math104-f21:hw15

# HW 15

This is for your practice only, not going to be graded. The solution will be release next Wednesday.

#### 1. Ross Ex 33.9

(uniform convergence and integral).

See Rudin Thm 7.16 (page 151)

#### 2. Ross Ex 33.5

(bound an integral by replacing an part of the integrand by something nice)

We want to show $$| \int_{-2\pi}^{2 \pi} x^2 \sin^8(e^x) dx | \leq 16 \pi^3 /3$$ since $|\sin^8(e^x)| \leq 1$, we have $$| \int_{-2\pi}^{2 \pi} x^2 \sin^8(e^x) dx | \leq \int_{-2\pi}^{2 \pi} |x^2 \sin^8(e^x)| dx \leq \int_{-2\pi}^{2 \pi} |x^2| dx = x^3/3|^{2\pi}_{-2\pi} = 16 \pi^3 /3$$

#### 3. Ross Ex 33.14 (a).

For any continuous $g(x) \geq 0$ on $[a,b]$ and continuous $f(x)$, we want to show $$\int_a^b f(t) g(t) dt = f(x) \int_a^b g(t) dt$$ for some $x \in [a,b]$.

Proof: If $g(t) =0$ for all $t \in [a,b]$ then both sides are zero, and there is nothing to prove. Otherwise, assume $g(t) \neq 0$ for some $t \in [a,b]$. By continuity, $g(t)$ is non-zero hence positive on an open subset, hence $\int_a^b g(t) dt > 0$.

let $M = \sup \{f(t): t \in [a,b]\}$ and $m= \inf \{f(t): t \in [a,b]\}$, then since $g(t) \geq 0$, we have $m g(t) \leq f(t) g(t) \leq M g(t)$, thus $$m \int_a^b g(t) dt \leq \int_a^b f(t) g(t) dt \leq M \int_a^b g(t) dt.$$ Thus, $$A = \frac{\int_a^b f(t) g(t) dt}{ \int_a^b g(t) dt} \in [m, M]$$ by intermediate value theorem, there is $x \in [a,b]$, such that $f(x) = A$. This finishes the proof.

#### note

For the following two problems, if $a > b$, then $\int_a^b f(t) dt = - \int_b^a f(t) dt$.

#### 4. Ross Ex 34.5,

Let $f$ be continuous on $\R$, and let $F(x)$ be defined by $$F(x) = \int_{x-1}^{x+1} f(t) dt$$ show that $F$ is differentiable on $\R$ and compute $F'(x)$.

Solution: we prove by definition. For any $\epsilon \neq 0$, we have $$F(x+\epsilon) - F(x) = \int_{x+\epsilon-1}^{x+\epsilon+1} f(t) dt - \int_{x-1}^{x+1} f(t) dt = \int_{x+1}^{x+\epsilon+1} f(t) dt - \int_{x-1}^{x+\epsilon-1} f(t) dt$$ Divide by $\epsilon$ and taking limit, we have $$\lim_{\epsilon \to 0} \frac{F(x+\epsilon) - F(x)}{\epsilon} = \lim_{\epsilon \to 0^+} \epsilon^{-1} \int_{x+1}^{x+\epsilon+1} f(t) dt - \epsilon^{-1} \int_{x-1}^{x+\epsilon-1} f(t) dt = f(x+1) - f(x-1)$$ where we used fundamental theorem of calculus.

Hence $$F'(x) = f(x+1) - f(x-1)$$

#### 34.6.

Let $f$ be continuous function on $\R$, let $$G(x) = \int^{sin(x)}_0 f(t) dt$$ prove that $G(x)$ is differentiable, and compute $G'(x)$.

Define $H(u) = \int^{u}_0 f(t) dt$, for any $u \in \R$, then $G(x) = H(\sin(x))$. Since $H(u)$ is differentiable, with $H'(u) = f(u)$, and $\sin(x)$ is differentiable, we have $G(x)$ is differentiable, with $$G'(x) = H'(\sin(x)) \cos(x) = f(\sin(x)) \cos(x)$$

#### 6, 35.3

See definition 35.2 for Stieljes integral. In particular, if $F$ has a jump on the integration domain's boundary, those points are considered in the integral. Hence $$\int_a^b f(t) d F(t) = \sum_{n \in [a,b] \cap \Z} f(n)$$ For example, $$\int_0^6 x dF(x) = 0 + 1 + \cdots + 6 = ..$$

#### 7, 35.4

Since $F(t)$ is differentiable and monotone over that range, we have $dF(x) = F'(x) dx$, with $F'(x) = \cos(x)$ for $t \in [-\pi/2, \pi/2]$. $$\int_0^{\pi/2} x d F(x) = \int_0^{\pi/2} x \cos(x) d x$$

Alternatively, one can compute using integration by part. If $f$ is also monotone and differentiable, then $$\int_a^b f dF = \int_a^b d(f F) - F df = f F|^b_a - \int_a^b F df(x)$$ Here, in this problem, we have $f(x) = x$, thus $$\int_0^{\pi/2} x d F(x) = x \sin(x)|^{\pi/2}_0 - \int_0^{\pi/2} \sin(x) dx = \pi/2 - (-\cos(x))|^{\pi/2}_0 = \pi/2 -1.$$