Re: Help Improving this integral calculation / solution
- To: mathgroup at smc.vnet.net
- Subject: [mg108103] Re: [mg108044] Help Improving this integral calculation / solution
- From: "David Park" <djmpark at comcast.net>
- Date: Mon, 8 Mar 2010 06:15:17 -0500 (EST)
- References: <14141119.1267954710922.JavaMail.root@n11>
This goes much better if you use exact values:
d = 7/10;
v = 1/20;
a = 1/10;
nk = (a + (1 - a) k v t)/(d + a + (1 - a) k v t) // Simplify
(20 + 9 k t)/(160 + 9 k t)
pg[k_] = Exp[1 - k];
gavg = Integrate[k pg[k], {k, 1, \[Infinity]}];(*This is just 2*)
result = 1/gavg Integrate[nk k pg[k], {k, 1, \[Infinity]},
Assumptions -> 0 < t < 1];
f[t_] = result
(9 t (-70 + 9 t) -
11200 E^(1 + 160/(9 t)) ExpIntegralEi[-1 - 160/(9 t)])/(81 t^2)
Plot[f[t] - t, {t, 0, 0.2},
PlotRange -> Automatic]
FindRoot[f[t] == t, {t, .15}]
{t -> 0.142045}
I tried solving the integral with a, d and v as parameters, and accepting
the conditions but that did not work.
David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/
From: DOD [mailto:dcodea at gmail.com]
I have an integral I need to calculate, with one variable left
symbolic, and then use the result to find a numerical solution to an
equation, and the result of the integration is not giving me what I
need:
------
d = .7;
v = .05;
a = .1;
nk = (a + (1 - a) k v t)/(d + a + (1 - a) k v t);
pg[k_] = Exp[1-k];
gavg = Integrate[k pg[k], {k, 1, \[Infinity]}]; (* This is just 2 *)
result = 1/gavg Integrate[nk k pg[k], {k, 1, \[Infinity]},
Assumptions -> 0 < t < 1];
test = FullSimplify[result]
FindRoot[test == t, {t, .1}]
-----
This code always gives up and stays at the initial guess. So, I look
at the result of the integration, "result" (or it's simplified
version, test2) and calculated it for various values of t, and it is
always zero.
----
test/.t->{.1,.2,.3}
----
Output:{3.36999*10^66, 0., 0.}
-----
So that's a problem. If I set t=.3, say, and the beginning, and
calculate the integral, I get 0.160047, which is clearly not zero.
-----
d = .7;
v = .05;
a = .1;
t = .3;
nk = (a + (1 - a) k v t)/(d + a + (1 - a) k v t);
pg[k_] = Exp[1 - k];
gavg =Integrate[k pg[k], {k, 1, \[Infinity]}];(*This is just 2*)
result = 1/gavg Integrate[nk k pg[k], {k, 1, \[Infinity]}]
----
Output=0.160047
-----
So there is a problem in the Integrate step. So I want to find a
point where the output of that integral (result), as a function of t,
satisfies result=t. Is there anyway to do this using built-in
functions?