Given that
`lim_(n to oo)sum_(r=1)^(n)(log(n^(2)+r^(2))-2logn)/(n)=log2+(pi)/(2)-2`, then
`lim_(n to oo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+r^(2))^(m)......(n^(2))^(m)]^(1//n)` is equal to

Options

A. `2^(m)e^(m((pi)/(2)-2))`
B. `2^(m)e^(m(2-(pi)/(2)))`
C. `e^m((pi)/(2)-2)`
D. `e^(2m((pi)/(2)-2))`

Answer

To solve the given problem step by step, we will analyze the limit and the summation involved.

### Step 1: Understand the Given Limit
We start with the limit:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{\log(n^2 + r^2) - 2\log n}{n} = \log 2 + \frac{\pi}{2} - 2 \]
This can be rewritten as:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} \left( \log\left(1 + \frac{r^2}{n^2}\right) \right) \]
This expression is a Riemann sum for the function \( f(x) = \log(1 + x^2) \) over the interval \([0, 1]\).

### Step 2: Convert the Sum to an Integral
Using the property of Riemann sums, we can convert the limit of the sum into an integral:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} \log\left(1 + \frac{r^2}{n^2}\right) = \int_0^1 \log(1 + x^2) \, dx \]
We know from the problem statement that this integral evaluates to:
\[ \int_0^1 \log(1 + x^2) \, dx = \log 2 + \frac{\pi}{2} - 2 \]

### Step 3: Analyze the Second Limit
Next, we need to evaluate the limit:
\[ \lim_{n \to \infty} \frac{1}{n^{2m}} \left[(n^2 + 1^2)^m (n^2 + 2^2)^m \cdots (n^2 + n^2)^m \right]^{\frac{1}{n}} \]
This can be rewritten as:
\[ \lim_{n \to \infty} \frac{1}{n^{2m}} \left[ \prod_{r=1}^{n} (n^2 + r^2)^m \right]^{\frac{1}{n}} \]

### Step 4: Simplify the Product
Taking the logarithm, we have:
\[ \log L = \lim_{n \to \infty} \left( \frac{1}{n} \sum_{r=1}^{n} m \log(n^2 + r^2) - 2m \log n \right) \]
This simplifies to:
\[ \log L = m \lim_{n \to \infty} \left( \frac{1}{n} \sum_{r=1}^{n} \log(n^2 + r^2) - 2 \log n \right) \]

### Step 5: Convert the Sum to an Integral Again
Using the Riemann sum representation again:
\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \log(n^2 + r^2) = \int_0^1 \log(1 + x^2) \, dx + 2 \log n \]
Thus, we have:
\[ \log L = m \left( \log 2 + \frac{\pi}{2} - 2 \right) \]

### Step 6: Exponentiate to Find L
Exponentiating both sides gives:
\[ L = e^{m \left( \log 2 + \frac{\pi}{2} - 2 \right)} = 2^m e^{m \left( \frac{\pi}{2} - 2 \right)} \]

### Final Answer
Thus, the limit evaluates to:
\[ \lim_{n \to \infty} \frac{1}{n^{2m}} \left[(n^2 + 1^2)^m (n^2 + 2^2)^m \cdots (n^2 + n^2)^m \right]^{\frac{1}{n}} = 2^m e^{m \left( \frac{\pi}{2} - 2 \right)} \]