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.