Exercise: substitution to show that \[\int_{[ka,kb]}f(x)=k\int_{[a,b]}f(kx)\]
33.2 Integration by Parts
Theorem 33.2 (Integration by Parts) Let \(f\) be continuous and \(g\) continuously differentiable on \([a,b]\). Then \[\int_{[a,b]}f(x)g(x) = F(x)g(x)\Bigg|_{[a,b]}-\int_{[a,b]}f(x)g^\prime(x)\] where \(F\) is an antiderivative of \(f\).
Corollary 33.2 (Iterated Integration by Parts)
33.3 Power Series
Proposition 33.1 Let \(\sum a_n x^n\) be a power series with radius of convergence \(r\). Then the series \(\sum \frac{a_n}{n+1}x^{n+1}\) also has radius of convergence \(r\).
Like previously we give two proofs here, a simple and memorable proof in the case that the ratio tests suffices to determine the convergence of the original series, and a more robust application of the root test for the general case.
Proof. Like for the differentiable case, we prove this here under the assumption that the Ratio test succeeds in computing the radius of convergence for the original series, so for any \(x\in(-R,R)\) \[\lim \left|\frac{a_{n+1}}{a_n}\right||x|<1\]
We now turn to compute the ratio test for our new series \(\sum_\frac{a_{n}}{n+1}x^{n+1}\): the ratio in question is
\[\frac{\frac{a_{n+1}}{n+2}x^{n+2}}{\frac{a_n}{n+1}x^{n+1}}=\left(\frac{a_{n+1}}{a_n}\right)\left(\frac{n+1}{n+2}\right)x\]
Since \((n+1)/(n+2)\to 1\) we can compute the overall limit using the limit theorems and see we end up with the exact same limit as for the original series! Thus integrating term by term does not change the radius of convergence at all.
Now for the general case: :::{.proof} :::
Having confirmed that \(\sum \frac{a_n}{n+1}x^n\) converges when the original series does, we can provide a direct proof of the term-by-term integrability of power series, avoiding the use of dominated convergence:
Theorem 33.3 (Integrating Power Series) Let \(f(x)=\sum a_n x^n\) be a power series with radius of convergence \(r\). Then \(f\) is integrable on \((-r,r)\) and for any \(0< x < r\) \[\int_{[0,x]}f = \sum_{n\geq 0} \frac{a_n}{n+1}x^{n+1}\]
Proof. The function \(\sum a_n x^n\) is continuous on \((-r,r)\) and thus integrable by Theorem 31.1. Define \(F(x)= \sum \frac{a_n}{n+1}x^n\). This converges on \((-r,r)\) by Proposition 33.1, and defines a differentiable function on this interval; whose derivative can be calculated term-by-term (Theorem 26.1), giving \[F^\prime(x)=\left(\sum_{n\geq 0}\frac{a_n}{n+1}x^{n+1}\right)^\prime = \sum_{n\geq 0} a_n x^n = f(x)\] Thus \(F\) is an antiderivative of \(f\), and by the fundamental theorem of calculus \[\int_{[0,x]}f = F(x)-F(0)\] Since \(F\) has no constant term, \(F(0)=0\) and so \(\int_{[0,x]}f = \sum_{n\geq 0} \frac{a_n}{n+1}x^{n+1}\) as claimed.
Theorem 33.4 (Integral form of Taylor Remainder)
33.4 \(\bigstar\) Differentiating Under the Integral
33.5 The \(dx\) Notation
Introduce the familiar calculus notation \(\int_a^b f dx\) and show how the \(dx\) works as a nice short-hand for u-substitution, and integration by parts. Also useful just to tell us what the variable of integration is in multivariate integrals.