Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Complex analysis, holomorphic functions, automorphic group actions and forms, pseudoconvexity, complex geometry, analytic spaces, analytic sheaves.
1
vote
Integrate Faddeeva function
For the Faddeeva function
$$F(z)=e^{-z^2}\, \text{erfc}(-i z)\tag{1}$$
the integral
$$I=\int\limits_{-\infty}^{\infty} |F(z)|^2 \, dz\tag{2}$$
can be evaluated using the Plancherel theorem
$$\int\limi …
1
vote
When is $\Re(\zeta(s)) - \Im(\zeta(s)) = 0 $ with $\Re(\zeta(s))\neq 0$ and $\Im(\zeta(s))\n...
Here are a couple of contour plots of
$$\Re(\zeta(\alpha+i \beta))=\Im(\zeta(\alpha+i \beta))\quad\tag{blue curve}$$
$$\Re(\zeta(\alpha+i \beta))=0\quad\tag{orange curve}$$
$$\Im(\zeta(\alpha+i \beta) …