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