Home MATH 255 Monday, March 11, 2013

Monday, March 11, 2013 CC-BY-NC
Antiderivatives and Integration by Parts

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$\DeclareMathOperator*{\sgn}{sgn}$

A few classes back the prof posed the following theorem, asking for a proof. Here is one:

Theorem: If $f : [0,1] \to \mathbb{R}$ is Riemann integrable and $f > 0$ then $\int_a^b f > 0$.

Proof: We know that $\int_a^b f \geq 0$. If $\int_a^b f = 0$ then $L(f) = 0$ and $L_f(P) = 0$ for any partition $P$. Thus if $P = \{ x_0, x_1, \ldots, x_n \}$ then $m_k = \inf\{ f(x) : x \in [x_{k-1}, x_k] \} = 0$. $L_f(P) \leq L(f)$. Thus if $f$ is continuous at a point $c$ since $f(c) > 0$ then there exists a $\delta$ neighborhood of $c$ such that $f(x) > 0$ in that neighborhood. Then $L_f(P) > 0$. Consequently, if $\int_a^b f = 0$ then $f$ must be discontinuous at every point. Consider $c = 1 / 2$. $f$ much be discontinuous at $1/2$. THus there exists $\epsilon_0 > 0$ such that for $\delta = 1$, $\exists x_0$ such that $|x_0 - 1/2| < \delta = 1$ but $|f(x_0) - f(1/2)| \geq \epsilon_0$. Thus on $[0,1]$, $M_1 = \sup\{f(x) : x \in [0,1]\}$ and $m_1 = \inf\{f(x) : x \in [0,1]\}$

At this point we get lost. We'll come back to this Wednesday.

1Antiderivatives

Definition: Let $I = [a,b]$ be a closed and bounded interval. If $f : I \to \mathbb{R}$ is a function then an antiderivative of $f$ is a function $F : I \to \mathbb{R}$ such that $F'(x) = f(x)$ for all $x \in I$. Also, if $f : I \to \mathbb{R}$ is integrable, than the function $F : I \to \mathbb{R}$ such that $F(x) = \int_a^x f$ is called the indefinite integral of $f$.

Recap: if $f$ is integrable on $I$ then the indefinite integral $F(x) = \int_a^x f$ is continuous on $I$. Also, if $f$ is continuous on $I$ then the indefinite integral is an antiderivative of $f$, and furthermore

$$ \int_a^b f = F(b) - F(a) $$

i.e. continuous functions have antiderivatives.

Claim: An integrable function may not have an antiderivative.

Proof:

$$ \sgn(x) = \left\{ \begin{array}{lr} -1 & : x < 0 \\ 0 & : x = 0 \\ 1 & : x > 0 \end{array} \right. $$

is integrable on $[-1,1]$ as it is monotone increasing. For a given $\epsilon > 0$, take WLOG $\epsilon < 1$. Consider

$$ P = \{ -1, -\frac{\epsilon}{4}, \frac{\epsilon}{4}, 1 \} $$

$M_1 = \sup\{\sgn(x) : x \in [-1, -\epsilon / 4]\} = -1 = m_1$ so $M_1 - m_1 = 0$. Similarly $M_3 - m_3 = 0$, and we can show $M_2 - m_2 < \epsilon$, so it follows that the function is indeed integrable. However, there does not exist a function $F$ such that $F'(x) = \sgn(x)$, as by Darboux's theorem $F'(-1/2) = -1$, $F'(0) = 0$, so there must exists some $c$ such that $F'(c) = \sgn(c) = -1/2$, which is impossible.

Claim: A function may have an antiderivative but not be integrable. Consider

$$ F(x) = \left\{ \begin{array}{lr} x^2\sin\left(\frac{\pi}{x^2}\right) & : x \neq 0 \\ 0 & : x = 0 \end{array} \right. $$

$F'$ is not integrable.

2Integration by parts

Theorem: Let $I = [a,b]$, $f, g : I \to \mathbb{R}$ integrable on $I$ and let $F, G$ be antiderivatives of $f$ and $g$ respectively. Then:

$$ \int_a^b F(x)g(x) dx = F(b)G(b) - F(a)G(a) - \int_a^b f(x)G(x) dx $$

Proof: Let $H(x) = F(x)G(x)$. $H$ is continuous (and thus integrable) on $I$ as $F$ and $G$ must be continuous. $H$ is also differentiable for the same reasoning. $H'(x) = F'(x)G(x) + F(x)G'(x) = f(x)G(x) + F(x)g(x)$ which is also integrable. By the FTC, $\int_a^b H' = H(b) - H(a)$

i.e. $\int_a^b F(x)g(x) + f(x)G(x) dx = F(b)G(b) - F(a)G(a)$.