Re: Novice/Integral
- To: mathgroup at smc.vnet.net
- Subject: [mg79374] Re: Novice/Integral
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Wed, 25 Jul 2007 02:22:50 -0400 (EDT)
- Organization: The Open University, Milton Keynes, UK
- References: <f84ih0$oie$1@smc.vnet.net>
Daniel wrote:
> Hi all,
>
> I'm new at this, but I'm trying to evalute the integral like so,
>
> g = {g0, g1, g2, g3, g4}
> fg[x_] = (1 - x)^g[[3]]*(1 + g[[4]]*x^g[[5]]) /. x -> E^(y)
> IntX[m_, y_] = Integrate[fg[x], y]
>
> I'm hoping there's standard tricks for manipulating integrals so mathematica can win out. Any advice is appreciated.
>
> Thanks,
> Danny
You could tell Mathematica that the values of g are real and transform
the exponentials into equivalent trigonometric functions. For instance,
In[1]:= $Version
Out[1]= "6.0 for Microsoft Windows (32-bit) (June 19, 2007)"
In[2]:= g = {g0, g1, g2, g3, g4};
fg[x_] = (1 - x)^g[[3]]*(1 + g[[4]]*x^g[[5]]) /. x -> E^y;
IntX[m_, y_] = Integrate[ExpToTrig[fg[x]], y,
Assumptions -> {Element[g, Reals]}]
Out[4]= ((1 - \[ExponentialE]^-y)^-g2 (1 - \[ExponentialE]^y)^
g2 Hypergeometric2F1[-g2, -g2,
1 - g2, \[ExponentialE]^-y])/g2 + ((\[ExponentialE]^y)^
g4 g3 Hypergeometric2F1[-g2, g4, 1 + g4, \[ExponentialE]^y])/g4
This works also with version prior to 6.0.
In[1]:=
$Version
Out[1]=
5.2 for Microsoft Windows (June 20, 2005)
In[2]:=
g = {g0, g1, g2, g3, g4};
fg[x_] = (1 - x)^g[[3]]*(1 + g[[4]]*x^g[[5]]) /. x -> E^y;
IntX[m_, y_] = Integrate[ExpToTrig[fg[x]], y,
Assumptions -> {Element[g, Reals]}]
Out[4]=
y g2
(1 - E ) Hypergeometric2F1[-g2, -g2, 1 - g2,
-y
1 E ]
-- ---------------------------------------------------\
g2 -y g2
(1 - E )
y g4 y
(E ) g3 Hypergeometric2F1[-g2, g4, 1 + g4, E ]
+ ------------------------------------------------
g4
Regards,
Jean-Marc