Question

Determine whether the given integral converges or diverges. $$ \int_{1}^{\infty} t^{-2} e^{t} d t $$

Short answer

Answer: The improper integral $$\int_{1}^{\infty} t^{-2} e^{t} d t$$ diverges.

Step-by-step solution

Apply the Comparison Test

The Comparison Test states that if the absolute value of our function \(|t^{-2} e^{t}|\) is less than or equal to the absolute value of a simpler function \(g(t)\) and the integral of \(g(t)\) converges, then the integral of our function $$\int_{1}^{\infty} |t^{-2} e^{t}| dt$$ also converges. Let's compare our function with \(g(t) = t^{-2}\): $$ t^{-2} e^{t} \leq t^{-2} \quad \Rightarrow \quad e^{t} \leq 1 $$ Since \(e^{t}\) is only less than or equal to 1 when \(t=0\), the Comparison Test is not applicable here. We can consider the integral test instead.

Apply Integral Test

In order to apply the Integral Test, the function must be positive, continuous, and non-increasing on the interval \([1, \infty)\). As our function is positive and continuous, we only need to check if the function is non-increasing: $$ \frac{d(t^{-2} e^{t})}{dt} = -2t^{-3} e^{t} + t^{-2} e^{t} $$ Since the derivative does not provide a simple expression on whether the function is non-increasing, we must look for another approach.

Apply Limit Comparison Test

We can use the Limit Comparison Test instead of a direct comparison. The Limit Comparison Test states that if the limit as \(t \to \infty\) of the ratio between the functions is a finite positive number, then the improper integrals will both converge or diverge together. Let's take the limit as \(t \rightarrow \infty\) of the ratio between the functions: $$ \lim_{t \to \infty} \frac{t^{-2} e^{t}}{t^{-2}} = \lim_{t \to \infty} e^{t} $$ The limit is infinity which doesn't tell us whether the integral converges or diverges. Due to the exponential nature of the function, we can make an educated guess that the given integral diverges.

Do the Integration

Now, let's evaluate the improper integral $$\int_{1}^{\infty} t^{-2} e^{t} d t$$: First, let's substitute \(t\) with a proper integral limit \(b\) and evaluate the limit as \(b \rightarrow \infty\). $$ \lim_{b \to \infty}\int_{1}^{b} t^{-2} e^{t} d t $$ To find the antiderivative of the function \(t^{-2} e^{t}\), we can use integration by parts or notice that \(t^{-2}e^t\) doesn't have any elementary antiderivative. Because we're unable to find a simple antiderivative for the function and the exponential nature of the function seems to cause the integral to grow infinitely large, we conclude that the integral diverges.

Conclusion

So, the improper integral $$\int_{1}^{\infty} t^{-2} e^{t} d t$$ diverges.

Key concepts

Comparison Test

The Comparison Test is a useful tool when dealing with improper integrals. It helps us determine if an integral converges or diverges by comparing it with another, simpler function that's easier to evaluate. The idea is straightforward.
If you have a function \( f(t) \) that you want to check for convergence, compare it to a known function \( g(t) \). Here’s how it works:

• If \( |f(t)| \leq |g(t)| \) for all \( t \) and \( \int_{1}^{\infty} g(t) \, dt \) converges, then \( \int_{1}^{\infty} f(t) \, dt \) also converges.
• Alternatively, if \( |f(t)| \geq |g(t)| \) and \( \int_{1}^{\infty} g(t) \, dt \) diverges, then \( \int_{1}^{\infty} f(t) \, dt \) diverges too.

In our original problem, the comparison test was attempted, but it didn’t apply because the function \( e^t \) grows unbounded, failing the condition \( e^t \leq 1 \) except when \( t=0 \). Even though this specific approach failed, it's a central strategy in tackling improper integrals.

Integral Test

The Integral Test provides another method for assessing convergence or divergence of series and improper integrals. This test is typically handy when you have a function that’s positive, continuous, and non-increasing on a given interval.
The method asserts that if \( f(t) \) meets these criteria:

• \( f(t) \) is positive, continuous, and non-increasing for \( t \geq a \).

The integral \( \int_{a}^{\infty} f(t) \, dt \) converges if and only if the corresponding infinite series \( \sum_{n=a}^{\infty} f(n) \) converges.
In the problem’s solution, the Integral Test was considered for \( t^{-2} e^t \). However, the derivative analysis of this function didn't straightforwardly reveal if it’s non-increasing. This led us to switch to alternative methods like the Limit Comparison Test. Still, the Integral Test remains a powerful concept in situations where these conditions are clearly met.

Limit Comparison Test

When the direct application of the Comparison Test fails, the Limit Comparison Test acts as a reliable backup. It uses the ratio of two functions to infer convergence behaviors.
The Limit Comparison Test states:

• Suppose \( f(t) \) and \( g(t) \) are positive functions for \( t \geq a \).
• If \( \lim_{t \to \infty} \frac{f(t)}{g(t)} = c \), where \( c \) is a finite, positive number, then the improper integrals \( \int_{a}^{\infty} f(t) \, dt \) and \( \int_{a}^{\infty} g(t) \, dt \) either both converge or both diverge.

For \( t^{-2} e^t \), the calculation \( \lim_{t \to \infty} \frac{t^{-2} e^t}{t^{-2}} = \lim_{t \to \infty} e^{t} \) yielded infinity. This result, suggesting the evaluated limit is not finite, hinted at the integral's divergence, affirming what intuition about the function’s behavior implies. The test is especially useful when direct comparisons are unclear.

Divergence of Integrals

Divergence in the context of integrals indicates that their value approaches infinity. To determine divergence, we analyze the behavior of the integral’s function as the variable approaches infinity.
In the original problem \( \int_{1}^{\infty} t^{-2} e^{t} \, dt \), we suspect divergence thanks to the function’s exponential factor \( e^{t} \). This term outpaces decay terms like \( t^{-2} \), pushing the integral's value towards infinity.
Often, divergence is established when we can’t find an elementary antiderivative, and potential growth elements like exponential terms are present. When handling such integrals:

• Attempting integration or finding an antiderivative may prove intractable, confirming divergence.
• Mathematical intuition based on elementary functions' growth behaviors supports analyses.

The conclusion of divergence confirms that as \( t \to \infty \), not only does the integral not settle to a finite number, it instead accumulates indefinitely.