RE: [Integrate] Why two results of same eq. are different?

*To*: mathgroup at smc.vnet.net*Subject*: [mg44748] RE: [mg44655] [Integrate] Why two results of same eq. are different?*From*: "Sung Jin Kim" <kimsj at mobile.snu.ac.kr>*Date*: Tue, 25 Nov 2003 00:45:16 -0500 (EST)*Reply-to*: <kimsj at mobile.snu.ac.kr>*Sender*: owner-wri-mathgroup at wolfram.com

> From: Vladimir Bondarenko [mailto:vvb at mail.strace.net] To: mathgroup at smc.vnet.net > Sent: Thursday, November 20, 2003 5:55 PM > Mathematica 3.0 (April 25, 1997) returns a valid value for your > integral > 1/10*(9*Exp[1/10]*ExpIntegralE[1, 1/10] + 10)/Log[2] //N > 4.05856 Thanks to your helps (Vladimir Bondarenko; Andrzej Kozlowski [akoz at mimuw.edu.pl]; David W. Cantrell [DWCantrell at sigmaxi.org]), I finally got the solution of my proposed integral problem as below: Integrate[ Log[2, 1 + a*x]*Exp[-x]*x, {x, 0, Infinity}] ==> (a-Exp[1/a]*(-1+a)*ExpIntegralEi[-1/a])/(a*Log[2]) Note that since I don't have older version, I did it with somehow manipulation of my finger, e.g. including usage of Limit[] function. To prohibit same failures again by others, I describe the procedures what I have had to end this integration briefly. Let us assume the original integral as *Integrate[ f, {x,0,Infinity}]*, Then its step to find is denoted by Step1: F = Integrate[ f, x] Step2: Find F0 = F/.x->0 Step3: To find limit of F (x->Infinity) where assuming F = A+B, I get limit values of A and B instead since limit of F is unavailable to find and add two values as FL. Therefore, I get final solution, which is FL - F0. At this moment, I have another question: why do I need such dumb step3 to find Limit value, at least why in this example? Regards, --- Sung Jin Kim A member of MCL in SNU: kimsj at mobile.snu.ac.kr, A MTS of i-Networking Lab in SAIT: communication at samsung.com