`I_(m,n)=int_0^1x^m(logx)^ndx =`

Options

A. `n/(n+1)I_(m,n-1)`
B. `-m/(n+1)I_(m,n-1)`
C. `-n/(m+1)I_(m,n-1)`
D. `m/(n+1)I_(m,n-1)`

Answer

To solve the integral \( I_{m,n} = \int_0^1 x^m (\log x)^n \, dx \), we will use integration by parts. Here are the steps:

### Step 1: Set up integration by parts
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = (\log x)^n \) (which we will differentiate)
- \( dv = x^m \, dx \) (which we will integrate)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now, we differentiate \( u \) and integrate \( dv \):
- \( du = n (\log x)^{n-1} \cdot \frac{1}{x} \, dx = \frac{n (\log x)^{n-1}}{x} \, dx \)
- \( v = \int x^m \, dx = \frac{x^{m+1}}{m+1} \)

### Step 3: Apply integration by parts
Now substitute into the integration by parts formula:
\[ I_{m,n} = \left[ (\log x)^n \cdot \frac{x^{m+1}}{m+1} \right]_0^1 - \int_0^1 \frac{x^{m+1}}{m+1} \cdot \frac{n (\log x)^{n-1}}{x} \, dx \]
This simplifies to:
\[ I_{m,n} = \left[ \frac{(\log x)^n \cdot x^{m+1}}{m+1} \right]_0^1 - \frac{n}{m+1} \int_0^1 x^m (\log x)^{n-1} \, dx \]

### Step 4: Evaluate the boundary terms
Now we evaluate the boundary terms:
- At \( x = 1 \): \( \log(1) = 0 \) so \( (\log 1)^n \cdot \frac{1^{m+1}}{m+1} = 0 \)
- At \( x = 0 \): \( \lim_{x \to 0} (\log x)^n \cdot \frac{x^{m+1}}{m+1} \) tends to \( 0 \) for \( m \geq 0 \) (as \( x^{m+1} \) goes to \( 0 \) faster than \( (\log x)^n \) diverges).

Thus, the boundary terms contribute \( 0 \):
\[ I_{m,n} = 0 - \frac{n}{m+1} I_{m,n-1} \]
This leads us to:
\[ I_{m,n} = -\frac{n}{m+1} I_{m,n-1} \]

### Final Result
The final expression for the integral is:
\[ I_{m,n} = -\frac{n}{m+1} I_{m,n-1} \]