Exercises

Exercises 4.3 Exercises

Suppose

0 \leq f \leq g

[a, b)

and that

lim_{x \to b^{-}} f (x) = + \infty .

Show that for improper integrals

\int_{a}^{b} f (x) d x

and

\int_{a}^{b} g (x) d x

the following relations hold (c.f. Proposition 4.12)

if $\int_{a}^{b} g (x) d x$ converges, so does $\int_{a}^{b} f (x) d x .$ Moreover, $\int_{a}^{b} f (x) d x \leq \int_{a}^{b} g (x) d x .$
if $\int_{a}^{b} f (x) d x$ diverges to $+ \infty,$ then so does $\int_{a}^{b} g (x) d x .$

Formulate and proof a similar statement about the improper integrals for functions

0 \leq f \leq g

(a, b]

with

lim_{x \to a^{+}} f (x) = + \infty .

Show that for all

n \geq 0,

\int_{0}^{\infty} x^{n} e^{- x} d x = n!

(Hint: choose

f (x) = x^{n}

Determine the values of

p

for which the integral

\int_{0}^{1} x^{p} \ln (x) d x

converges. And for those values of

p,

find the integral. (Hint: use integration by parts.)

Use the comparison test to determine the convergence of the integral

\int_{0}^{π / 2} \frac{1}{x \sin (x)} d x

For

0 \leq b < 2,

show that

\int_{0}^{1} \frac{d x}{\sqrt{x^{b} - x^{2}}} = \frac{π}{2 - b}

Hint.

Factor

x^{b}

out from the square root. Then make a suitable substitution to convert the integral into one that can be handled by trigonometric substitution.