Solution

Both functions are continuous on the positive domains in question, so they are measurable. For the first function, the decreasing substitution

t = 1/ x

and

∣ d x / d t ∣ = t^{- 2}

give

\int_{0}^{1} \frac{s i n ( 1/ x )}{x} d x = \int_{1}^{\infty} \frac{∣ s i n t ∣}{t} d t .

For every integer

k \geq 1

\int_{kπ}^{(k + 1) π} \frac{∣ s i n t ∣}{t} d t \geq \frac{1}{( k + 1 ) π} \int_{kπ}^{(k + 1) π} ∣ sin t ∣ d t = \frac{2}{( k + 1 ) π} .

Summing this lower bound gives a divergent harmonic series. Thus $f$ is not Lebesgue integrable on $(0, 1)$ .

∣ sin x ∣ \leq x

(0, 1)

and

∣ sin x ∣ \leq 1

everywhere give

\int_{0}^{\infty} \frac{∣ s i n x ∣}{x ^{5/4}} d x \leq \int_{0}^{1} x^{- 1/4} d x + \int_{1}^{\infty} x^{- 5/4} d x = \frac{4}{3} + 4 < \infty.

Therefore $g$ is Lebesgue integrable on $(0, \infty)$ . The sine power kernel integrability threshold is

1 < p < 2

: one endpoint requires

p < 2

, and the other requires

p > 1

. The signed improper integral for

f

does converge by the Dirichlet test for improper integrals, but conditional improper convergence does not give a finite integral of

∣ f ∣

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