How to solve this integral in Mathematica

Is it possible to solve it numerically?

$$\begin{align} \int_{-1}^1 \cos\big(&(.75-.75x^2)\cdot y_1 + (.25 - .5x - .75x^2) \cdot y_2 \\ + &(-.5x + .5x^3) \cdot y_3 + (.125 - .75 x^2 + 0.625 x^4) \cdot y_4 \big) dx \end{align}$$

  • 4
    $\begingroup$ Please put some effort into making the question readable. $\endgroup$ – Mikael Öhman Dec 13 '11 at 14:01

The short answer is No.
You can't have undefined symbols/variables in numerical integrals.

If you want a purely numerical solution, you could make a 4-dimensional grid that covers the range of $y_i$ that you're interested in, do the integrals at those points and interpolate between the results.
Or, more simply, just define a numerical integral function that you can call when ever you need the result (it's a simple integral, so it is quite fast).

nInt[y1_?NumericQ, y2_?NumericQ, y3_?NumericQ, y4_?NumericQ] := 
 NIntegrate[Cos[(.75 - .75 x^2) y1 + (.25 - .5 x - .75 x^2) y2
    + (-.5 x + .5 x^3) y3 + (.125 - .75 x^2 + 0.625 x^4) y4], 
   {x, -1, 1}]

You can then plot any section of the 4D function that you want. For example and no particular reason,

Plot[nInt[x, 2 x, 0, 1], {x, -1, 1}]


More interesting (to me) is how far you can get with this integral analytically. If you didn't have any $x^3$ or $x^4$ terms, then integral can be done completely in terms of Fresnel C and S functions (W|A).

Sqrt[2c/Pi] Integrate[Cos[a + b x + c x^2], {x, -1, 1}]//TraditionalForm


In general, it is a little more tricky. You could try a series expansion:

sc = Simplify[SeriesCoefficient[
  Cos[a + b x + c x^2 + d x^3 + e x^4], {x, 0, n}], n >= 0]

(* Integrate term by term. Sum can't be done by Mma *)
(*Sum[(1+(-1)^n)/(1+n) sc, {n, 0, Infinity}]*) 

(* but can expand up to any order *)
Table[(1 + (-1)^n)/(1 + n) sc, {n, 0, 10}] // ExpToTrig // 
  Simplify[#, Trig -> False] &

The SeriesCoefficient returns a DifferenceRoot expression - i.e., it can't be given in "closed form". Mathematica is then unable to compute the commented out Sum. However, you can get the expansion of the result to any order, the Table command returns

{2 Cos[a], 0, 1/3 (-b^2 Cos[a] - 2 c Sin[a]), 0, ...

If you think about a being dimensionless, b having dimension one, ..., e having dimension 4, then each term is of a fixed dimension. Note that odd terms in the series vanish (because of the symmetry of the integral).

This result can then be specialized to your particular values, e.g., a = .75 y1 + .25 y2 + .125 y4, etc...

  • 2
    $\begingroup$ I must try Simon. Really, answer is professional, better 100 times then my question. $\endgroup$ – George Dec 13 '11 at 22:59

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