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Re: Integrate with PrincipalValue->True

• To: mathgroup at smc.vnet.net
• Subject: [mg79974] Re: Integrate with PrincipalValue->True
• From: dimitris <dimmechan at yahoo.com>
• Date: Fri, 10 Aug 2007 01:40:17 -0400 (EDT)
• References: <f9em11$k0v$1@smc.vnet.net>

I guess the post of Chris Chiasson (link below)
might be interesting for you.

http://groups.google.gr/group/comp.soft-sys.math.mathematica/browse_thread/thread/d9791517f8fe1e82/081c097d7948de33?hl=el#081c097d7948de33

At least in version 5.2 the integral is evaluated with the convolution
method.

Dimitris

PS

Note in direct connection with your message but,
I think you may find useful the following

In[1]:=
$Version

Out[1]=
"5.2 for Microsoft Windows (June 20, 2005)"

In[2]:=
f[x_] := Sqrt[x]/(1 - x^2)

In[3]:=
Plot[f[x], {x, 0, 5}]

In[5]:=
(Limit[f[x], x -> 1, Direction -> #1] & ) /@ {-1, 1} (*simple pole at
x=1*)

Out[5]=
{-Infinity, Infinity}

Let's take some numerical estimations.

In[6]:=
lst = Range[2, 10];

In[7]:=
o = 10^(-10); (*I know about the CaucyPrincipalValue package!*)
(NIntegrate[f[x], {x, 0, 1 - o}, MaxRecursion -> 10, WorkingPrecision -
> 30, PrecisionGoal -> 10] +
    NIntegrate[f[x], {x, 1 + o, #1}, MaxRecursion -> 10,
WorkingPrecision -> 30, PrecisionGoal -> 10] & ) /@ lst

Out[8]=
{-0.0739430311049670432204230836`8.939604229977633,
-0.3887186027341901733414810109`9.432530200498425,
  -0.5578425734600364471`9.068583719613938,
-0.6690501664513288361`9.285122568660134,
-0.7496922768944342099`8.870518013789683,
  -0.8117464722762367919`8.956342895479304,
-0.8614604454149061598`9.006467398351345,
-0.9024721821182825622`9.082584062920915,
  -0.9370688073879695108`9.154292337156724}

Let's see how someone could obtain the integral.

In[9]:=
F[x_] = Integrate[f[x], x]

Out[9]=
-ArcTan[Sqrt[x]] - (1/2)*Log[-1 + Sqrt[x]] + (1/2)*Log[1 + Sqrt[x]]

In[10]:=
myres = (Limit[#1, e -> 0] & )[Limit[F[x], x -> a] - Limit[F[x], x ->
1 + e, Assumptions -> 0 < e < 1] +
    Limit[F[x], x -> 1 - e, Assumptions -> 0 < e < 1] - Limit[F[x], x -
> 0, Direction -> 1]]

Out[10]=
(1/2)*(-2*ArcTan[Sqrt[a]] - Log[-1 + Sqrt[a]] + Log[1 + Sqrt[a]])

In[11]:=
(N[#1, 10] & )[FullSimplify[(myres /. a -> #1 & ) /@ lst]]

Out[11]=
{-0.0739430311049662529312477775`10.,
-0.3887186027341893918416912875`10.,
-0.5578425734600356573194428417`10.,
  -0.6690501664513280438666761403`10.,
-0.7496922768944334136540147013`10.,
-0.8117464722762359983781875487`10.,
  -0.861460445414905367905256551`10.,
-0.9024721821182817711426768946`10.,
-0.9370688073879687198265482121`10.}

Agreement with the numerical estimations.

Let's see what Mathematica 5.2 does

In[12]:=
Integrate[f[x], {x, 0, a}, Assumptions -> a > 1, PrincipalValue ->
True]
res52 = (FullSimplify[#1, a > 1] & )[FunctionExpand[%]]

Out[12]=
-(Pi/2) + (2*Hypergeometric2F1[1/4, 1, 5/4, 1/a^2])/Sqrt[a]

Out[13]=
(1/2)*(-Log[1 - 1/Sqrt[a]] + Log[1 + 1/Sqrt[a]] + I*(Log[I - Sqrt[a]]
- Log[I + Sqrt[a]]))

In[14]:=
(N[#1, 10] & )[FullSimplify[(res52 /. a -> #1 & ) /@ lst]]

Out[14]=
{-0.0739430311049662529312477775`10.,
-0.3887186027341893918416912875`10.,
-0.5578425734600356573194428417`10.,
  -0.6690501664513280438453002623`10.,
-0.7496922768944334136540147013`10.,
-0.8117464722762359983781875488`10.,
  -0.8614604454149053679052565509`10.,
-0.9024721821182817711426768945`10.,
-0.937068807387968719826548212`10.}

Again correct.

How about Mathematica 6?

In[15]:=
res6 = ArcCoth[Sqrt[a]] - ArcTan[Sqrt[a]];

In[16]:=
(N[#1, 10] & )[(res6 /. a -> #1 & ) /@ lst]

Out[16]=
{-0.0739430311049662529312477776`10.,
-0.3887186027341893918416912874`10.,
-0.5578425734600356573194428417`10.,
  -0.6690501664513280438453002623`10.,
-0.7496922768944334136540147013`10.,
-0.8117464722762359983781875487`10.,
  -0.8614604454149053679052565509`10.,
-0.9024721821182817711426768945`10.,
-0.937068807387968719826548212`10.}

Correct!

So it appears that Mathematica 6 is one step forward from version 5.2
since
not only it gets the correct symbolic answer but it also it returns a
much more
simplified answer!

The funny is that Mathematica needs a little help in order to show
that the three expressions (i e res52, myres, res6) are equal.

In[17]:=
(FullSimplify[#1, a > 1] & )[{res6 == res52, res52 == myres, res6 ==
myres}]
Apply[Plus[#1 - #2] & , Drop[%, {3}], {1}]
% /. a -> 0
Together[D[%%, a]]

Out[17]=
{Log[I - Sqrt[a]] + Log[1 - I*Sqrt[a]] == Log[1 + I*Sqrt[a]] + Log[I +
Sqrt[a]],
  (-I)*(Log[I - Sqrt[a]] - Log[I + Sqrt[a]]) == 2*ArcTan[Sqrt[a]],
True}

Out[18]=
{Log[I - Sqrt[a]] + Log[1 - I*Sqrt[a]] - Log[1 + I*Sqrt[a]] - Log[I +
Sqrt[a]],
  -2*ArcTan[Sqrt[a]] - I*(Log[I - Sqrt[a]] - Log[I + Sqrt[a]])}

Out[19]=
{0, 0}

Out[20]=
{0, 0}







On 9    , 12:16, chuck009 <dmili... at comcast.com> wrote:
> Can someone explain to me how Integrate calculates the following improper integral?
>
> In[42]:= $Version
> Integrate[Sqrt[x]/(1 - x^2), {x, 0, a},
>    PrincipalValue -> True, Assumptions ->
>      a > 1]
>
> Out[42]= "6.0 for Microsoft Windows (32-bit) (June 19, 2007)"
>
> Out[43]= ArcCoth[Sqrt[a]] - ArcTan[Sqrt[a]]
>
> Thanks,
> Charles