Hey there! Great job working through that factorial limit and connecting it to Riemann sums—your LaTeX looks solid for a first post too, nice work! Let's tackle your questions step by step.

First, let's recap your awesome derivation for reference:

To solve $\lim_{n \to \infty}\frac{\sqrt[n]{n!}}{n}$, you took the natural log, rewrote the expression as a sum, recognized it as a Riemann sum for $\int_{0}^{1}\ln(x)\ dx$, evaluated the integral to $-1$, then exponentiated to get $\frac{1}{e}$. Perfect stuff.

Now, onto your goal of building a similar limit for $e^\pi$ using the improper integral $\int_{0}^{{\infty}\ln\left(1+\frac{1}{x}2}\right)\ dx=\pi$. The key here is that we can map infinite intervals to finite ones with a variable substitution, which lets us convert the improper integral into a finite-interval integral that we can express as a Riemann sum.

Converting the infinite integral to a finite one (and thus a Riemann sum) #

Let's split the integral at $x=1$ to handle the infinite upper bound:
$$\int_{0}^{{\infty}\ln\left(1+\frac{1}{x}2}\right)\ dx = \int_{0}^{{1}\ln\left(1+\frac{1}{x}2}\right)\ dx + \int_{1}^{{\infty}\ln\left(1+\frac{1}{x}2}\right)\ dx$$

For the second integral, use the substitution $x = \frac{1}{t}$ (so $dx = -\frac{1}{t^2}dt$). When $x=1$, $t=1$; when $x\to\infty$, $t\to0$. This transforms the second integral into:
$$\int_{1}^{{\infty}\ln\left(1+\frac{1}{x}2}\right)\ dx = \int_{0}^{{1}\ln\left(1+t}2\right)\cdot\frac{1}{t^2}\ dt$$

Now combine both parts to get a finite-interval integral:
$$\int_{0}^{{\infty}\ln\left(1+\frac{1}{x}2}\right)\ dx = \int_{0}^{{1}\left[\ln\left(1+\frac{1}{x}2}\right) + \frac{\ln(1+x^2)}{x2}\right]\ dx$$

This is now an integral over $[0,1]$, so we can write it as a Riemann sum! If we partition $[0,1]$ into $n$ equal subintervals with sample points $x_k = \frac{k}{n}$ for $k=1,2,...,n$, the Riemann sum is:
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n\left[\ln\left(1+\frac{n}2}{k^2}\right) + \frac{\ln\left(1+\left(\frac{k}{n}\right)^{2\right)}{\left(\frac{k}{n}\right)}2}\right]$$

Since this limit equals $\pi$, exponentiating both sides gives us the expression you're after:
$$e^{{\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}}n\left[\ln\left(1+\frac{n^2}{k2}\right) + \frac{\ln\left(1+\left(\frac{k}{n}\right)^{2\right)}{\left(\frac{k}{n}\right)}2}\right]} = e^\pi$$

Alternative integrals to build $e^\pi$ #

If you want simpler integrals to work with, here are a couple of options that are easier to convert to Riemann sums:

• Option 1: Use $\int_{0}^{{1}\frac{2}{\sqrt{1-x}2}}\ dx = \pi$. This is just twice the standard arcsine integral. Its Riemann sum is:
  $$\lim_{n\to\infty}\frac{2}{n}\sum_{k=1}^{n\frac{1}{\sqrt{1-\left(\frac{k}{n}\right)}2}} = \pi$$
  Exponentiating gives:
  $$e^{{\lim_{n\to\infty}\frac{2}{n}\sum_{k=1}}n\frac{1}{\sqrt{1-\left(\frac{k}{n}\right)^2}}} = e^\pi$$

• Option 2: Use $\int_{0}^{{\infty}\frac{2}{1+x}2}\ dx = \pi$. Again, split at $x=1$ and substitute $x=1/t$ for the infinite part, or use $x=\tan\theta$ to map to $[0,\pi/2]$. The Riemann sum form would look like:
  $$\lim_{n\to\infty}\frac{2}{n}\sum_{k=1}^{n\left[\frac{1}{1+\left(\frac{k}{n}\right)}2} + \frac{1}{1+\left(\frac{n}{k}\right)^{2}\cdot\frac{n}2}{k^2}\right] = \pi$$
  And exponentiating gives $e^\pi$.

Quick LaTeX tips for your future posts #

You're already off to a great start! A couple of small tweaks to make your formulas even cleaner:

• For inline sums (like $\sum_{i=1}^n$), use \sum\limits_{i=1}^n to force the limits above and below the sum symbol instead of to the side (this makes it easier to read in paragraphs).
• If you want to align multiple equations vertically, use the align environment (wrap in $$):

  $$
  \begin{align*}
  \ln\left(\frac{\sqrt[n]{n!}}{n}\right) &= \frac{1}{n}\ln\left(\frac{n!}{n^n}\right) \\
  &= \frac{1}{n}\sum_{i=1}^n\ln\left(\frac{i}{n}\right)
  \end{align*}
  $$
  

  This automatically lines up the equals signs for a polished look.

Keep exploring—this kind of connection between sums, integrals, and special constants is really fun to dig into!

备注：内容来源于stack exchange，提问作者bob