Kouznetsov Wrote:Yes I mean that paper. In (3.2) there is not yet the logarithm but in the following formulas. Its just the 1 in the denominator \( 1+ip-z \) of the second summand that has to be changed into a -1. There are also some minor corrections for the description of the path on page 6 A.,C.: instead of just writing \( t=-iA \) or \( t=iA \) you have to add the corresponding 1's, as you mention before \( \Re(t)=\pm 1 \). But that didnt injure the readability.bo198214 Wrote:In formulas (3.2), (3.3), (3.4), (3.6), (4.4) you always omit the minus sign in front of the 1 below the log. Only in the computation formula 4.2 the minus sign is at the right place.Do you mean paper at http://www.ils.uec.ac.jp/~dima/PAPERS/2008analuxp.pdf ?
There is no log in formula (3.2), is there?
Quote:Now I have doubts about (3.3); I remember I had to play with phases manually combinig the logarithms; both Mathematica and Maple failed to do it as I wanted.
Can you reproduce Figure 2? Could you show it?
Unfortunately I am not the deep into the matter and also not familiar with complex function visualization. Hm, does that mean that I can not just take the normally cutted \( 0\dots -\infty \) logarithm but have to use something else?
I dint get you algorithm to work, it does always diverge. Though I use not Laguerre Gauss integration but simple equidistant integration this shouldnt stop the convergence. Can you send me a sample of your code?
Quote:Yes, I assume \( F(z^*)=F(z)^* \), but it follows from the analyticity and the assumption that \( F(z) \) is real at z>-2.Oh I wasnt aware of this.
