From $\lim_{n\to\infty}n\sum_{k=1}^\infty\int_0^1\frac{dx}{(1+n^2x^2)(k+x)^3}$ to a multiple of $\zeta(3)$ This afternoon I tried do the specialisation of another problem due to Furdui. THat is PROBLEMA 103, La Gaceta de la RSME, Volumen 12, número 2 (2009), see the first identity in page 317 of the solution (in spanish). 
Thus I know that one can state an identity from the following limit with a multiple of $\zeta(3)$ following the quoted identity, $$\lim_{n\to\infty}n\sum_{k=1}^\infty\int_0^1\frac{dx}{(1+n^2x^2)(k+x)^3}.$$ 
My attempt to try show such identity was write $\frac{1}{1+n^2x^2}$ as $$\frac{-i}{2n}\frac{1}{x-\frac{i}{n}}+\frac{i}{2n}\frac{1}{x+\frac{i}{n}}.$$
THen after the change of variables $u=(x+k)^3$, I can write $$\int_0^1\frac{dx}{(1+n^2x^2)(k+x)^3}=\int_{k^3}^{(k+1)^3}\frac{n/3du}{u^{5/3}((nu^{1/3}-nk)^2+1)}.$$
Also I've calculated with help of Wolfram Alpha previous indefinite integral, with the code 

integrate x^(-5/3)/(n^2(x^(1/3)-k)^2+1) dx.

On the other hand I tried calculate the previous indefinite integral by parts 
$$\int\frac{nx^{-5/3}dx}{(n^2(x^{1/3}-k)^2+1)}=\frac{\arctan n(x^{1/3}-k)}{x}+\int \frac{\arctan n(x^{1/3}-k)}{x^2}dx+\text{cte}.$$
I don't know how evaluate (the limits of integration and after take the series and the limit) previous calculations to get an identity involving $\zeta(3)$ without using Furdui's result. I don't know if there are mistakes in my calculations. 

Question. What's is a right approach and set of calculations, to show that $$\lim_{n\to\infty}n\sum_{k=1}^\infty\int_0^1\frac{dx}{(1+n^2x^2)(k+x)^3}$$
  is related with $\zeta(3)$ (neccesarly thus agree with Furdui's identity) without using PROBLEMA 103? Thus you need take the limit, sum the series and compute the integral. Thanks in advance.

 A: Consider 
$$I(n) = \int_0^1 \frac{dx}{(1+n^2 x^2) (k+x)^3} = \int_0^1 dx \, (k+x)^{-3} e^{-\log{(1+n^2 x^2)}}$$
We note that the integral is dominated by contributions about $x=0$ as $n \to \infty$.  Thus, we consider only an interval near $x=0$; that is $0 \le x \le \epsilon$ for some $\epsilon \gt 0$.  An approximation to the integral then takes the form
$$I(n) \approx \frac1{k^3} \int_0^{\epsilon} dx \, e^{-n^2 x^2} $$
However, the integral can be further approximated with exponentially small error as an integral out to infinity.  Thus, an approximation to the integral is
$$I(n) \approx \frac{\sqrt{\pi}}{2 n k^3} $$
In this case, the above limit turns out to be $\frac{\sqrt{\pi}}{2} \zeta(3)$.  
The error in the integral may be evaluated by Taylor expansion; it is easy to show that the error is in fact $O(1/n)$.  
A: Using partial fraction decomposition $$\frac{1}{(1+n^2x^2)(k+x)^3}$$ write $$\frac{n^2 \left(3 k^2 n^2-1\right)}{\left(k^2 n^2+1\right)^3 (k+x)}+\frac{2 k
   n^2}{\left(k^2 n^2+1\right)^2 (k+x)^2}+\frac{1}{\left(k^2 n^2+1\right)
   (k+x)^3}+\frac{\left(k^3 n^6-3 k n^4\right)+x \left(n^4-3 k^2 n^6\right)}{\left(k^2 n^2+1\right)^3
   \left(n^2 x^2+1\right)}$$ which is not to bad (and easy to integrate).
So computing $$I_k=\int_0^1\frac{1}{(1+n^2x^2)(k+x)^3}\,dx$$ does not make any specific problem (the formulae are not reported here because of their length using MathJax).
Expanding as Taylor series for large values of $k$ leads to $$I_k=\frac{\tan ^{-1}(n)}{k^3 n}-\frac{3 \log \left(n^2+1\right)}{2 k^4 n^2}+\frac{6 \left(n-\tan ^{-1}(n)\right)}{k^5 n^3}
+O\left(\frac{1}{k^6}\right)$$ which makes $$n I_k=\frac{\tan ^{-1}(n)}{k^3 }-\frac{3 \log \left(n^2+1\right)}{2 k^4 n}+\frac{6 \left(n-\tan ^{-1}(n)\right)}{k^5 n^2}
+O\left(\frac{1}{k^6}\right)$$ $$\sum_{k=1}^\infty n I_k=\zeta (3) \tan ^{-1}(n)-\left( \frac{1}{60} \pi ^4 \log \left(n^2+1\right)+\frac{\pi ^6}{189}-6 \zeta (5)\right)\frac{1}{ n}+\cdots$$ and hence the limit and also how it is approached when $n \to \infty$.
A: Letting $x = y/n,$ we see the expression equals
$$\tag 1\sum_{k=1}^{\infty}\int_0^n\frac{1}{(1+y^2)(k+y/n)^3}\, dy.$$
A straightforward DCT argument shows that as $n\to \infty,$ the $k$th summand increases to $(1/k^3)\int_0^\infty\frac{1}{(1+y^2)}\, dy = (1/k^3)(\pi/2).$ Another application of the DCT then shows the limit of $(1)$ as $n\to \infty$ is $\zeta (3) \frac{\pi}{2}.$
