To evaluate the integral

$\lambda \int_{0}^{\infty} e^{x(t - \lambda)} \, dx,$

we will proceed step-by-step.

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Step 1: Interpret the Integral

The integral is dependent on the variables $\lambda$ and $t$. We assume $\lambda > 0$ and $t - \lambda \neq 0$ to ensure convergence.

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Step 2: Solve the Exponential Integral

The integral is of the form:

$\int_{0}^{\infty} e^{x(t - \lambda)} \, dx.$

• Let $A = t - \lambda$.
• Rewrite the integral:

$\int_{0}^{\infty} e^{xA} \, dx.$

This converges only when $A < 0$, i.e., $t - \lambda < 0$ or $t < \lambda$. When $A < 0$, we can compute the integral as:

$\int_{0}^{\infty} e^{xA} \, dx = \frac{1}{-A} = \frac{1}{\lambda - t}.$

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Step 3: Multiply by $\lambda$

Including the prefactor $\lambda$, the result becomes:

$\lambda \int_{0}^{\infty} e^{x(t - \lambda)} \, dx = \frac{\lambda}{\lambda - t},$

valid for $\lambda > t$.

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Final Answer:

$\boxed{\lambda \int_{0}^{\infty} e^{x(t - \lambda)} \, dx = \frac{\lambda}{\lambda - t}, \quad \text{for } \lambda > t.}$

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Would you like a detailed explanation of each step or further exploration of related topics?

Follow-up Questions:

1. What is the physical interpretation of this integral in an applied context?
2. How does the result change if $t = \lambda$?
3. Can you compute the same integral over a finite interval, say, from 0 to $a$?
4. What happens to the solution when $\lambda$ approaches $t$ from above?
5. Could this integral be evaluated using Laplace transforms?

Tip:

When dealing with exponential integrals, always verify convergence criteria before attempting to compute the integral.