7
$\begingroup$

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?

  • 0
    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...2012-08-28

1 Answers 1

6

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".

  • 0
    I would be very surprised if there was one. Of course, I've been surprised before.2012-08-30