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I would like to sum the series $$ \sum_{n=0}^\infty \frac{1}{(1+a^2 (n+1/2)^2) ^{3/2}} . $$ It arose when trying to perform a calculation on superconductivity. In particular I am interested in its diverging behaviour for $a\to 0$.

I have tried turning it into a contour integral by defining $$ f(z)=tan(\pi z)(1+a^2 z^2) ^{-3/2} $$

The integral along the real axis gives (modulo factors of $\pi$ and 2) the desired sum. The integral over a great cicle in the upper half plane should vanish, but there is a branch cut to take care of. When I choose it to lie from $\frac i a$ to $+i \infty$ my integral along this branch cut from $i \infty$ to the pole diverges and I can get no meaningful answer. Maybe I am using the wrong technique for evaluating this sum. Mathematica also refuses to solve it.

Any suggestion on this problem would be much appreciated.

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This series is a good candidate for the Poisson summation formula. Let $f(x) = \frac{1}{(1+a^2 x^2)^{3/2}}$ and $f_{\frac{1}{2}}(x) = f(x+1/2)$. Define the sum $J = \sum_{n\in \mathbb{Z}} f_{\frac{1}{2}}(n)$. It is related to the desired sum $I = \sum_{n=0}^{\infty} f_{\frac{1}{2}}(n)$ by the formula $I=J/2$, because $f_{\frac{1}{2}}(-n-1) = f_{\frac{1}{2}}(n)$.

By the Poisson summation formula, we have the equality $$ J = \sum_{n\in\mathbb{Z}} f_{\frac{1}{2}}(n) = \sum_{p\in\mathbb{Z}} F_{\frac{1}{2}}(2\pi p) , $$ where $F(k) = \int_{-\infty}^{\infty} dx\, e^{-ikx} f(x)$ and $F_\frac{1}{2}(k) = \int_{\infty}^{\infty} dx\, e^{-ikx} f_{\frac{1}{2}}(x) = e^{ik/2} F(k)$. The idea is that for each larger value of $p$, the terms of the sum in Fourier space will be of larger and larger subleading order in $a$, so that only the first few terms of that serious would be needed to get a good asymptotic estimate.

Now, we can use some integration by parts and complex contour integration methods to evaluate the actual Fourier transform $F(k)$. \begin{align} F(k) &= \int_{-\infty}^{\infty} dx\, e^{-ikx} \frac{1}{(1+a^2 x^2)^{3/2}} \\ &= \int_{-\infty}^{\infty} dx\, e^{-ikx} \frac{d}{dx}\frac{x}{(1+a^2 x^2)^{1/2}} \\ &= ik \int_{-\infty}^{\infty} dx\, e^{-ikx} \frac{x}{(1+a^2 x^2)^{1/2}} \\ \end{align} The integration by parts helps make the subsequent deformed contour integrals finite. For $k>0$ the integral can be deformed into the lower half complex plane, while for $k<0$ it can be deformed into the upper half. In each case, the new contour will run along both sides of a branch cut, extending from either $x=i/a$ or $x=-i/a$ outward to infinity. Since $f(x)$ is real and symmetric, so is $F(k)$. So it is enough to consider only $k>0$ and the lower branch cut. Note that the imaginary part of the square root will be negative along the branch of that contour running to infinity.

Letting $x=-i(z+1)/a$ and summing the contributions to the contour from either side of the branch cut we get \begin{align} F(k) &= (ik) (-2i/a) e^{-k/a} \int_0^\infty dz\, e^{-kz/a} \frac{-i(z+1)/a}{-i\sqrt{(z+1)^2-1}} \\ &= \frac{2k}{a^2} e^{-k/a} \int_0^\infty dz\, \frac{e^{-kz/a}}{\sqrt{z}} \frac{1+z}{\sqrt{2+z}} \\ &\sim \frac{2k}{a^2} e^{-k/a} \int_0^\infty dz\, \frac{e^{-kz/a}}{\sqrt{z}} \frac{1}{\sqrt{2}} \left(1+ \frac{3}{4}z - \frac{5}{32} z^2 + \cdots \right) \\ &\sim \frac{\sqrt{2\pi}}{a} e^{-k/a} \frac{k^{1/2}}{a^{1/2}} \left(1 + \frac{3}{8}\frac{a}{k} - \frac{15}{128}\frac{a^2}{k^2} + \cdots \right), \end{align} where the last two steps constitute an asymptotic series in $a/k$ obtained by expanding the integrand in a non-everywhere uniformly converging power series. Note that the singularity at $z=0$ is of integrable type. It would have been non-integrable without the integration by parts done before deforming the contour. Note also that the above formulas don't work for $k=0$. Fortunately, we can obtain that value directly, $$ F(0) = \int_{-\infty}^{\infty} dx\, \frac{1}{(1+a^2x^2)^{3/2}} = \int_{-\infty}^{\infty} dx\, \frac{d}{dx} \frac{x}{(1+a^2x^2)^{1/2}} = \frac{2}{a} . $$

So, finally, an $a\sim 0$ asymptotic formula for the desired sum is \begin{align} I &= \frac{1}{2}\left(F(0) + 2\cos(\pi)F(2\pi) + \cdots \right) \\ &\sim \frac{1}{a} - \frac{2\pi}{a^{3/2}} e^{-2\pi/a} + \cdots , \end{align} where more terms could be computed as needed.

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Even for $a=1$ it looks sort of zeta-like series and existence of a closed form is not evident. (Say, is there a closed form for $\sum_{n=0}^\infty\frac1{(1+n^2)^{3/2}}$?) As for the asymptotic near the origin, changing the sum to integral gives $$ \int_0^{\infty } \frac{1}{\left(a^2\left(x+\frac{1}{2}\right)^2+1\right)^{3/2}} \, dx =\frac{1}{a}-\frac{1}{\sqrt{a^2+4}}=\frac{1}{a}-\frac{1}{2}+O(a^2). $$ For the series numerical evidence suggests (using NSum in Mathematica) $$ \sum_{n=0}^\infty \frac{1}{(1+a^2 (n+1/2)^2) ^{3/2}}=\frac{1}{a}+o(1), $$ where $o(1)$ seems to decay pretty fast as $a\to0^+$.

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  • $\begingroup$ I don't know whether $\sum_{n=0}^\infty (1+n^2)^{-3/2}$ has a closed form solution. When I try to do that as a contour integral I essentially encounter the same problem as in my inital question. In the end maybe numerics is all one can do. $\endgroup$ – Jonathan Edge Jan 24 '14 at 13:37
  • $\begingroup$ @JonathanEdge still the asymptotic as written perhaps can be obtained. Say via the Euler–Maclaurin summation formula. Or taking the difference between the series and integral and directly proving that it converges to zero as $a\to0^+$. $\endgroup$ – Andrew Jan 24 '14 at 13:58
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Here is an idea. Try to bound your series by simple expressions. For $a>0,k \ge 1$ define

$$ f(a,k)= \sum_{n=0}^\infty \frac{1}{(1+a^2 (n+1/2)^2) ^{k}} $$

Your series is $f(a,\frac32)$. All terms are positive and for all partial sums we have $ f(a,1) > f(a,\frac32) > f(a,2)$. This will hold for the infinite series assuming they converge.

$$ f(a,1) = 1/2\,\pi \,\tanh \left( {\frac {\pi }{a}} \right) {a}^{-1}$$ $$ f(a,2) = -1/4\,\pi \, \left( \pi \, \left( \cot \left( 1/2\,{\frac {\pi \, \left( 2\,i-a \right) }{a}} \right) \right) ^{2}-a\tanh \left( { \frac {\pi }{a}} \right) +\pi \right) {a}^{-2}$$

These bounds appear easier to analyze.

I believe the bounds as $a \to 0^+$ are $ \frac{\pi}{2 a} > f(a,\frac32) > \frac{\pi}{4a}$.

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