# 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}