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Chapter 6: Problem 22

Determine whether the given integral converges or diverges. $$ \int_{0}^{\infty} t e^{-t} d t $$

Short Answer

Expert verified
Answer: The improper integral converges, and its value is -1.

Step by step solution

01

Identify the improper integral

We have an improper integral with infinite limits of integration: $$ \int_{0}^{\infty} t e^{-t} dt $$ We can rewrite this integral as a limit: $$ \lim_{R\to\infty} \int_{0}^{R} t e^{-t} dt $$
02

Find the antiderivative of the integrand

To find the antiderivative of the function \(te^{-t}\), we can use integration by parts. Let \(u = t\) and \(dv = e^{-t} dt\). Then, \(du = dt\) and \(v = -e^{-t}\). According to the integration by parts formula, we have: $$ \int u dv = uv - \int v du $$ Substituting our expressions for \(u\), \(dv\), \(v\), and \(du\), we get: $$ \int t\cdot e^{-t}\,dt = -te^{-t} - \int(-e^{-t})\,dt $$ Now we integrate \(-e^{-t}\,dt\): $$ \int(-e^{-t})\,dt = e^{-t} + C $$ So, the antiderivative of the integrand is: $$ -te^{-t} + e^{-t} + C $$
03

Evaluate the limit

Now, we will substitute the limits of integration and apply the limit that makes the integral improper: $$ \lim_{R\to\infty} \left[-te^{-t} + e^{-t}\right]_0^R $$ First, we will substitute in the limits of integration: $$ \lim_{R\to\infty}\left( -Re^{-R} + e^{-R} - \left(-0e^0 + e^0\right)\right) $$ Simplify the expression within the parentheses: $$ \lim_{R\to\infty}\left( -Re^{-R} + e^{-R} - 1\right) $$ To find the limit as \(R\) approaches infinity, we can separately find the limits of each term: $$ \lim_{R\to\infty}(-Re^{-R}) = 0\\ \lim_{R\to\infty}(e^{-R}) = 0\\ \lim_{R\to\infty}(1) = 1 $$ Therefore, the limit converges: $$ \lim_{R\to\infty}\left( -Re^{-R} + e^{-R} - 1\right) = -1 $$
04

Determine convergence/divergence

Since the limit converges, the improper integral converges as well. Thus, the given improper integral converges, and the final answer is: $$ \int_{0}^{\infty} t e^{-t} dt = -1 $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Parts
Integration by parts is a technique used to evaluate integrals where the integrand is a product of two functions for which standard integration methods don't apply directly. This method stems from the product rule in differentiation and is formally expressed as:
\[\begin{equation}\int u dv = uv - \int v du\end{equation}\].
In the context of our exercise, we applied integration by parts to the integrand \(t e^{-t}\). Here, we let \(u = t\) and \(dv = e^{-t} dt\), yielding \(du = dt\) and \(v = -e^{-t}\). This decomposition allows us to transform the original integral into a simpler form:
\[\begin{equation}\int t\text{\textbullet} e^{-t}\,dt = -te^{-t} - \int(-e^{-t})\,dt\end{equation}\].
Using this technique, we aim to reduce the complexity of the integral by breaking it down into parts that are easier to handle. It's beneficial to choose \(u\) and \(dv\) such that \(dv\) is easily integrable and \(du\) simplifies the original integrand.
Antiderivative
An antiderivative, also known as an indefinite integral, is a function whose derivative is the original function we started with. Identifying the antiderivative is a critical step when solving definite and improper integrals. In our exercise, the antiderivative of \(te^{-t}\) was found using integration by parts, which provided us with:
\[\begin{equation}-te^{-t} + e^{-t} + C\end{equation}\].
The constant \(C\) represents the family of functions that differ by a constant, since the process of differentiation removes any constant term. When faced with integrals involving products or complex expressions, finding the antiderivative often involves techniques such as substitution, integration by parts, or even looking up common antiderivatives.
Limit of an Integral
The limit of an integral assesses the behavior of a function as the input approaches a specific value or infinity. This concept is especially pertinent in our exercise, dealing with improper integrals, which extend over an infinite interval or approach a singularity. To determine the convergence of an improper integral like \(\int_{0}^{\infty} t e^{-t} dt\), we express it as a limit:
\[\begin{equation}\lim_{R\to\infty} \int_{0}^{R} t e^{-t} dt\end{equation}\].
By doing so, we calculate the integral over a finite range and then investigate the behavior as that range expands to infinity. We then evaluate this limit by substituting into the antiderivative and calculating:
\[\begin{equation}\lim_{R\to\infty} \left[-te^{-t} + e^{-t}\right]_0^R\end{equation}\].
If the limit exists and is finite, the integral converges; if not, it diverges. For \(te^{-t}\), we observed that each term in the antiderivative approaches zero as \(R\) approaches infinity, except for a constant term. Hence, the integral converges, revealing the power of the concept of limits in solving improper integrals and understanding the behavior of functions at infinity.

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Most popular questions from this chapter

In this problem we show how a general partial fraction expansion can be used to calculate many inverse Laplace transforms. Suppose that $$ F(s)=P(s) / Q(s) $$ where \(Q(s)\) is a polynomial of degree \(n\) with distinct zeros \(r_{1} \ldots r_{n}\) and \(P(s)\) is a polynomial of degree less than \(n .\) In this case it is possible to show that \(P(s) / Q(s)\) has a partial fraction cxpansion of the form $$ \frac{P(s)}{Q(s)}=\frac{A_{1}}{s-r_{1}}+\cdots+\frac{A_{n}}{s-r_{n}} $$ where the coefficients \(A_{1}, \ldots, A_{n}\) must be determined. \(\begin{array}{ll}{\text { (a) Show that }} & {} \\ {\qquad A_{k}=P\left(r_{k}\right) / Q^{\prime}\left(r_{k}\right),} & {k=1, \ldots, n}\end{array}\) Hint: One way to do this is to multiply Eq. (i) by \(s-r_{k}\) and then to take the limit as \(s \rightarrow r_{k}\) (b) Show that $$ \mathcal{L}^{-1}\\{F(s)\\}=\sum_{k=1}^{n} \frac{P\left(r_{k}\right)}{Q^{\prime}\left(r_{k}\right)} e^{r_{k} t} $$

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