Henri . Cohen on Sat, 06 Feb 2010 00:42:57 +0100

 Re: PARI stable release 2.3.5 (for real now)

```No, not quite. There are indeed some of problems with the method, but here you
do not have to do anything: the [[1],-3/2] does not at all mean a decrease in
Z^(-3/2) (which is in fact in Z^(-2)), but a decrease like exp(-(3/2)Z). So
it must simply be omitted:
intnum(Z=[0,-1/2],[1],1/sqrt(Z)*(1+Z)^(-3/2))
2.000000000000000000000000000

Henri Cohen

============================

Bill Allombert <Bill.Allombert@math.u-bordeaux1.fr> a écrit :

```
```On Wed, Feb 03, 2010 at 12:14:57AM -0800, Ilya Zakharevich wrote:
```
```  *) intnum() is still borken (same as in 3.0):

? intnum(Z=[0,-1/2],[[1],-3/2],1/sqrt(Z)*(1+Z)^(-3/2))
%1 = 7.150393303262919781690330041 + 1018.215229301285883563073358 I

(I presume this is integral from 0 to infinity...)
```
```
You need to split the integral:

```
? intnum(Z=[0,-1/2],1,1/sqrt(Z)*(1+Z)^(-3/2))+intnum(Z=1,[[1],-3/2],1/sqrt(Z)*(1+Z)^(-3/2)) %36 = 2.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
```
This belongs to a class of problem inherent to the method.

Bill.

```
```

```