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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
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