To solve the integral \( \int_0^\infty x^n e^{-ax} \, dx \), given that \( \int_0^\infty e^{-ax} \, dx = \frac{1}{a} \), we can use integration by parts and the properties of the gamma function. Here’s the step-by-step solution:
### Step 1: Set Up the Integral
We start with the integral we want to evaluate:
\[
I_n = \int_0^\infty x^n e^{-ax} \, dx
\]
### Step 2: Use Integration by Parts
We will use integration by parts, where we let:
- \( u = x^n \) (thus \( du = n x^{n-1} \, dx \))
- \( dv = e^{-ax} \, dx \) (thus \( v = -\frac{1}{a} e^{-ax} \))
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:
\[
I_n = \left[ -\frac{1}{a} x^n e^{-ax} \right]_0^\infty + \frac{n}{a} \int_0^\infty x^{n-1} e^{-ax} \, dx
\]
### Step 3: Evaluate the Boundary Term
Now we evaluate the boundary term \( \left[ -\frac{1}{a} x^n e^{-ax} \right]_0^\infty \):
- As \( x \to \infty \), \( e^{-ax} \) approaches 0 faster than \( x^n \) approaches infinity, so this term approaches 0.
- As \( x \to 0 \), \( x^n e^{-ax} \) approaches 0.
Thus, the boundary term is 0:
\[
\left[ -\frac{1}{a} x^n e^{-ax} \right]_0^\infty = 0
\]
### Step 4: Substitute Back into the Integral
Now we have:
\[
I_n = \frac{n}{a} I_{n-1}
\]
### Step 5: Recursively Define the Integral
We can express \( I_n \) in terms of \( I_{n-1} \):
\[
I_n = \frac{n}{a} I_{n-1}
\]
Continuing this process, we can express \( I_{n-1}, I_{n-2}, \ldots, I_0 \):
\[
I_{n-1} = \frac{n-1}{a} I_{n-2}
\]
\[
I_{n-2} = \frac{n-2}{a} I_{n-3}
\]
Continuing this until we reach \( I_0 \):
\[
I_0 = \int_0^\infty e^{-ax} \, dx = \frac{1}{a}
\]
### Step 6: Combine the Results
Now we can substitute back:
\[
I_n = \frac{n}{a} \cdot \frac{n-1}{a} \cdot \frac{n-2}{a} \cdots \frac{1}{a} \cdot I_0
\]
This gives us:
\[
I_n = \frac{n!}{a^{n+1}}
\]
### Final Result
Thus, the final result for the integral \( \int_0^\infty x^n e^{-ax} \, dx \) is:
\[
\int_0^\infty x^n e^{-ax} \, dx = \frac{n!}{a^{n+1}}
\]
