Questions: Let G(x) = ∫ from e^(x^2) to e^(e^(z^2)) ln(t)/t dt. What is the domain of G?

Solution Steps

To determine the domain of \( G(x) \), we need to ensure that the limits of the integral are valid and that the integrand is defined for all \( t \) in the range of integration. Specifically, we need to check the conditions under which \( e^{x^2} \) and \( e^{e^{z^2}} \) are valid and ensure that \( \frac{\ln t}{t} \) is defined for all \( t \) in the interval \([e^{x^2}, e^{e^{z^2}}]\).

1. Identify the range of \( e^{x^2} \) and \( e^{e^{z^2}} \).
2. Ensure that \( \frac{\ln t}{t} \) is defined for all \( t \) in the interval \([e^{x^2}, e^{e^{z^2}}]\).
3. Determine the values of \( x \) for which the above conditions hold.

The limits of the integral are given by: \[ \text{lower limit} = e^{x^2} \] \[ \text{upper limit} = e^{e^{z^2}} \]

The integrand is: \[ \frac{\ln t}{t} \] This function is defined for \( t > 0 \). Since both \( e^{x^2} \) and \( e^{e^{z^2}} \) are always positive for all real values of \( x \) and \( z \), the integrand is defined over the interval \([e^{x^2}, e^{e^{z^2}}]\).

Since \( e^{x^2} > 0 \) and \( e^{e^{z^2}} > 0 \) for all \( x, z \in \mathbb{R} \), the domain of \( G(x) \) is: \[ x \in \mathbb{R}, \quad z \in \mathbb{R} \]

Final Answer