How can I numerically evaluate a non-trivial integral in MuPAD?

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Tommaso Isolabella 2015-9-2

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评论： Torsten 2015-9-4

I am trying to evaluate an integral in MuPAD using numeric::int(), numeric::quadrature and float(hold(int)) but none of these methods seems to work. The function I want to integrate is a parametric function defined as (h and c1 are numbers I defined before)

 Psi: (M_p,R_p,x)-> -(h^3/c1)*(M_p/R_p)*(cos(x)/(R_p^2*(1-`&epsiv;`*cos(x))^2-h^2))

I then put the parameters I want into the function, leaving x as the only variable (thus defining a new function):

psi_(venus): x->Psi(4.867*10^24,108*10^9,x)

and then when I try to integrate it, with x from 0 to 2*PI, MuPAD doesn't evaluate it with any of the routines listed above.

Thanks in advance to whoever can help me solve this problem.

Tommaso

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Tommaso Isolabella 2015-9-3

as I wrote in my post, c1,h and epsilon are constants I defined before integrating

Torsten 2015-9-3

No poles in between 0 and 2*pi ?

Then it seems that your integral is equal to zero:

http://integrals.wolfram.com/index.jsp?expr=Cos%5Bx%5D%2F%28a%5E2*%281-e*Cos%5Bx%5D%29%5E2-h%5E2%29&random=false

Best wishes

Torsten.

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Answer 1

Walter Roberson 2015-9-4

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You wrote, and I quote, "h and c1 are numbers I defined before". Nothing about epsilon there.

You also wrote "I then put the parameters I want into the function, leaving x as the only variable (thus defining a new function):"

That tells us that you do not intend anything other than x to be a free variable, but it does not tell us that epsilon has been given a value.

The values of h and epsilon are important, as they determine whether there are any poles in the integration. With the values you gave for the parameters, there is a pole when the denominator is 0, and that occurs when

x = pi - arccos((1/108000000000)*(h-108000000000)/epsilon)
x = arccos((1/108000000000)*(h+108000000000)/epsilon)

both of which are potentially in the range 0 to 2*pi.

We can insure that there are no real-valued roots by making the arccos() complex, which would be the case if abs() of the interior expression is greater than 1 since arccos() of a value outside -1 to +1 is complex. For any given h, the epsilon that are on the border are +/- ((1/108000000000)*h-1) and +/- (1/108000000000)*h+1 : epsilon larger in absolute value give you problems.

If there is a pole or two, and we have not been given reason to know otherwise, then the integral involves a division by 0 and becomes either infinite or undefined depending on circumstances.

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Tommaso Isolabella 2015-9-4

You are right; I didn't intend to be rude and I apologize if I answered inappropriately-I just didn't check my question to see if I mentioned epsilon (which I tought I did). Anyways thank you for the answer; epsilon is indeed in the allowed interval, so the problem is not there.