0
|
14341 Views
|
2 Replies
|
1 Total Likes
View groups...
Share
GROUPS:

# Integral question (new to Mathematica)

Posted 10 years ago
 Hello,I've just begun using WA recently and am very fond of it. At the moment I am working on solving an integral and I want to know if WA can help me calculate it - I'm a bit stuck and can't seem to figure out why things aren't working... Many thanks in advance for your input!Integrate[E^(-2 I \[Pi] t (-(1/6)       Cos[\[Theta]]^2 + (-(5/12) -        1/8 (1 + rinner^2/r^2) Cos[2 \[Phi]] +        3/4 Log[routter/r]) Sin[\[Theta]]^2 +     1/2 (-(1/3) + Cos[\[Theta]]^2 - (       rinner^2 Cos[2 \[Phi]] Sin[\[Theta]]^2)/r^2))), {\[Phi], 0,   2 \[Pi]}]As you can see, I'd like to take the integral of a complex function. In this step I am taking the integral with respect to phi. There is another integral outside of this with respect to r. There are some variables here such as theta, rinner, routter, that are not integrated over. Right now when I run this, it just spits out the input. And I can't figure out why...
2 Replies
Sort By:
Posted 10 years ago
 As Bill Simpson said, Integrate returns unevaluated if Mathematica couldn't work out what the value of the integral is. However, you can use NIntegrate to get a numerical approximation (if that's helpful). In that case, you will need to set values for all the variables which are not integrated over (since it's impossible to do a numerical integration when there are symbolic quantities in there).
Posted 10 years ago
 When Mathematica or WolframAlpha cannot find a solution they often display the original problem and assume you understand what this means.If I study your problem and assume that the variables do not have special values which make the problem much simpler and I try to make a simpler problem that is similar to your problem to see if Mathematica or WolframAlpha is able to solve thisIntegrate[E^(a*Cos[2*phi]+b),phi]then there appears to be no solution in ordinary functions. Thus I believe there is no solution in ordinary functions for your problem.