-5
$\begingroup$

1.$$\int_{0}^{1} \frac{1}{1+e^{-(x+\ln(u/(1-u)))/\tau}}\, du$$

2.$$\frac{1}{\sqrt{2}\pi}\int_{-\infty}^{+\infty}\frac{e^{-u^{2}/2}}{1+e^{-(x-u)/\tau}}\,du$$

please help me. I tried to use MATLAB but I failed. The background of the first one is the expectation of gumbel-softmax.

Improve this question
$\endgroup$
4
  • $\begingroup$ Would you give me more hints? $\endgroup$
    – Haoyu Wang
    Dec 23, 2018 at 11:39
  • $\begingroup$ The original one is a little wrong. I modified it. The new can be solved in the same way? $\endgroup$
    – Haoyu Wang
    Dec 23, 2018 at 11:51
  • 1
    $\begingroup$ The question is better suited to MSE: math.stackexchange.com, and is likely to be closed here briefly. $\endgroup$
    – Josiah Park
    Dec 23, 2018 at 11:53
  • $\begingroup$ I transfer it to MSE $\endgroup$
    – Haoyu Wang
    Dec 23, 2018 at 12:03

1 Answer 1

Reset to default
1
$\begingroup$

Neither integral has a closed form expression. For $\tau=1$ the first integral evaluates to $$\int_{0}^{1} \frac{1}{1+e^{-(x-\ln(-\ln u))}}\, du=-e^{x+e^x} \text{Ei}\left(-e^x\right).$$

Improve this answer
$\endgroup$
3
  • $\begingroup$ I think the second doesn't diverge because it is smaller than the integral of normal distribution. $\endgroup$
    – Haoyu Wang
    Dec 23, 2018 at 0:20
  • $\begingroup$ you can take the $e^{-x^2/2}$ in the numerator out of the integral, since it does not depend on $u$, and then you can change variables from $x-u$ to $y$, so the integral you are evaluating is $\int_{-\infty}^\infty(1+e^{-y/\tau})^{-1}dy$ which diverges no matter what the sign is of $\tau$. $\endgroup$
    – Carlo Beenakker
    Dec 23, 2018 at 8:11
  • $\begingroup$ I am sorry that I made a mistake. I revised my question. Please help me. $\endgroup$
    – Haoyu Wang
    Dec 23, 2018 at 8:42

Not the answer you're looking for? Browse other questions tagged or ask your own question.