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The image you've provided shows the following expression:

\[
\frac{\int_0^\infty v^n e^{-v \sum x_i} dv}{\int_0^\infty v^{n+1} e^{-v \sum x_i} dv}
\]

This is a ratio of two integrals involving an exponential function. The integrals are of the form:

\[
\int_0^\infty v^k e^{-a v} dv
\]

where \( k \) is an exponent (in this case, \( n \) or \( n+1 \)) and \( a = \sum x_i \).

### Solving the integrals:
This type of integral is a standard gamma function integral. Specifically, we use the result:

\[
\int_0^\infty v^k e^{-a v} dv = \frac{\Gamma(k+1)}{a^{k+1}}
\]

Thus, applying this to both integrals:

1. For the numerator: 
   \[
   \int_0^\infty v^n e^{-v \sum x_i} dv = \frac{\Gamma(n+1)}{(\sum x_i)^{n+1}}
   \]
   
2. For the denominator:
   \[
   \int_0^\infty v^{n+1} e^{-v \sum x_i} dv = \frac{\Gamma(n+2)}{(\sum x_i)^{n+2}}
   \]

### Simplifying the ratio:
Now, divide the numerator by the denominator:

\[
\frac{\frac{\Gamma(n+1)}{(\sum x_i)^{n+1}}}{\frac{\Gamma(n+2)}{(\sum x_i)^{n+2}}}
\]

This simplifies to:

\[
\frac{\Gamma(n+1)}{\Gamma(n+2)} \cdot (\sum x_i)
\]

Using the property of the gamma function \( \Gamma(n+2) = (n+1)\Gamma(n+1) \), the ratio simplifies to:

\[
\frac{1}{n+1} \cdot (\sum x_i)
\]

### Final result:
The expression evaluates to:

\[
\frac{\sum x_i}{n+1}
\]

This is the simplified result of the given ratio of integrals.

Would you like further details on any part of this solution? Let me know if you'd like a step-by-step breakdown or have specific questions.

Here are 5 related questions to explore:
1. What is the gamma function, and how is it defined?
2. How can we generalize this result to other types of exponential integrals?
3. What happens to this ratio when \( n = 0 \)?
4. How does this apply to probability distributions, particularly the Gamma distribution?
5. Can this method be extended to multi-dimensional integrals?

**Tip:** The Gamma function generalizes the factorial function to non-integer values, and it often appears in integrals involving exponential terms.

Evaluate the ratio of integrals: 
 \(\frac{\int_0^\infty v^n e^{-v \sum x_i} dv}{\int_0^\infty v^{n+1} e^{-v \sum x_i} dv}\)

Explore the solution to the ratio of integrals involving an exponential term: \(\frac{\int_0^\infty v^n e^{-v \sum x_i} dv}{\int_0^\infty v^{n+1} e^{-v \sum x_i} dv}\). Learn how the Gamma function is used to simplify this integral for higher-level calculus students.

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