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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.

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    $\begingroup$ The change of variable $$(x,y)=(\tau\xi-\sigma\eta,\sigma\xi+\tau\eta)$$ allows to write this double integral as the product of two simple integrals, one of whose is zero by symmetry. $\endgroup$ – Did Dec 16 '17 at 10:05
  • $\begingroup$ @Did : How did you get a product of two simple integrals? $\endgroup$ – 193381 Dec 16 '17 at 20:58
  • $\begingroup$ @Did : After your change of variable, I need to show this: $$\frac{1}{\tau^2 + \sigma^2} \int_{-\infty}^{\infty} \int_{-\infty}^{0} x \frac{\exp(v + c)}{[1 + \exp(v + c)]^2} \exp\left(\frac{- (\tau u + \sigma v) ^2}{2(\tau^2 + \sigma^2)^2}\right) \exp\left(\frac{- (\tau u - \sigma v) ^2}{2(\tau^2 + \sigma^2)^2}\right) dx dy \ = \\ \frac{1}{\tau^2 + \sigma^2} \int_{-\infty}^{\infty} \int_{0}^{\infty} x \frac{\exp(v + c)}{[1 + \exp(v + c)]^2} \exp\left(\frac{- (\tau u + \sigma v) ^2}{2(\tau^2 + \sigma^2)^2}\right) \exp\left(\frac{- (\tau u - \sigma v) ^2}{2(\tau^2 + \sigma^2)^2}\right) dx dy $$ $\endgroup$ – 193381 Dec 16 '17 at 21:18
  • $\begingroup$ @Did : I think the $c$ is the issue... fuks things up $\endgroup$ – 193381 Dec 16 '17 at 21:18
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    $\begingroup$ Simply develop and sum the exponents in the exponentials... to see what you think is an "issue" disappear. $\endgroup$ – Did Dec 16 '17 at 21:20
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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 $\xi$.

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  • $\begingroup$ ah, rotation. I need to brush up on that. Thank you for answering. $\endgroup$ – 193381 Dec 16 '17 at 21:21

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