Question

Show that $$ \int_{0}^{\infty} e^{-s t} e^{t^{2}} d t=\infty $$ for every real number \(s\).

Short answer

Question: Show that the value of the given integral is infinity for every real number \(s\): $$ \int_{0}^{\infty} e^{-st} e^{t^{2}} dt $$ Answer: The value of the given integral is infinity for every real number \(s\) because, as \(t\) approaches infinity, the integrand function \(e^{-st} e^{t^{2}} = e^{(-st + t^2)}\) also approaches infinity. This means that the integral does not converge to a finite value for any real number \(s\), and thus, the integral is equal to infinity for all \(s\).

Step-by-step solution

Analyzing the integrand

The given integral equation is $$ \int_{0}^{\infty} e^{-st} e^{t^{2}} dt $$ Our goal is to analyze the behavior of the integrand function \(e^{-st} e^{t^{2}}\) as \(t\) approaches infinity.

Simplify the integrand

Combine the exponents using the product rule for exponents which states that \(a^{m} \cdot a^{n} = a^{m+n}\). The integrand simplifies to $$ e^{(-st + t^2)} $$

Analyze the behavior of \(-st + t^2\) as \(t \rightarrow \infty\)

Let's analyze the behavior of the exponent \(-st + t^2\) as \(t\) approaches infinity. To find the limit of the exponent as \(t \rightarrow \infty\), we have $$ \lim_{t \rightarrow \infty} (-st + t^2). $$ To analyze this limit, it's helpful to look at two cases: 1. When \(s > 0\) 2. When \(s < 0\) Case 1: When \(s > 0\), the exponent is dominated by the quadratic term \(t^2\). Therefore, $$ \lim_{t \rightarrow \infty} (-st + t^2) = \infty. $$ Case 2: When \(s < 0\), we can factor out a negative \(t\) from the exponent: $$ \lim_{t \rightarrow \infty} -(t(s - t)) = \infty $$ since \(t\) is still approaching infinity and the expression \(s - t\) becomes more negative as \(t\) increases. So, $$ \lim_{t \rightarrow \infty} (-st + t^2) = \infty. $$

Evaluate the limit of the integrand

Now that we've established that the limit of the exponent \((-st + t^2)\) is infinity as \(t\) approaches infinity, we can evaluate the limit of the integrand: $$ \lim_{t \rightarrow \infty} e^{(-st + t^2)} = e^{\infty} = \infty. $$

Conclusion

Since the integrand function \(e^{-st} e^{t^{2}} = e^{(-st + t^2)}\) approaches infinity as \(t\) approaches infinity, the integral does not converge to a finite value for any real number \(s\). Hence, we can conclude that $$ \int_{0}^{\infty} e^{-st} e^{t^{2}} dt = \infty $$ for every real number \(s\).

Key concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. The basic form of an exponential function is \(f(x) = a^{x}\), where \(a\) is a constant and \(x\) is the exponent. They appear in various real-world contexts like population growth, radioactive decay, and interest calculations.
Exponential functions have unique properties:

• If the base \(a > 1\), the function grows rapidly as \(x\) increases.
• If \(0 < a < 1\), the function decays towards zero as \(x\) increases.
• These functions have a constant rate of change, which makes them suitable for modeling continuous growth or decay.

In the integral \(\int_{0}^{\infty} e^{-st} e^{t^2} dt\), the base \(e\) (approximately 2.718) is raised to a complex exponent, \(-st + t^2\), which combines both linear and quadratic terms. This combination makes the behavior of the integrand more complicated and requires careful analysis.

Limit of a Function

The limit of a function describes the behavior of the function as it approaches a particular point or heads towards infinity. Knowing the limit helps predict the function's behavior without evaluating it at extremely large or small values.

• If \( \lim_{x \to a} f(x) = L \), it means as \(x\) gets closer to \(a\), \(f(x)\) gets closer to \(L\).
• Limits are crucial for understanding continuity, derivatives, and integrals.

In the integral from the exercise, we look at the limit of the exponent \(-st + t^2\) as \(t\) approaches infinity. This calculation shows whether the integrand grows without bound and, as a result, whether the integral converges or diverges. Analyzing the limit is especially important since it defines the integral's behavior over an infinite range.

Convergence and Divergence

Convergence and divergence are key concepts in calculus, especially concerning integrals and series. An integral or series converges if it approaches a specific value as it extends towards infinity. In contrast, it diverges if it doesn't approach a specific limit.
To test for convergence:

• Evaluate if the sequence of partial sums of an infinite series approaches a finite number.
• Check if an improper integral has a finite upper limit towards infinity.

For the exercise integral \(\int_{0}^{\infty} e^{-st} e^{t^2} dt\), the behavior of the integrand \(e^{(-st + t^2)}\) when \(t\) approaches infinity is key. Since the limit of the exponent is infinity for any real \(s\), the integrand itself grows without bounds, indicating that the integral diverges, or doesn't converge to a finite value.

Real Number

Real numbers include all the numbers on the continuous number line, encompassing both rational numbers (e.g., 1/2, 3) and irrational numbers (e.g., \(\sqrt{2}\), \(\pi\)). They are fundamental in calculus and provide the basis for measuring distance and quantities.
Properties of real numbers include:

• They are ordered, meaning any two real numbers can be compared and arranged in sequence.
• They are dense, indicating there are infinitely many real numbers between any two distinct real numbers.
• The real numbers form the real-valued function's domain when defining behavior over all possible input values.

In the context of the exercise, \(s\) represents any real number, which is important because it demonstrates that no matter the value (positive, negative, or zero), the behavior of the integral remains unchanged—it diverges.