Integral of this exponential function: does it have a solution?


by fchopin
Tags: exponential, function, integral, solution
fchopin
fchopin is offline
#1
Oct31-09, 01:29 PM
P: 9
Hi all,

I'm trying to solve the definite integral between 0 and inf of:

exp(a*x^2 + b*x + c)
--------------------- dx
1 + exp(m*x + n)

with a,b,c,m,n real numbers and a < 0 (negative number so it converges).

I've read in the forum's rules that I have to post the work that I have done to get an answer but I have nothing reasonable to post (I have tried many alternatives but I didn't suceed, sorry)

A way to obtain the exact solution would be perfect but an approximate result, even an upper/lowerbound would be fine as well.

Any idea or help, please?

Thanks in advance,
FC.
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g_edgar
g_edgar is offline
#2
Oct31-09, 08:00 PM
P: 608
Can you do the special case...
[tex]
\int _{0}^{\infty }\!{\frac {{{\rm e}^{-{x}^{2}}}}{1+{{\rm e}^{x}}}}{d
x}
[/tex]
Pere Callahan
Pere Callahan is offline
#3
Nov1-09, 05:21 AM
P: 588
Quote Quote by g_edgar View Post
Can you do the special case...
[tex]
\int _{0}^{\infty }\!{\frac {{{\rm e}^{-{x}^{2}}}}{1+{{\rm e}^{x}}}}{d
x}
[/tex]
Mathematica can't...

Gib Z
Gib Z is offline
#4
Nov2-09, 03:18 AM
HW Helper
Gib Z's Avatar
P: 3,353

Integral of this exponential function: does it have a solution?


Maybe hes alluding to the idea that if you can't do one of the basic cases you wouldn't be able to do the general case either.
fchopin
fchopin is offline
#5
Nov3-09, 03:29 AM
P: 9
Hi guys,

thank you for your answers. I have found that if m<0 and n<0 then 1/(1 + exp(m*x + n)) can be expanded into Maclaurin series, which yields something like 1 + e^() - e^2*(). The terms can be multiplied by the numerator of the original integral, thus finally obtaining three integrals of the form exp^(a*x^2 + b*x + c), which have closed-form solutions in terms of the error function.

Thank you for your interest!


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