To solve the integral \( I = \int_0^1 (\ln x)^n \, dx \) where \( n \) is a positive integer, we can use integration by parts. Here’s a step-by-step solution:
### Step 1: Set up the integration by parts
We can use the integration by parts formula:
\[
\int u \, dv = uv - \int v \, du
\]
Let:
- \( u = (\ln x)^n \)
- \( dv = dx \)
Then, we differentiate and integrate:
- \( du = n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx \)
- \( v = x \)
### Step 2: Apply integration by parts
Now we apply the integration by parts:
\[
I = \left[ x (\ln x)^n \right]_0^1 - \int_0^1 x \cdot n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx
\]
This simplifies to:
\[
I = \left[ x (\ln x)^n \right]_0^1 - n \int_0^1 (\ln x)^{n-1} \, dx
\]
### Step 3: Evaluate the boundary term
Now we evaluate the boundary term:
- At \( x = 1 \): \( \ln(1) = 0 \) so \( 1 \cdot (\ln 1)^n = 0 \).
- At \( x = 0 \): \( \ln(0) \) approaches \(-\infty\), but \( x \cdot (\ln x)^n \) approaches \( 0 \) as \( x \) approaches \( 0 \) (since \( x \) goes to \( 0 \) faster than \( (\ln x)^n \) goes to \(-\infty\)).
Thus, the boundary term evaluates to \( 0 - 0 = 0 \).
### Step 4: Substitute back into the equation
Now we have:
\[
I = 0 - n \int_0^1 (\ln x)^{n-1} \, dx
\]
This gives us:
\[
I = -n I_{n-1}
\]
where \( I_{n-1} = \int_0^1 (\ln x)^{n-1} \, dx \).
### Step 5: Recursion relation
We can express this as:
\[
I_n = -n I_{n-1}
\]
### Step 6: Base case
To find \( I_n \), we need to find the base case:
For \( n = 1 \):
\[
I_1 = \int_0^1 \ln x \, dx
\]
Using integration by parts again:
Let \( u = \ln x \) and \( dv = dx \):
- Then \( du = \frac{1}{x} \, dx \) and \( v = x \).
\[
I_1 = \left[ x \ln x \right]_0^1 - \int_0^1 x \cdot \frac{1}{x} \, dx = 0 - \int_0^1 1 \, dx = -1
\]
### Step 7: Calculate \( I_n \)
Now we can calculate \( I_n \) using the recursion:
\[
I_2 = -2 I_1 = -2(-1) = 2
\]
\[
I_3 = -3 I_2 = -3(2) = -6
\]
\[
I_4 = -4 I_3 = -4(-6) = 24
\]
Continuing this pattern, we can generalize:
\[
I_n = (-1)^n n!
\]
### Final Result
Thus, the final result is:
\[
I = \int_0^1 (\ln x)^n \, dx = (-1)^n n!
\]
