Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
1997
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 1997

[Date Index] [Thread Index] [Author Index]

Search the Archive

RE: [Q] Why Integrate[1/x,x] <>

  • To: mathgroup at smc.vnet.net
  • Subject: [mg8778] RE: [mg8699] [Q] Why Integrate[1/x,x] <>
  • From: Ersek_Ted%PAX1A at mr.nawcad.navy.mil
  • Date: Thu, 25 Sep 1997 12:26:12 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

Dennis wrote:
|
|it is well known that Integrate[1/x,x] = Log[Abs[x]], but Mma 2.2 returns
|Log[x].
|
|
Mma 3.0 gives the same result.
I have spent several years pondering this.  At one time I even sent WRI a 
suggestion to
make the same change you discuss.  Then about a year ago I learned that the 
answer given by Mma is more correct.  I consider it a great injustice that 
few of us are ever taught that Log[x] is the more correct answer.

Recall  Exp[ Pi * I ] = -1    ,   This gives  Log[-1]=Pi*I
So  Log[-5]=Log[(5)(-1)]=Log[5]+Log[-1]=Log[5] + Pi*I
Likewise Log[-3] = Log[3] + Pi*I
Mma gives the same for   Log[-5]  and  Log[-3]

Now if we say Integrate[1/x, x] = Log[x]
What will happen if we use this to integrate (dx/x) from -5 to -3

Area=Integral of  (dx/x)  from (lower limit) = -5   to   (upper limit) = -3
Area= Log[-3] - Log[-5]
Area= (Log[3] + Pi*I ) - (Log[5] + Pi*I )
Area= Log[3] - Log[5]
The same answer you get using Log[Abs[x]].

Now consider the following integral along a line in the complex plane.

In[1]:= answer =  Integrate[1/x, { x,  (1 - I) , (1 + I)  } ]
Out[1]=  -Log[1-I] + Log[1+I]

In[2]:= FullSimplify[ answer ]
Out[1]=  Pi * I / 2

I think if you estimate this integral using a Riemann Sum along the path of 
integration you will get approximately  ( Pi * I / 2 ).

Now if you use Integrate[1/x, x]=Log[ Abs[x] ]  to do the same problem you 
will get zero.

______________________________________
The Handbook of Mathematical Functions
by Milton Abramowitz and Irene Stegun
gives a very precise statement on this subject.
They say (using their notation as best I can with ASC characters ):

The integral of (dt/t) from (lower limit)=1 to (upper limit)=z
equals  ln(z).

Where the path of integration does not cross through the origin, and does 
not cross the negative real axis.

____________________________________
As I recall the  ln(z)  is defined as follows:
   If       z=r * E^(I * theta)        where  r, and theta are real,   (r > 
0) ,  and  ( -Pi < theta <= Pi )
Then   ln(z)= ln(r) + I * theta
I think this is the same as Log[z] in Mma.

____________________________________
The statement by Abramowitz and Stegun doesn't  address an integral
along the negative real axis.  However, I am confident that
Integrate[1/x, { x, a , b} ] = Log[b] - Log[a]
when (a) and (b) are real and negative.
To be precise this is provided the path of integration is the straight line 
connecting (a) and (b).

     Ted Ersek



  • Prev by Date: MatrixForm
  • Next by Date: installation HELP!
  • Previous by thread: Re: MatrixForm
  • Next by thread: installation HELP!