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Does there exist an explicit expression for

$$\sum_{k=0}^{\infty }\frac{\left( -1\right) ^{k}e^{-\lambda \left( 2k+1\right) ^{2}}}{\left( 2k+1\right) ^{3}}\;,$$

where $\lambda$ is a positive scalar?

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This is not "a sum of infinite series", but just a series (which in itself is an sum of infinite terms) – leonbloy Aug 28 '12 at 20:04
@leonbloy Actually, the OP is asking about the sum of a series. (But "infinite" is useless, I agree.) – Did Aug 28 '12 at 20:05
Try differentiating with respect to $\lambda$ for a start. Maybe do it twice, and then use facts about the geometric series. I mean, $x=e^{-\lambda (2k+1)^2} < 1$ is true... – nayrb Aug 28 '12 at 20:34
up vote 6 down vote accepted

The Jacobi theta function $$\theta_1(z,e^{-4\lambda}) = 2 \sum_{k=0}^\infty (-1)^k e^{-\lambda (2k+1)^2} \sin((2k+1) z)$$ To get a factor of $1/(2k+1)^3$, we can do some integration: $$ \int_0^{\pi} \left( \frac{\pi^2}{16} - \frac{z^2}{8} \right) \theta_1(z,e^{-4\lambda})\ dz = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^3} e^{-\lambda (2k+1)^2}$$ since $\int_0^{\pi} \left(\frac{\pi^2}{8} - \frac{z^2}{4} \right) \sin((2k+1)z)\ dz = 1/(2k+1)^3$.

I rather doubt that the integral can be done in "closed form".

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Can I say that there is either no closed form for $$\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^5} e^{-\lambda (2k+1)^2}$$ – Xiangyu Meng Aug 30 '12 at 5:50
I would be very surprised if there was one. Of course, I've been surprised before. – Robert Israel Aug 30 '12 at 6:51

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