91Ó°ÊÓ

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.

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

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, $$ \lim_{t \rightarrow \infty} (-st + t^2) = \infty. $$

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: $$ \lim_{t \rightarrow \infty} e^{(-st + t^2)} = e^{\infty} = \infty. $$

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 \(s\).