Question

Let \(h\) be continuous on \([0, \infty) .\) Prove: (a) If \(\int_{0}^{\infty} e^{-s_{0} x} h(x) d x\) converges absolutely, then \(\int_{0}^{\infty} e^{-s x} h(x) d x\) converges absolutely if \(s>s_{0}\) (b) If \(\int_{0}^{\infty} e^{-s_{0} x} h(x) d x\) converges, then \(\int_{0}^{\infty} e^{-s x} h(x) d x\) converges if \(s>s_{0}\).

Short answer

In summary, for any given \(s > s_0\), we have proven the absolute convergence of the improper integral \(\int_{0}^{\infty} e^{-sx} h(x) dx\) using the Comparison Test in part (a). Then, we used integration by parts to establish the convergent behavior of the integral for \(s > s_0\) in part (b). As a result, we have shown the convergence properties of the integral as required by the problem statement.

Step-by-step solution

Part (a): Prove Absolute Convergence

To prove absolute convergence, we need to show that \(\int_{0}^{\infty} |e^{-sx} h(x)| dx < \infty\) when \(s > s_{0}\). Since \(s > s_{0}\), we know that \(e^{-(s-s_{0})x} > 1\) for all \(x \geq 0\). Then, we have: \(|e^{-sx} h(x)| = e^{-s_{0}x} |e^{-(s-s_{0})x} h(x)| \leq e^{-s_{0}x} |h(x)|\). Now, we can use the Comparison Test: \(\int_{0}^{\infty} |e^{-sx} h(x)| dx \leq \int_{0}^{\infty} e^{-s_{0}x} |h(x)| dx.\) Since we are given that \(\int_{0}^{\infty} e^{-s_{0}x} h(x) dx\) converges absolutely, the right-hand side is finite. So, \(\int_{0}^{\infty} |e^{-sx} h(x)| dx\) also converges, proving that the integral converges absolutely when \(s > s_{0}\).

Part (b): Prove Convergence

To prove convergence, we can make use of integration by parts. First, let \(u = e^{-sx}\) and \(dv = h(x) dx\). Then, \(du = -se^{-sx} dx\) and \(v = \int h(x) dx\). Applying integration by parts to the improper integral, we get: \(\int_{0}^{\infty} e^{-sx} h(x) dx = \left[ e^{-sx} \int h(x) dx \right]_{0}^{\infty} + s \int_{0}^{\infty} e^{-sx} \int h(x) dx\). First, consider the first term on the right-hand side: \(\lim_{t\to\infty} e^{-st} \int_0^t h(x) dx = 0\) since \(e^{-st} \to 0\) as \(t \to \infty\) and \(\int_0^t h(x) dx\) is finite for all \(t\) as \(h(x)\) is continuous. Now, we need to prove that the second term also converges: \(s \int_{0}^{\infty} e^{-sx} \int h(x) dx\). Since \(\int_{0}^{\infty} e^{-s_{0}x} h(x) dx\) converges, we know that \(\int_{0}^{\infty} e^{-s_{0}x} \int h(x) dx < \infty\). Then, we can apply the result from part (a), which guarantees that: \(s \int_{0}^{\infty} e^{-sx} \int h(x) dx < \infty\) for \(s > s_{0}\). Thus, both terms on the right-hand side of the integration by parts equation converge, so we can conclude that \(\int_{0}^{\infty} e^{-sx} h(x) dx\) also converges for \(s > s_{0}\).

Key concepts

Absolute Convergence

When we talk about 'absolute convergence,' particularly in the context of improper integrals, we mean that an integral remains finite when we consider the absolute value of the integrand over the interval of integration. In simpler terms, an integral \( \int_{a}^{b} f(x) dx \) is said to exhibit absolute convergence if \( \int_{a}^{b} |f(x)| dx \) is finite.

As highlighted in the exercise, if we have an integral \( \int_{0}^{\text{\infty}} e^{-s_{0} x} h(x) dx \) that converges absolutely, it sets a benchmark. If we increase the decay rate, represented by a larger \( s \) than \( s_{0} \) in the exponential function, the absolute value of the integral becomes even more likely to stay finite. This is due to the fact that increasing \( s \) causes the exponential term to drop off more quickly, resulting in a smaller product with \( h(x) \), and therefore a smaller overall integral value.

Evaluating absolute convergence is crucial because it helps insure against the potential pitfalls of regular convergence, where positive and negative areas under a curve might undesirably cancel each other out. This concept ensures the integral 'behaves well' regardless of any sign changes in the integrand.

Integration by Parts

The technique known as 'integration by parts' is essentially the reverse process of the product rule in differentiation. It's particularly handy in solving integrals where the integrand is a product of two functions that are not easily integrable in their current form. The formula for integration by parts is \( \int u dv = uv - \int v du \).

In our problem, by selecting \( u = e^{-sx} \) and \( dv = h(x) dx \) we cleverly split the original problem into parts that can be managed more easily. After computing \( du \) and \( v \) we can reframe the integral in terms of these new functions, simplifying the original problem and making it solvable.

Integration by parts is especially useful for improper integrals because it can transform a difficult-to-evaluate integral into a simpler form or even break it down into terms that can each be evaluated for convergence separately, as seen in step 2 of our solution.

Comparison Test

The Comparison Test is a powerful tool for establishing the convergence or divergence of improper integrals. The test compares our integral to another integral whose convergence behavior is already known. Specifically, if we have two functions \( f(x) \) and \( g(x) \) where \( 0 \leq f(x) \leq g(x) \) for all \( x \) in the interval, and if \( \int g(x) dx \) is known to be finite, then \( \int f(x) dx \) must also be finite.

In our case, we knew that the integral \( \int_{0}^{\text{\infty}} e^{-s_{0} x} h(x) dx \) converges absolutely. By using the fact that \( |e^{-sx} h(x)| \leq e^{-s_{0}x} |h(x)| \) for \( s > s_{0} \) and all \( x \geq 0 \) (as \( e^{-(s-s_{0})x} \geq 1 \) ), we established that \( \int_{0}^{\text{\infty}} |e^{-sx} h(x)| dx \) must converge as well. This is because it's, in essence, smaller than a convergent integral, hence the exercise used the Comparison Test to prove absolute convergence of this new integral.