How to evaluate this series using fourier series?

With the help of Hermite's Integral,I got $$\sum_{n=1}^{\infty }\frac{1}{n}\int_{2\pi n}^{\infty }\frac{\sin x}{x}\mathrm{d}x=\pi-\frac{\pi}{2}\ln(2\pi)$$ I'd like to know can we solve this one using fourier series?

• use the change of variable $x = 2 \pi ny$ – reuns Apr 18 '16 at 7:46

We begin with the expression

$$S=\sum_{n=1}^\infty \frac{1}{n}\int_{2\pi n}^\infty \frac{\sin (x)}{x}\,dx \tag 1$$

Enforcing the substitution $x \to 2\pi n x$ in $(1)$ yields

\begin{align} \sum_{n=1}^\infty \frac{1}{n}\int_{2\pi n}^\infty \frac{\sin (x)}{x}\,dx&=\sum_{n=1}^\infty \frac{1}{n} \int_{1}^\infty \frac{\sin (2\pi nx)}{x}\,dx\\\\ &=\int_{1}^\infty \frac 1x\sum_{n=1}^\infty \frac{\sin (2\pi nx)}{n}\,dx \tag 2\\\\ &=\sum_{k=1}^\infty \int_{k}^{k+1}\frac1x \left(\frac{\pi}{2}(1-2(x-k))\right)\,dx \tag 3\\\\ &=\frac{\pi}{2} \sum_{k=1}^\infty \left((2k+1)\log\left(\frac{k+1}{k}\right)-2\right) \tag 4\\\\ \end{align}

where we recognized the Fourier sine series for $\frac{\pi}{2}(1-2x)$ for $x\in(0,1)$ to go from $(2)$ to $(3)$.

To evaluate the sum in $(4)$, we write the partial sum $S_K$

\begin{align} S_K&=\sum_{k=1}^K \left((2k+1)\log\left(\frac{k+1}{k}\right)-2\right)\\\\ &=-2K+\sum_{k=1}^K \left((2k+1)\log(n+1)-(2k-1)\log(n)\right)-2\sum_{k=1}^K\log(k) \\\\ &=-2K+(2K+1)\log(K+1)-2\log(K!) \\\\ &=-2K+(2K+1)\log(K)+(2K+1)\log\left(1+\frac1K\right)-2\log\left(K!\right) \tag 5\\\\ &=-2K+(2K+1)\log(K)+2-2\log\left(\sqrt{2\pi K}\left(\frac{K}{e}\right)^K\right)+O\left(\frac1K\right) \tag 6\\\\ &=2-\log(2\pi)+O\left(\frac1K\right) \end{align}

In going from $(5)$ to $(6)$, we used the asymptotic expansion for the logarithm function, $\log(1+x)=x+O(x^2)$ and Stirling's Formula

$$K!=\left(\sqrt{2\pi K}\left(\frac{K}{e}\right)^K\right)\left(1+O\left(\frac1K\right)\right)$$

Finally, putting it all together, we find that

\begin{align} S&=\sum_{n=1}^\infty \frac{1}{n}\int_{2\pi n}^\infty \frac{\sin (x)}{x}\,dx\\\\ &=\frac{\pi}{2}\lim_{K\to \infty}S_K\\\\ &=\pi-\frac{\pi}{2}\log(2\pi) \end{align}

and therefore

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{1}{n}\int_{2\pi n}^\infty \frac{\sin (x)}{x}\,dx=\pi-\frac{\pi}{2}\log(2\pi)}$$

as was to be shown!

Hint: Observe that \begin{align} \dfrac{\sin x}{x} &= \dfrac1x\sum_{k=0}^{\infty} \dfrac{\left(-1\right)^{k}}{\left(2k+1\right)!}\,x^{2k+1} = \sum_{k=0}^{\infty} \dfrac{\left(-1\right)^{k}}{\left(2k+1\right)!}\,x^{2k} \end{align} Then \begin{align} \sum_{n=1}^{\infty }\dfrac{1}{n}\int_{2\pi n}^{\infty}\!\!\dfrac{\sin x}{x}\,dx &= \sum_{n=1}^{\infty }\dfrac{1}{n}\int_{2\pi n}^{\infty }\!\!\bigg(\sum_{k=0}^{\infty}\dfrac{\left(-1\right)^{k}}{\left(2k+1\right)!}\,x^{2k}\bigg)\,dx \\ &= \sum_{n=1}^{\infty }\dfrac{1}{n}\bigg(\sum_{k=0}^{\infty} \dfrac{\left(-1\right)^{k}}{\left(2k+1\right)!}\int_{2\pi n}^{\infty}\!\!\!x^{2k}\,dx \bigg) \\ &= \sum_{n=1}^{\infty }\dfrac{1}{n}\bigg(\sum_{k=0}^{\infty} \dfrac{\left(-1\right)^{k}}{\left(2k+1\right)!} \!\cdot\!\left.\dfrac{x^{2k+1}}{2k+1}\bigg)\right\rvert_{2\pi n}^{\infty} \\ &= \sum_{n=1}^{\infty }\dfrac{1}{n}\Bigg[\bigg(\sum_{k=0}^{\infty} \left.\dfrac{1}{\left(2k+1\right)!}\bigg)\arctan x\,\right\rvert_{2\pi n}^{\infty}\Bigg] \end{align}

The last equality is due to the series representation of inverse of a tangent function:

$$\bbox[1ex, border:solid 1.5pt #e10000]{ \arctan x = \sum_{k=0}^{\infty} \dfrac{\left(-1\right)^{k}}{2k+1}\,x^{2k+1}}$$

Hope you can proceed from here on.

• This development has serious flaws. It is incorrect to use a power series $\sum_{n=1}^\infty a_nx^n$ to evaluate the integral $\int_a^\infty \sum_{n=1}^\infty a_nx^n$ by interchanging the integral with the summation. And $\sum_{k=0}^\infty \frac{(-1)^k\,x^{2k+1}}{(2k+1)(2k+1)!} \ne \sum_{k=0}^\infty \frac{\arctan(x)}{(2k+1)!}$. – Mark Viola May 3 '16 at 14:37
• @Dr.MV Why is the arctan formula incorrect? Also, what would be a good way to establish or disprove switching order of summation and integration? – Vlad May 3 '16 at 23:36
• Vlad, the series for the arctangent is indeed $\sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{2k+1}$. But one cannot write $\sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1(2k+1)!}=\left(\sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{2k+1}\right)\sum_{k=0}^\infty \frac{1}{(k+1)!}$. And in the case of this solution, the interchange of integration and summation is not justifiable. -Mark – Mark Viola May 4 '16 at 2:09
• @Dr.MV Thank you for the explanation! Indeed I made a dumb mistake with arctan series. Just for my own record, why exactly can't we interchange integration and summation signs? Isn't Riemannian integration countably additive? What would be the proper formal argument against changing order of operations? – Vlad May 4 '16 at 4:12
• If the summation were finite, then of course the interchange of operations is justified. However, with a series and an integral, the interchange is not generally permissible. Under sufficient conditions (e.g. dominated convergence, uniform convergence), the interchange of the order of the series and integration is permitted. In this case, it is obviously not legitimate since prior to the interchange, the improper integral of the sinc function converges. But after the interchange, none of the terms of the term-by-term integration converge (the upper limit is causing the problem). – Mark Viola May 4 '16 at 4:20