23 Integration: Techniques and Theorems

23.1 Elementary Theorems

23.1.1 Integration of Even Functions about Zero

Suppose \(f\) is an integratable function, and let \(F(x)=\int\limits_{0}^{x_0}f(x)dx\).

Then \(\int\limits_{-x_0}^{0}f(x)dx=\int\limits_{0}^{x_0}f(x)dx\) if and only if \(f\) is an even function.\

Proof:

First, let \(f\) be an even function. Then, by Theorem 15.4.5, the anti-derivative \(F\) is an odd function.

\[\begin{aligned} \int\limits_{-x_0}^{0}f(x)dx &= F(0) - F(-x_0) \\ &= F(0) + F(x_0) \\ ^{[1]} &= F(x_0) \\ \\ \\ \int\limits_{0}^{x_0}f(x)dx &= F(x_0) - F(0) \\ &= F(x_0) \end{aligned}\]

  1. \(F(0)=\int\limits_{0}^{0}f(x)dx=0\).

So \[\begin{aligned} \int\limits_{-x_0}^{0}f(x)dx &= F(x_0) \\ &= \int\limits_{0}^{x_0}f(x)dx \end{aligned}\]

Now suppose

\[ \int\limits_{-x_0}^{0}f(x)dx = \int\limits_{0}^{x_0}f(x)dx \]

Then

\[\begin{aligned} \int\limits_{-x_0}^{0}f(x)dx &= F(0) - F(-x_0) \\ &= -F(-x_0) \end{aligned}\]

and

\[\begin{aligned} \int\limits_{0}^{x_0}f(x)dx = F(x_0) - F(0) \\ = F(x_0) \end{aligned}\]

So \[\begin{aligned} -F(-x_0) &= F(x_0) \\ \Rightarrow F(-x_0) &= -F(x_0) \end{aligned}\]

This satisfies the definition of an odd function. So by Theorem 15.4.4, \(f\) must be an even function.

23.1.2 Corollary

If \(f\) is a continuous and even function and \(t\in\Re\), then

\[ \int\limits_{-t}^{t}f(x)dx = 2\int\limits_{0}^{t}f(x)dx \]

Furthermore,

\[\int\limits_{-\infty}^{\infty}f(x)dx = 2\int\limits_{0}^{\infty}f(x)dx\].

Proof: Since \(f(x)\) is even and by Theorem 23.1.1

\[\begin{aligned} \int\limits_{-t}^{0}f(-x)dx &= \int\limits_{-t}^{0}f(x)dx \\ &= \int\limits_{0}^{t}f(x)dx \end{aligned}\]

It follows that

\[\begin{aligned} \int\limits_{-t}^{t}f(x)dx &= \int\limits_{-t}^{0}f(-x)dx + \int\limits_{0}^{t}f(x)dx \\ &= \int\limits_{0}^{t}f(x)dx + \int\limits_{0}^{t}f(x)dx \\ &= 2\int\limits_{0}^{t}f(x)dx \end{aligned}\]

The second statement is proven by taking the limits as \(t\rightarrow\infty\).

23.1.3 Integrals of Horizontal Translations

Let \(x\) be any real number and \(a,b,\) and \(c\) be constants. Also, let \(f(x)\) be continuous on the interval \((a,b)\). Then

\[\int\limits_{a}^{b}f(x)dx = \int\limits_{a+c}^{b+c} f(x+c)dx\]

Proof:

The proof of this theorem is completed by applying a change of variable to

\[\int\limits_{a}^{b}f(x)dx\]

We let

\[\begin{aligned} y &= x+c \\ \Rightarrow x &= y-c \end{aligned}\]

So \(dx=dy\).

\[\begin{aligned} x &= a & \Rightarrow \ \ \ \ y &= a+c\\ x &= b & \Rightarrow \ \ \ \ y &= b+c \end{aligned}\].

Thus

\[\begin{aligned} \int\limits_{a}^{b}f(x)dx &= \int\limits_{a+c}^{b+c}f(y)dy \\ &= \int\limits_{a+c}^{b+c}f(x+c)dx \end{aligned}\]