Let m,n be two positive real numbers and define `f(n)=int_(0)^(oo)x^(n-1)e^(-x)dx` and `g(m,n)=int_(0)^(1)x^(m-1)(1-m)^(n-1)dx`.
It is known that f(n) for n gt 0 is finite and g(m, n) = g(n, m) for m, n gt 0.
`int_(0)^(1)x^(m)(log_(e).(1)/(x))dx=`

Options

A. `(f(n+1))/((m+1)^(n))`
B. `(f(n))/((m+1)^(n+1))`
C. `(f(n+1))/((m+1)^(n+1))`
D. `g(m+1),n+1)`

Explanation

To solve the given problem, we need to evaluate the integral:

\[ I = \int_0^1 x^m \log\left(\frac{1}{x}\right) dx \]

### Step 1: Rewrite the logarithmic term
We can rewrite the logarithmic term as follows:

\[ \log\left(\frac{1}{x}\right) = -\log(x) \]

Thus, the integral becomes:

\[ I = -\int_0^1 x^m \log(x) dx \]

### Step 2: Use integration by parts
To evaluate the integral \(\int_0^1 x^m \log(x) dx\), we can use integration by parts. Let:

- \(u = \log(x)\) \(\Rightarrow du = \frac{1}{x} dx\)
- \(dv = x^m dx\) \(\Rightarrow v = \frac{x^{m+1}}{m+1}\)

Now, applying integration by parts:

\[ \int u \, dv = uv - \int v \, du \]

Substituting our values:

\[ \int_0^1 x^m \log(x) dx = \left[ \log(x) \cdot \frac{x^{m+1}}{m+1} \right]_0^1 - \int_0^1 \frac{x^{m+1}}{m+1} \cdot \frac{1}{x} dx \]

### Step 3: Evaluate the boundary term
Evaluating the boundary term:

At \(x = 1\):

\[ \log(1) \cdot \frac{1^{m+1}}{m+1} = 0 \]

At \(x = 0\), we need to consider the limit:

\[ \lim_{x \to 0} \log(x) \cdot \frac{x^{m+1}}{m+1} \]

As \(x \to 0\), \(\log(x) \to -\infty\) and \(x^{m+1} \to 0\). The product approaches \(0\) since \(x^{m+1}\) approaches \(0\) faster than \(\log(x)\) approaches \(-\infty\).

Thus, the boundary term evaluates to \(0\).

### Step 4: Evaluate the integral
Now we need to evaluate the remaining integral:

\[ -\int_0^1 \frac{x^{m+1}}{m+1} \cdot \frac{1}{x} dx = -\frac{1}{m+1} \int_0^1 x^m dx \]

The integral \(\int_0^1 x^m dx = \frac{1}{m+1}\).

Thus,

\[ -\frac{1}{m+1} \cdot \frac{1}{m+1} = -\frac{1}{(m+1)^2} \]

### Step 5: Combine results
Putting it all together, we have:

\[ I = -\left(-\frac{1}{(m+1)^2}\right) = \frac{1}{(m+1)^2} \]

### Final Result
Thus, the value of the integral is:

\[ \int_0^1 x^m \log\left(\frac{1}{x}\right) dx = \frac{1}{(m+1)^2} \]