Let $ a$ be a positive constant number. For a positive integer $ n$, define a function $ I_n(t)$ by $ I_n(t)\equal{}\int_0^t x^ne^{\minus{}ax}dx$. Answer the following questions.

Note that you may use $ \lim_{t\rightarrow \infty} t^ne^{\minus{}at}\equal{}0$ without proof.

(1) Evaluate $ I_1(t)$.

(2) Find the relation of $ I_{n\plus{}1}(t),\ I_n(t)$.

(3) Prove that there exists $ \lim_{t\to\infty} I_n(t)$ for all natural number $ n$ by using mathematical induction.

(4) Find $ \lim_{t\to\infty} I_n(t)$.
1. **Evaluate \( I_1(t) \):**

   We start with the definition of \( I_1(t) \):
   \[
   I_1(t) = \int_0^t x e^{-ax} \, dx
   \]
   To evaluate this integral, we use integration by parts. Let \( u = x \) and \( dv = e^{-ax} \, dx \). Then, \( du = dx \) and \( v = -\frac{1}{a} e^{-ax} \). Applying integration by parts:
   \[
   \int u \, dv = uv - \int v \, du
   \]
   we get:
   \[
   I_1(t) = \left. -\frac{x}{a} e^{-ax} \right|_0^t + \int_0^t \frac{1}{a} e^{-ax} \, dx
   \]
   Evaluating the boundary terms:
   \[
   \left. -\frac{x}{a} e^{-ax} \right|_0^t = -\frac{t}{a} e^{-at} + 0 = -\frac{t}{a} e^{-at}
   \]
   Now, we evaluate the remaining integral:
   \[
   \int_0^t \frac{1}{a} e^{-ax} \, dx = \frac{1}{a} \left[ -\frac{1}{a} e^{-ax} \right]_0^t = \frac{1}{a^2} \left( 1 - e^{-at} \right)
   \]
   Combining these results, we have:
   \[
   I_1(t) = -\frac{t}{a} e^{-at} + \frac{1}{a^2} \left( 1 - e^{-at} \right) = \frac{1}{a^2} - \frac{1}{a^2} e^{-at} - \frac{t}{a} e^{-at}
   \]

2. **Find the relation of \( I_{n+1}(t) \) and \( I_n(t) \):**

   Using integration by parts again, we start with:
   \[
   I_{n+1}(t) = \int_0^t x^{n+1} e^{-ax} \, dx
   \]
   Let \( u = x^{n+1} \) and \( dv = e^{-ax} \, dx \). Then, \( du = (n+1)x^n \, dx \) and \( v = -\frac{1}{a} e^{-ax} \). Applying integration by parts:
   \[
   I_{n+1}(t) = \left. -\frac{x^{n+1}}{a} e^{-ax} \right|_0^t + \int_0^t \frac{n+1}{a} x^n e^{-ax} \, dx
   \]
   Evaluating the boundary terms:
   \[
   \left. -\frac{x^{n+1}}{a} e^{-ax} \right|_0^t = -\frac{t^{n+1}}{a} e^{-at} + 0 = -\frac{t^{n+1}}{a} e^{-at}
   \]
   The remaining integral is:
   \[
   \int_0^t \frac{n+1}{a} x^n e^{-ax} \, dx = \frac{n+1}{a} I_n(t)
   \]
   Combining these results, we get:
   \[
   I_{n+1}(t) = -\frac{t^{n+1}}{a} e^{-at} + \frac{n+1}{a} I_n(t)
   \]

3. **Prove that there exists \( \lim_{t \to \infty} I_n(t) \) for all natural numbers \( n \) by using mathematical induction:**

   We start by noting that:
   \[
   \lim_{t \to \infty} t^n e^{-at} = 0
   \]
   For \( n = 1 \), we have:
   \[
   I_1(t) = \frac{1}{a^2} - \frac{1}{a^2} e^{-at} - \frac{t}{a} e^{-at}
   \]
   Taking the limit as \( t \to \infty \):
   \[
   \lim_{t \to \infty} I_1(t) = \frac{1}{a^2}
   \]
   Assume \( \lim_{t \to \infty} I_n(t) = L_n \) exists. Then:
   \[
   I_{n+1}(t) = -\frac{t^{n+1}}{a} e^{-at} + \frac{n+1}{a} I_n(t)
   \]
   Taking the limit as \( t \to \infty \):
   \[
   \lim_{t \to \infty} I_{n+1}(t) = 0 + \frac{n+1}{a} L_n = \frac{n+1}{a} L_n
   \]
   By induction, \( \lim_{t \to \infty} I_n(t) \) exists for all \( n \).

4. **Find \( \lim_{t \to \infty} I_n(t) \):**

   From the induction step, we have:
   \[
   L_{n+1} = \frac{n+1}{a} L_n
   \]
   Starting with \( L_0 = \lim_{t \to \infty} I_0(t) = \lim_{t \to \infty} \int_0^t e^{-ax} \, dx = \frac{1}{a} \):
   \[
   L_1 = \frac{1}{a} L_0 = \frac{1}{a^2}
   \]
   \[
   L_2 = \frac{2}{a} L_1 = \frac{2}{a^3}
   \]
   Continuing this pattern, we get:
   \[
   L_n = \frac{n!}{a^{n+1}}
   \]

The final answer is \( \boxed{ \lim_{t \to \infty} I_n(t) = \frac{n!}{a^{n+1}} } \)