- **Physics Forum**
(*http://www.molecularstation.com/forum/physics-forum/*)

- - **Integration Problem / Finding Contour lines Problem**
(*http://www.molecularstation.com/forum/physics-forum/32462-integration-problem-finding-contour-lines-problem.html*)

Integration Problem / Finding Contour lines ProblemHello! I am trying to find out if the following integral (given in mathematica's notation) can be solved analytically (using elementary or transcendental functions). If anyone knows the answer to this I'd appreciate the help very much! F[A_,B_] := Integrate[Exp[-x^2/2]/Sqrt[2*Pi] * Log[1 + Exp[A*x+B]], {x, -Infinity, Infinity}] where both A,B are REAL numbers. I'd like to find the relationship between A and B which causes this integral to be constant (finding the contour lines of F(A,B)). Knowing the solution to this integration problem would make finding this relationship easier, probably, however can anyone think of a way to discover this relationship without actualy having to solve the integral ? I'm completly out of ideas. Mathematica couldn't help me, I couldn't find this integral or similar ones in any of the integral table books I've searched through, I also had no success in trying to expand to a power series and then integrating. It's important to me to find the answer so any suggestions will be immensely appreciated. Thanks in advance. |

Integration Problem / Finding Contour lines ProblemOn Wed, 06 Aug 2003 18:24:47 -0400, blah12 wrote: Where did this integral come from? What range of A and B are important? |

Integration Problem / Finding Contour lines ProblemOn Wed, 6 Aug 2003 [Only registered and activated users can see links. Click Here To Register...] wrote:hmmm. check this: (sorry about not matching your notation.) change of variable, y=a*x+b, x=(y-b)/a (A=a,B=b 'cus I already typed it) F[a_,b_] := Integrate[Exp[-(y-b)^2/(2*a^2)/Sqrt[2*Pi]*log[1+exp(x)] now series expand the log (not really legal. a bit more on this later) F[a_,b_] := Integrate[Exp[-(y-b)^2/(2*a^2)/Sqrt[2*Pi]*sum[(-1)^(n-1) *exp(n*y),n=1,2...] swap the sum & integration sum[ Integrate[Exp[-(y-b)^2/(2*a^2)]/Sqrt[2*Pi]*(-1)^(n-1)n*exp(n*y), {y, -Infinity, Infinity}]),n=1,2...] The integral can be done in closed form, it's just a shifted Gaussian, and you have a series. Hopefully one you can put in some closed or otherwise useful form. The problem: The log series fails (may fail?) to converge when exp(y)>1. Is this patchable? Can you intergrate to exp(y)=.5, getting an sum of erf()s, re-expand around 1, go exp(y)=.5->1.5. You'd get a nasty double sum. Or is this an asymptotic series that works anyway, at least for a few terms? I forget how to tell. _or_: there is "clearly" little contribution from exp(y)>200 => integrate[f(x,a,b), {y, -Infinity, Infinity}] ~= integrate[f(x,a,b), {y, -Infinity, log(200)}] so if you expand the log series around 100 you would get a radius of convergance of 101 (100 to the pole @ -1), so you could integrate to y=log(200). maybe do it again for an expansion point of 200 or 1000 and compare. Or maybe even better, expand around M and let M tend to infinity? <fredrick frankenstein>It... Could... work.</fredrick frankenstein> I think I haven't used this. with the change of variable above the integrand is a Gaussian at -B with a variace of (maybe a constant times) A (which can be taken as positve without loss of generality) times a function g(A,B,y) ~= 0 for y << 0 ~= y for y >> 0 with a curved interpolation region near y=0. So for b >> 0, F will be ~0, independant of A and B (tho tiny variations may be vital to your problem) and F(A,B) ~= canstant + constant*B for B << 0 (the center of the Gaussian is at -B, largely independant of A. Only near B=0 will there be large variations for both. And I guess any time A << 1, the integral will be ~= g(A,B,-B) Derive didn't like it either 8^) Numerical recipes? They might have ideas/methods. You might look at g(A,B,y)= 0 for y<0 = y for y>=0 integrate[exp{-(y-b)^2/(2*a^2)}*g(A,B,x), y=-inf...inf] to get a feel for things, as it can be done in clsed form. Note that this integral is strictly less than you integral. 'tis late CC p.s. if any of this works you must tell us what the project is... |

All times are GMT. The time now is 12:29 AM. |

Powered by vBulletin® Version 3.8.4

Copyright ©2000 - 2015, Jelsoft Enterprises Ltd.

Copyright 2005 - 2012 Molecular Station | All Rights Reserved