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