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MathGroup Archive 2007

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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>



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