Sample Integration Questions Solutions Math 127B. Winter, 2005

1. Give an example of a function f : [0, 1] → ℝ such that f^{2} is Riemann integrable, but f is not.

Solution.

• The function

f(x) = \{ \begin{matrix}1 & for x ∈ ℚ, \\ −1 & for x ∉ ℚ, \end{matrix}

is not Riemann integrable on [0, 1] since the upper Darboux sums are all equal to 1, and the lower darboux sums are all equal to −1. The function f^{2} is the constant function equal to 1 which is Riemann integrable.

2. Suppose that f : [a, b] → ℝ is a bounded, Riemann integrable function. Define F : [a, b] → ℝ by

F (x) = ∫_{a}^{x} f(t) dt.

Prove that there exists a constant M such that

|F (x) − F (y)| ≤ M|x − y| for all x, y ∈ [a, b].

Is F necessarily differentiable in (a, b)?

Solution.

• Since f is bounded, there is a constant M such that

|f(x)| ≤ M for all x ∈ [a, b].

It follows that for any x, y ∈ [a, b]

|F (x) − F (y)| = |∫_{x}^{y} f(t) dt|
≤ |∫_{x}^{y} |f(t)| dt|
≤ |∫_{x}^{y} M dt|
≤ M |x − y| .

• F need not be differentiable if f is not continuous. For example, if

f(x) = \{ \begin{matrix}1 & for x > 0, \\ −1 & for x < 0, \end{matrix}

then

F (x) = ∫_{0}^{x} f(t) dt = |x|

is not differentiable at x = 0.

3. Suppose that g : ℝ → ℝ is continuous. Define f : ℝ → ℝ by

f(x) = ∫_{0}^{x}(x − t)g(t) dt.

Prove that f satisfies the following equations:

f′′(x) = g(x), f(0) = f′(0) = 0.

Solution.

• We can rewrite f as

f(x) = x ∫_{0}^{x} g(t) dt − ∫_{0}^{x} tg(t) dt.

Differentiating this equation with respect to x using the product rule and the fundamental theorem of calculus (which applies since both g(t) and tg(t) are continuous), we get

f′(x) = x · g(x) + 1 · ∫_{0}^{x} g(t) dt − xg(x)
= ∫_{0}^{x} g(t) dt.

Differentiating this equation with respect to x and using the fundamental theorem of calculus again, we get

f′′(x) = g(x).

• We have

f(0) = ∫_{0}^{0}(x − t)g(t) dt = 0,
f′(0) = ∫_{0}^{0} g(t) dt = 0.

4. Define the improper integral

∫_{0}^{∞} \frac{sin x}{x} dx

as a limit of proper integrals, and prove that it converges.

Solution.

• Note that

lim_{x→0} \frac{sin x}{x} = 1

so sin x/x extends to a continuous function on [0, ∞). Its Riemann integral therefore exists on any bounded interval [0, b] with b > 0. Hence we define

∫_{0}^{∞} \frac{sin x}{x} dx = lim_{b→∞} ∫_{0}^{b} \frac{sin x}{x} dx.

• Let

I(b) = ∫_{0}^{b} \frac{sin x}{x} dx.

Then for any b, c > 0, we have (using an integration by parts)

|I(b) − I(c)| = |∫_{b}^{c} \frac{sin x}{x} dx|
= |[\frac{− cos x}{x}]_{b}^{c} − ∫_{b}^{c} \frac{cos x}{x^{2}} dx|
≤ |\frac{cos b}{b} − \frac{cos c}{c}| + |∫_{b}^{c} \frac{cos x}{x^{2}} dx|
≤ \frac{1}{b} + \frac{1}{c} + |∫_{b}^{c} \frac{1}{x^{2}} dx|
≤ \frac{1}{b} + \frac{1}{c} + |[−\frac{1}{x}]_{b}^{c}|
≤ \frac{1}{b} + \frac{1}{c} + |\frac{1}{b} − \frac{1}{c}|
≤ \frac{2}{b} + \frac{2}{c}.

Given any ϵ > 0, let N = 4/ϵ. It follows from this inequality that if b, c > N then

|I(b) − I(c)| < ϵ.

Hence, by the Cauchy criterion, the limit of I(b) as b → ∞ exists, and the improper integral converges.

• One can show that

∫_{0}^{∞} \frac{| sin x|}{x} dx = ∞,

so the integral is not absolutely convergent. This is why we had to integrate by parts (to improve the convergence of the integrals) before taking absolute values inside the integral.

• In fact, one can show using methods from complex analysis that

∫_{0}^{∞} \frac{sin x}{x} dx = \frac{π}{2} .

(This question is more difficult than the ones that will be on the final exam.)

5. Suppose that

F (x) = \{ \begin{matrix}x^{2} & for 0 ≤ x < 2, \\ x^{3} & for 2 ≤ x ≤ 3. \end{matrix}

Evaluate the Riemann-Stieltjes integral

∫_{0}^{3} xdF (x),

briefly justifying your computations.

Solution.

• Note that F has a jump discontinuity of size 4 at x = 2. We separate F into a ‘jump part’ F_{1} and a ‘continuous part’ F_{2} as:

F (x) = F_{1}(x) + F_{2}(x),
F_{1}(x) = \{ \begin{matrix}0 & for 0 ≤ x < 2, \\ 4 & for 2 ≤ x ≤ 3, \end{matrix}
F_{2}(x) = \{ \begin{matrix}x^{2} & for 0 ≤ x < 2, \\ x^{3} − 4 & for 2 ≤ x ≤ 3. \end{matrix}

Then, using a standard property of the Riemann-Stieltjes integral, we have

∫_{0}^{3} x dF (x) = ∫_{0}^{3} x dF_{1}(x) + ∫_{0}^{3} x dF_{2}(x).

The Riemann-Stietjes integral with respect to the jump-function F_{1} is given by

∫_{0}^{3} f(x) dF_{1}(x) = f(2) · 4,

so

∫_{0}^{3} x dF_{1}(x) = 8.

If F is continuously differentiable, then

∫_{a}^{b} f dF = ∫_{a}^{b} fF′ dx

Hence, for the continuous part, using the standard property that we can split up the domain of integration, we have

∫_{0}^{3} x dF_{2}(x) = ∫_{0}^{2} x dF_{2}(x) + ∫_{2}^{3} x dF_{2}(x)
= ∫_{0}^{2} x d(x^{2}) + ∫_{2}^{3} x d(x^{3} − 4)
= ∫_{0}^{2} x · 2x dx + ∫_{2}^{3} x · 3x^{2} dx
= ∫_{0}^{2} 2x^{2} dx + ∫_{2}^{3} 3x^{3} dx
= [\frac{2x^{3}}{3} ]_{0}^{2} + [\frac{3x^{4}}{4} ]_{2}^{3}
= \frac{16}{3} + \frac{243}{4} − 12
= \frac{649}{12}.

Hence

∫_{0}^{3} xdF (x) = 8 + \frac{649}{12} = \frac{745}{12}.