MHB What is the limit of Problem #2 using double integration?

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The limit of the expression using double integration is derived through the relationship between logarithms and integrals. The limit is expressed as \( L = 2 + e^{\frac{\pi}{2} - 2} \), with the integral \( \int_0^1 \log(1+x^2) \, dx \) simplifying to \( \frac{\pi}{2} - 2 + \ln(2) \). The discussion emphasizes the importance of recognizing Riemann sums and applying integration techniques, including integration by parts, to arrive at the solution efficiently. Alternative methods, such as double integration, are also explored but deemed more complex. Ultimately, the integration by parts approach is highlighted as the most straightforward method to achieve the limit.
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Find the limit.

$$\lim_{n \to \infty} \left\{ \left( 1+\frac{1^2}{n^2}\right)\left( 1+\frac{2^2}{n^2}\right)\left( 1+\frac{3^2}{n^2}\right) \cdots \left( 1+\frac{n^2}{n^2}\right)\right\}^{\dfrac{1}{n}}$$
 
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If $L$ is the limit, then $\log L=\displaystyle\lim_{n \to{+}\infty}\dfrac{1}{n}\;\displaystyle\sum_{k=1}^n \log \left(1+(k/n)^2\right)=\displaystyle\int_0^1 \log (1+x^2)\;dx$ .
 
Fernando Revilla said:
If $L$ is the limit, then $\log L=\displaystyle\lim_{n \to{+}\infty}\dfrac{1}{n}\;\displaystyle\sum_{k=1}^n \log \left(1+(k/n)^2\right)=\displaystyle\int_0^1 \log (1+x^2)\;dx$ .

Yeah that's right. The integral can be further simplified. \( \displaystyle \int_{0}^{1}\ln(1+x^2) dx= \cdots=\frac{\pi}{2}-2+\ln(2)\).

Therefore \( \displaystyle L=2+e^{\dfrac{\pi}{2}-2}\).
 
The integral is just boring old integration by parts (the real key part was realizing that it's a Riemann sum). I'm sure that's why Dr Revilla didn't bother to provide us with the calculation. But we can make it interesting in such a way that the given problem is in the end done by writing it as sum then as an integral, then as a sum, and then finally as an integral. You don't believe me? Watch this:

$$\displaystyle \begin{aligned} & \begin{aligned} \log (\ell) & = \lim_{n \to{+}\infty}\dfrac{1}{n}\;\displaystyle\sum_{k=1} ^n \log \left(1+(k/n)^2\right) \\& = \int_0^1 \log (1+x^2)\;dx \\& = \int_0^1\sum_{k \ge 0}\frac{(-1)^kx^{2k+2}}{k+1}\;{dx} \\& = \sum_{k \ge 0}\int_{0}^{1}\frac{(-1)^kx^{2k+2}}{k+1}\;{dx} \\& = \sum_{k \ge 0}\bigg[\frac{(-1)^kx^{2k+3}}{(k+1)(2k+3)}\bigg]_0^1 \\& = \sum_{k \ge 0}\frac{(-1)^k}{(k+1)(2k+3)} \\& = \sum_{k \ge 0}\frac{(-1)^k(2k+3)-2(-1)^k(k+1)}{(k+1)(2k+3)} \\& = \sum_{k\ge 0}\bigg(\frac{(-1)^k}{k+1}-\frac{2(-1)^k}{2k+3}\bigg) \\& = \sum_{k\ge 0}\int_0^1\bigg((-1)^kx^{k}-2(-1)^kx^{2k+2}\bigg)\;{dx} \\& = \int_0^1\sum_{k\ge 0}\bigg( (-1)^kx^{k}-2(-1)^kx^{2k+2}\bigg)\;{dx} \\&= \int_0^1\bigg(\frac{1}{1+x}-\frac{2x^2}{1+x^2}\bigg)\;{dx} \\& = \int_0^1\bigg(\frac{1}{1+x}-\frac{2x^2+2-2}{1+x^2}\bigg)\;{dx} \\& = \int_0^1\bigg(\frac{1}{1+x}-2+\frac{2}{1+x^2}\bigg)\;{dx} \\& = \ln|1+x|-2+2\tan^{-1}{x}\bigg|_{0}^{1} \\& = \ln(2)-2+\frac{\pi}{2} \end{aligned} \\& \therefore ~ \ell = \exp\left(\ln{2}-2+\frac{\pi}{2}\right). \end{aligned}$$
 
I like the double integration approach:

$\begin{aligned} \int_{0}^{1}{\ln (1+{{x}^{2}})\,dx}&=\int_{0}^{1}{\int_{0}^{{{x}^{2}}}{\frac{dt\,dx}{1+t}}} \\
& =\int_{0}^{1}{\int_{\sqrt{t}}^{1}{\frac{dx\,dt}{1+t}}} \\
& =\int_{0}^{1}{\frac{1-\sqrt{t}}{1+t}\,dt} \\
& =2\int_{0}^{1}{\frac{t-{{t}^{2}}}{1+{{t}^{2}}}\,dt} \\
& =-2+\frac{\pi }{2}+\ln 2.
\end{aligned}
$

Now, integration by parts it's faster than any of these methods, perhaps you may consider it a boring way, but it's still valid and saves a lot of time.
 
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