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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: ownerwrimathgroup 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[1I] + 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
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