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Re: Simple Trigonometric Integrals

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  • Subject: [mg32358] Re: Simple Trigonometric Integrals
  • From: "Allan Hayes" <hay at>
  • Date: Wed, 16 Jan 2002 03:30:13 -0500 (EST)
  • References: <a20n0a$4i4$>
  • Sender: owner-wri-mathgroup at

This often hits people working with Fourier series.
One solution is to integrate the summands separately and add the answers

Joe=a c Cos[t]/(g s)+b q Cos[2 t]/(c f)+c Cos[3 t]/(d a)+d f Cos[4 t]/(h a
    e q Cos[5t]/(g a)+f l Cos[6 t]/(w r m)+g b Cos[7 t]/(o n x)+
    h Sin[t]/(b c)+i Sin[2 t]/(h e r)+j y Sin[3 t]/(l p)+d k Sin[4 t]/(j c)+
    l m a Sin[5 t]/(f s b h)
+m p Sin[6 t]/(k j)+q n Sin[7 t]/(x c);


{3.29 Second,0}

(this was on a 233 MHz laptop)

> ... does Mathematica have a built in Fourier Series
> expansion?

Yes, in the Package  Calculus`FourierTransform`


Allan Hayes
Mathematica Training and Consulting
Leicester UK
hay at
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"Joe Helfand" <jhelfand at> wrote in message
news:a20n0a$4i4$1 at
> Hey,
>     I have a thing about Mathematica.  Sometimes I have a real long
> expression that involves the integral of the sum of lots of cosines and
> sines of some variable let's say 't'.  But having done some fancy maths
> on my own to reduce it and get into a simple integral from 0 to 2 Pi,
> and the sines and cosines all involve some integer multiple of t, the
> integration takes for ever, it basically hangs.  Now, although the
> expression is long, and there are a lot of terms in it, it still just
> becomes a simple periodic integral from zero to 2 pi, and all the
> trigonometric terms involving t should just drop out.  Kind of like what
> sometimes can happen if you are playing around with a Fourier series
> expansion (by the way, does Mathematica have a built in Fourier Series
> expansion?  I mean something like Series[], but returns fourier
> coefficients?).  Uptill now, I have been able to get by with something
> like using
> periodicIntegral={Cos[t] -> 0, Cos[2 t] -> 0, Cos[3 t] -> 0, Cos[4 t] ->
> 0, Cos[5t] -> 0, Cos[6 t] -> 0, Cos[7 t] -> 0, Sin[t] -> 0, Sin[2 t] ->
> 0, Sin[3 t] -> 0, Sin[4 t] -> 0, Sin[5t] -> 0, Sin[6 t] -> 0, Sin[7 t]
> -> 0};
> and then doing a replace on the expresion, multiplying the result by 2
> Pi.  But now I am in a bind where no amount of TrigReduce, TrigExpand,
> TrigFactor, etc. will get this big ass expression into the desired form
> where the above is approriate (because there are other sines and cosines
> of other variables that get put into the terms and stand by
> themselves).  Still, the expression should be easy to do for the
> computer, even I can go through and set these terms to zero, but it will
> just take me a long time.  An example of what I am talking about, just
> try the following:
> In[687]:=
> Joe = a c Cos[t]/(g s) + b q Cos[2 t]/(c f) + c Cos[3 t]/(d a) +
>       d f Cos[4 t]/(h a n) + e q Cos[5t]/(g a) + f l Cos[6 t]/(w r m) +
>       g b Cos[7 t]/(o n x) + h Sin[t]/(b c) + i Sin[2 t]/(h e r) +
>       j y Sin[3 t]/(l p) + d k Sin[4 t]/(j c) + l m a Sin[5 t]/(f s b h)
> +
>       m p Sin[6 t]/(k j) + q n Sin[7 t]/(x c);
> In[688]:=
> Integrate[Joe, {t, 0, 2 Pi}]
> and you willl see it takes a long time to integrate.  (It will
> eventually get done.)  I know this is just zero, but why does it take so
> long for the computer to figure out?  It is true that my expression is
> even longer than this one, so essentially it hangs, but basically it is
> the same problem.  I do not what to be hunting through my equation from
> hell setting all the relevant trigonometric terms to zero when the
> computer should be able to do this.  Well, sorry for the harangue but I
> greatly appreciate you reading down so far, really.  If you have any
> suggestions or comments, point out I am an idiot there is some simple
> thing in Mathematica, please send it.
> Joe

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