<s>[INST] Find f(x) using concept of Definite-Integral-as a limit of sum $$ f(x) = \lim_{n\to\infty} \left[\frac{n^n\left(x^2+n^2\right)\left(x^2+\frac{n^2}{4}\right)\cdots\left(x^2+\frac{n^2}{n^2}\right)}{n!\left(x^3+n^3\right)\left(x^3+\frac{n^3}{8}\right)\cdots\left(x^3+\frac{n^3}{n^3}\right)}\right]^\frac{x}{n},\qquad x>0$$
My approach:
I took $\log$ on both sides..
$$
\log(f(x)) =  x \lim_{n\to\infty}\left[ \sum_{r=1}^n  \frac{1}{n} \log\left(x^2+\frac{1}{\left(\frac{r}{n}\right)^2}\right)  - \sum_{r=1}^n \frac{1}{n} \log\left(x^3 + \frac{1}{(\frac{r}{n})^3}\right) + \frac{1}{n} \log\left(\frac{n^n}{n!}\right)\right]$$
The first two terms inside the square brackets become..
$$
\log(f(x)) =  x \left[\int_0^1 \log\left(x^2+\frac{1}{t^2}\right) dt + \int_0^1 \log\left(x^3+\frac{1}{t^3}\right) dt + \lim_{n\to\infty} \frac{1}{n} \log\left(\frac{n^n}{n!}\right)\right]$$
I was unable to solve the integrals and the last term..Any Help ??
 [/INST] Rather than first using riemann sum lets simplify some things. Take out $n^2,\frac {n^2}{4},...\frac {n^2}{n^2} $  from the brackets of  numerator and $n^3,\frac {n^3}{8},....\frac {n^3}{n^3} $  from brackets of denominator. This converts the function as $$(\frac {n^n.n^{2n}.\frac {1}{1.4.9...n^2}(\frac {x^2}{n^2}+1)(\frac{2^2}{n^2}x^2+1)...(\frac {n^2}{n^2}x^2+1)}{n!.n^{3n}.\frac {1}{1^3.2^3...n^3}(\frac{x^3}{n^3}+1)(\frac {2^3}{n^3}x^3+1)... (\frac {n^3}{n^3}x^3+1)})^{x/n}$$ . Now remember that $\pi _1^n \frac {n^3}{n^2}=n!$  thus the function becomes $\frac {(\frac {x^2}{n^2}+1)(\frac {4x^2}{n^2}+1).... (\frac {n^2x^2}{n^2}+1)}{(\frac {x^3}{n^3}+1)(\frac {2^3x^3}{n^3}+1).. (\frac {n^3x^3}{n^3}+1)}^{\frac {x}{n}} $ now take logs and the $f (x)=\int _0 ^1 x\log (\frac {(xr)^2+1}{(xr)^3+1})dr $ now just put $rx=k $ note that x is constant and then integration can be done by separating logs and then by parts indivdually.
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