How to show that this integral is $0$? Is there a way to show that 
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} (\tau \xi - \sigma \eta) \frac{\exp(\sigma\xi + \tau\eta + c)}{[1 + \exp(\sigma\xi + \tau\eta + c)]^2} \exp(\frac{-\xi^2}{2}) \exp(\frac{-\eta^2}{2}) d\xi d\eta = 0$$
where $\sigma, \tau > 0$ and $c \in \mathbb{R}$?
Numerically, with special cases, I can see it's true. I wonder if there's some sleek trick. You can even see that the integral is probably 0 due to the symmetry in the plot of the integrand, where z4 is the area and $x = \xi, y = \eta$:
Plot of the integrand when $\sigma = 1, \tau = 2, c = 7$
Thus, I wonder if the trick is to split up the positive parts (when $\tau \xi > \sigma \eta$) and the negative parts (when $\tau \xi < \sigma \eta$), and show that 
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\frac{\sigma \eta}{\tau}} (\tau \xi + \sigma \eta) \frac{\exp(\sigma\xi + \tau\eta + c)}{[1 + \exp(\sigma\xi + \tau\eta + c)]^2} \exp(\frac{-\xi^2}{2}) \exp(\frac{-\eta^2}{2}) d\xi d\eta \ + \\ \int_{-\infty}^{\infty} \int_{\frac{\sigma \eta}{\tau}}^{\infty} (\tau \xi + \sigma \eta) \frac{\exp(\sigma\xi + \tau\eta + c)}{[1 + \exp(\sigma\xi + \tau\eta + c)]^2} \exp(\frac{-\xi^2}{2}) \exp(\frac{-\eta^2}{2}) d\xi d\eta = 0, $$ where the first part on the left-hand side is the positive part and the second part on the left-hand side is the negative part.
So we must show:
$$\int_{-\infty}^{\infty} \int_{-\infty}^{\frac{\sigma \eta}{\tau}} (\tau \xi + \sigma \eta) \frac{\exp(\sigma\xi + \tau\eta + c)}{[1 + \exp(\sigma\xi + \tau\eta + c)]^2} \exp(\frac{-\xi^2}{2}) \exp(\frac{-\eta^2}{2}) d\xi d\eta \ = \\ -\int_{-\infty}^{\infty} \int_{\frac{\sigma \eta}{\tau}}^{\infty} (\tau \xi + \sigma \eta) \frac{\exp(\sigma\xi + \tau\eta + c)}{[1 + \exp(\sigma\xi + \tau\eta + c)]^2} \exp(\frac{-\xi^2}{2}) \exp(\frac{-\eta^2}{2}) d\xi d\eta$$
Any ideas? This is definitely not a HW question. It's something I came across while working out a derivation.
 A: Write $\mathrm{a} = (\tau, \sigma)$ and $\mathrm{x} = (\eta, \xi)$. Also let $R_{\theta}$ be the rotation by $\theta$-radian and $J = R_{\pi/2}$. Then your integral can be reformulated as
$$ \int_{\mathbb{R}^2} \frac{4\langle J\mathrm{a}, \mathrm{x} \rangle}{\cosh^2(\frac{1}{2}\langle \mathrm{a}, \mathrm{x} \rangle + \frac{c}{2})} e^{-|\mathrm{x}|^2} \, \mathrm{d}\mathrm{x}$$
Recognizing this integral as function of $\mathrm{a}$ and denoting this by $I(\mathrm{a})$, we find that for any $\theta \in \mathbb{R}$,
\begin{align*}
I(\mathrm{a})
&= \int_{\mathbb{R}^2} \frac{4\langle JR_{\theta}\mathrm{a}, R_{\theta}\mathrm{x} \rangle}{\cosh^2(\frac{1}{2}\langle R_{\theta} \mathrm{a}, R_{\theta}\mathrm{x} \rangle + \frac{c}{2})} e^{-|R_{\theta}\mathrm{x}|^2} \, \mathrm{d}\mathrm{x} \\
&= \int_{\mathbb{R}^2} \frac{4\langle JR_{\theta}\mathrm{a}, \mathrm{x}' \rangle}{\cosh^2(\frac{1}{2}\langle R_{\theta} \mathrm{a}, \mathrm{x}' \rangle + \frac{c}{2})} e^{-|\mathrm{x}'|^2} \, \mathrm{d}\mathrm{x}' \qquad (\mathrm{x}' = R_{\theta}\mathrm{x})\\
&= I(R_{\theta}\mathrm{a})
\end{align*}
In the first line, we utilized the fact that rotations are isometries, i.e., $\langle \mathrm{x}, \mathrm{y} \rangle = \langle R_{\theta}\mathrm{x}, R_{\theta}\mathrm{y} \rangle$ always holds, and the fact that any two 2D rotations commute, i.e. $R_{\theta}R_{\psi} = R_{\psi}R_{\theta}$.
So by applying a rotation, we may assume that $\mathrm{a} = (a, 0)$, where $a = \sqrt{\tau^2+\sigma^2}$. Then
$$ I(a,0)
= \int_{\mathbb{R}^2} \frac{4 a \xi }{\cosh^2(\frac{1}{2}a\eta + \frac{c}{2})} e^{-\eta^2-\xi^2} \, \mathrm{d}\eta\mathrm{d}\xi
= 0 $$
since the integrand is odd with respect to $\xi$.
