How to calculate the leading order solution to $\sum_{n=-\infty}^\infty\frac{1}{[(x+n\epsilon)^2+1]^{3/2}}$? I wish to calculate a series solution to the following sum,
$$S(x)=\sum_{n=-\infty}^\infty\frac{1}{[(x+n\epsilon)^2+1]^{3/2}},$$
where $x$, $\epsilon$ $\in$ $\mathbb{R}$ and $\epsilon \ll 1$.
Do you have any ideas for how I could do this?
My idea is to turn this infinite sum into an integral using the Fourier transform and then use an asymptotic method to evaluate this integral.
Here is what I have done so far.
Let
$$f(z) = \frac{1}{(z^2+1)^{3/2}}.$$
Hence,
$$S(x) = \sum_{n=-\infty}^\infty f(x+n\epsilon).$$
Taking the Fourier transform (in the sense of distributions) gives
$$\hat{S}(\xi)=\hat{f}(\xi)\sum_{n=-\infty}^{\infty}\exp(i2n\pi\xi).$$
Let
$$w = \exp(i2\pi\xi).$$
Hence,
$$\hat{S}(\xi)=\hat{f}(\xi)\lim_{N\rightarrow\infty}\left(\sum_{n=-N}^{N}w^n\right).$$
Using the Geometric series formula gives
$$\hat{S}(\xi) = \hat{f}(\xi)\lim_{N\rightarrow\infty}\left(\frac{w^{N+1}-w^{-N}}{w-1}\right).$$
Hence,
$$S(x)=\int_{-\infty}^\infty\hat{f}(\xi)\lim_{N\rightarrow\infty}\left(\frac{w^{N+1}-w^{-N}}{w-1}\right)\exp(i2\pi\xi x)d\xi.$$
Now we have turned the infinite sum into an integral. Does this help? Do you know any asymptotic techniques that can be used to progress from here? Maybe Cauchy's integral theorem can be somehow used to deal with the $1/(w-1)$ factor?
For those curious, I am interested in $S$ because it gives the magnetic field along the axis of an infinite series of current loops (see Field on Axis of Current Loop).
 A: The series you want is a Fourier series
$$S(x) = \sum_{n=-\infty}^\infty s_n e^{in\omega x}$$
and we directly compute the coefficients of this series with its known formula. Over a period $\epsilon \implies \omega = \frac{2\pi}{\epsilon}$ we have
$$s_m = \frac{1}{\epsilon}\int_0^\epsilon\sum_{n=-\infty}^\infty \frac{e^{\frac{i2\pi m x}{\epsilon}}}{((x+n\epsilon)^2+1)^{\frac{3}{2}}}\:dx$$
Geometrically, consider the contribution from each shifted peak to the interval $[0,\epsilon]$ - they each contribute further and further, but continguous, sections of the tails of the function. Therefore by this reasoning (and the $\epsilon$-periodicity of the integrand) we can simplify this expression to
$$s_m = \frac{1}{\epsilon}\int_{-\infty}^\infty\frac{e^{\frac{i2\pi m x}{\epsilon}}}{(x^2+1)^{\frac{3}{2}}}dx = \begin{cases}\frac{2}{\epsilon}\left|\frac{2\pi m}{\epsilon}\right|K_1\left(\left|\frac{2\pi m}{\epsilon}\right|\right) & m\neq 0 \\ s_0 = \frac{2}{\epsilon} & m = 0\end{cases}$$
where $K_n$ is the modified Bessel function of the second kind. Simplifying a bit and assuming $\epsilon > 0$ we get
$$S(x) = \frac{2}{\epsilon} + \sum_{n=1}^\infty \frac{8\pi n}{\epsilon^2}K_1\left(\frac{2\pi n}{\epsilon}\right) \cos\left(\frac{2\pi n x}{\epsilon}\right)$$
the cosine Fourier series form.
