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Topic: An analytical solution to an integral not currently
Replies: 1   Last Post: Jul 16, 2013 5:55 AM

 Matthias Bode Posts: 84 Registered: 3/16/05
Re: An analytical solution to an integral not currently
Posted: Jul 16, 2013 5:55 AM

Hola:

Input: (Sqrt(Log[x])^-1 + a*x + b)

Result:

Integrate[Sqrt[Log[x]]^(-1) + a*x + b, x] = (x*(2*b + a*x + 4*DawsonF[Sqrt[Log[x]]]))/2

From:

http://integrals.wolfram.com/index.jsp?expr=%28Sqrt%28Log[x]%29^-1+%2B+a*x+%2B+b%29&random=false

Muchos saludos,

MATTHIAS BODE
LVSABA@HOTMAIL.COM

> From: rprogrammer@gmail.com
> Subject: An analytical solution to an integral not currently in Mathematica?
> To: mathgroup@smc.vnet.net
> Date: Sun, 14 Jul 2013 01:48:46 -0400
>
> Question: Integral dx of 1/sqrt(Log[x] + a*x + b)
> (sorry if my notation is off; I just used the online integrator and don't have Mathematica proper, http://integrals.wolfram.com/index.jsp?expr=1%2Fsqrt%28Log%5Bx%5D+%2B+a*x+%2B+b%29)
> (the online integrator returned this as of the time of writing this (2013-07-13): "Mathematica could not find a formula for your integral. Most likely this means that no formula exists." )
>
>
> Another system's unconfirmed answer (in that notation; sorry) (version 5.27.0): -sqrt(%pi)*%i*%e^(-a*x-b)*erf(%i*sqrt(log(x)+a*x+b))
>
> Strangely, the other system only produces this result when given, say, x(t) in all places for x (including variable of integration).
>
> I can't seem to get the other system to verify its result symbolically, but when I try random numerical sampling, it does seem to agree, albeit horribly plagued by floating point errors for large x.
>
>
> Can anyone offer insight, or possibly prove it's correctness or incorrectness? :)
>
>
> (P.S. I just joined this group, so apologies if it's the wrong one or I'm not following guidelines)
>