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?
 A: 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!
A: 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.
