Assume that f is integrable over [a, t] for all t ≥ a and there exists a constant M > 0 such that for all t ≥ a, |∫_a^t f(x) dx| ≤ M. Further suppose that g(x) is monotonic decreas... Assume that f is integrable over [a, t] for all t ≥ a and there exists a constant M > 0 such that for all t ≥ a, |∫_a^t f(x) dx| ≤ M. Further suppose that g(x) is monotonic decreasing and g(x) → 0 as x → ∞. Prove that the integral ∫_a^∞ f(x) g(x) dx is convergent. Using this convergence criterion, show that the integral ∫_a^∞ sin(x) / x^p dx is convergent for all x ≥ a > 0 and p > 0. Read more

Steps to Solve

1. Prove the convergence of the integral $\int_a^\infty f(x) g(x) , dx$

We apply the properties of integrals with absolute convergence. Given that $$ | \int_a^t f(x) , dx | \leq M $$ for all $t \geq a$, which suggests that $f(x)$ is bounded in terms of its integral, we reinterpret the integral:

Consider the integral $I(t) = \int_a^t f(x) g(x) , dx$. By properties of absolutely convergent integrals and the Lebesgue Dominated Convergence Theorem, we note that:

$$ |I(t)| = \left| \int_a^t f(x) g(x) , dx \right| \leq \int_a^t |f(x)| |g(x)| , dx $$

2. Use properties of $g(x)$

Since $g(x)$ is monotonically decreasing and tends to $0$ as $x \to \infty$, we can find a constant $C > 0$ such that for sufficiently large $x$, $g(x) < C/x$ for some $C$.

Thus, replace $g(x)$ in the integral:

For large enough $x$, we have $$ |f(x) g(x)| \leq |f(x)| \frac{C}{x} $$

3. Apply the convergence criterion

We can now evaluate:

$$ \int_a^\infty |f(x)| g(x) , dx \leq C \int_a^\infty \frac{|f(x)|}{x} , dx $$

Since the integral $\int_a^\infty \frac{|f(x)|}{x} , dx$ converges (as we assume $f(x)$ is integrable), the integral $\int_a^\infty f(x) g(x) , dx$ converges.

4. Show the convergence of $\int_a^\infty \frac{\sin x}{x^p} , dx$ for $p > 0$

Using the result from the earlier proof, notice that:

Since $g(x) = \frac{1}{x^p}$ is decreasing and approaches $0$ as $x \to \infty$, we can use similar reasoning:

$$ |g(x)| = \left| \frac{\sin x}{x^p} \right| $$

Using boundedness of $|\sin x| \leq 1$, we integrate:

$$ \int_a^\infty \frac{1}{x^p} , dx $$

This integral converges for $p > 1$.

5. Summarize the results

Thus, we conclude that both integrals converge under the given conditions.