# 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.

• I know how evaluate the limits (limits of integration and $\lim_{n\to\infty}$) involving the arctangent function, but I don't know if my previous calculations were right. Thanks. – user243301 Aug 29 '16 at 0:28

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 $n \to \infty$.

• Very thanks much I accept your answer. On the other hand when I was trying solve the problem was hard to try calculate (and now evaluate) the fraction decomposition. Thanks one more time. – user243301 Aug 29 '16 at 6:13
• You are very welcome ! Very often, fraction decomposition can resuire some patience ... plus a few tricks ! I prefered to keep $(1+n^2x^2)$ intact. – Claude Leibovici Aug 29 '16 at 6:16

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 $O(1/n)$.

• I understand the first step, first the approximation $\log(1+\delta)\approx\delta$ near the origin. But I don't know how one calculate the approximation to the integral (I know by comparison that $\int_0^\infty e^{-x^2}dx=\sqrt{\pi}/2$). If you can explain it is the best. Your answer had the best score and seems that it is also the best techniq as previous answers. Thus very thanks much to share with us. – user243301 Aug 29 '16 at 6:28
• look up ''Gaussian Integral'' – tired Aug 29 '16 at 10:17
• It's a sort of 'Laplace Method'. Nice. – Felix Marin Aug 30 '16 at 4:42

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 $\zeta (3) \frac{\pi}{2}.$

• Very thanks much for your clear proof, and use of the dominated convergence theorem with the integration for the arctangent function on $[0,\infty]$. I tried combine such theorem but with my previous other calculations. – user243301 Aug 29 '16 at 6:17