[Date Index]
[Thread Index]
[Author Index]
Re: Integrate modified in version 6?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg78142] Re: Integrate modified in version 6?
*From*: "Dana DeLouis" <dana.del at gmail.com>
*Date*: Sat, 23 Jun 2007 07:23:04 -0400 (EDT)
> The good old (?) Mathematica 5.2 returns
> Integrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}]
> -1 - 2*Catalan + Pi*(1/4 + Log[2])
Hi. I get -Infinity also using 6.0. It appears if one uses "TrigToExp",
one gets the same answer as 5.2's.
equ = z*(ArcSin[z]/(1 + z)^2);
Integrate[equ, {z, 0, 1}]
-Infinity
NIntegrate[equ, {z, 0, 1}]
0.13105306534661287
(* Using TrigToExp to get same answer as 5.2 *)
Integrate[TrigToExp[equ], {z, 0, 1}]
-1 - 2*Catalan + Pi*(1/4 + Log[2])
N[%]
0.13105306534661243
(* Redoing calculation which "appears" to indicate they are similar *)
N[NIntegrate[equ, {z, 0, 1}, WorkingPrecision -> 20]]
0.1310530653466124
--
Dana DeLouis
Using 6.0, but having to use 5.2 Help files.
"dimitris" <dimmechan at yahoo.com> wrote in message
news:f5asi1$9t6$1 at smc.vnet.net...
>I don't have 6 to be more rigorous but
> based on some integrals post in another
> forum by Vladimir Bondarenko I am quite
> sure that something has change in the integration
> algorithm for definite integrals...
>
> Say for example the integral Integrate[z ArcSin[z]/(1+z)^2, {z, 0,
> 1}].
>
> In Mathematica 6 we have
>
> Integrate[z ArcSin[z]/(1+z)^2, {z, 0, 1}]
> -Infinity
>
> which is far away from truth.
>
> The good old (?) Mathematica 5.2 returns
>
> In[3]:=
> Integrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}]
> {N[%], NIntegrate[z*(ArcSin[z]/(1 + z)^2), {z, 0, 1}]}
>
> Out[3]=
> -1 - 2*Catalan + Pi*(1/4 + Log[2])
> Out[4]=
> {0.13105306534661265, 0.1310530653479215}
>
<snip>
Prev by Date:
**Re: Combination List**
Next by Date:
**Re: Re: bad performance of Reduce (5.2)**
Previous by thread:
**Re: Integrate modified in version 6?**
Next by thread:
**Re: Integrate modified in version 6?**
| |