Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm using Gauss-Hermite quadrature to integrate

$$ \int_{-\infty}^{\infty} \! e^{-x^2} \cos x\,\mathrm{d}x $$

The exact solution is evidently $\sqrt{\pi\,\text{exp}(1/4)}$, but to be honest I don't even understand what this value is supposed to represent. How is $\cos x$ from ($-\infty,\infty$) a small, finite number? I've written code to apply the weights and abscissas for $2\text{ to } 16$ points, but the numbers I've gotten do not approach the true value and do not even converge on anything as I increase the number of points.

Would appreciate any guidance.

share|cite|improve this question
Please use TeX formatting next time. – nbubis Apr 23 '12 at 13:18

Your integral is supposed to have the exact value $\dfrac{\sqrt \pi}{\sqrt[4]{e}}$; I did my own Gauss-Hermite tests and they do just fine. Here's my Mathematica run:

(* Golub-Welsch algorithm *)
golubWelsch[d_?VectorQ, e_?VectorQ] := 
  MapAt[(First[e] Map[First, #]^2) &, 
    SparseArray[{Band[{1, 1}] -> d, Band[{1, 2}] -> Sqrt[Rest[e]], 
      Band[{2, 1}] -> Sqrt[Rest[e]]}, {Length[d], Length[d]}]], {2}]]

(* generate nodes and weights for Gauss-Hermite quadrature *)
ghq[n_Integer, prec_: MachinePrecision] := 
  Sort[golubWelsch[ConstantArray[0, n], 
    N[Prepend[Range[n - 1]/2, Sqrt[Pi]], prec]]]]

(* number of good digits in successive approximations *)
   Abs[Total[MapThread[#2 Cos[#1] &, ghq[n, 20]]] - Sqrt[Pi/Sqrt[E]]]/
    Sqrt[Pi/Sqrt[E]]], {n, 2, 10}]

{1.622937662555359724, 2.92393460116332523, 4.371438080373419,
 5.92810888858934, 7.571888699825, 9.2881713033, 11.06655943,
 12.8992701, 14.78026}

This says for instance that the ten-point quadrature rule gives fourteen or so accurate digits for your integral; that isn't bad in my book. Check your implementation and report back.

share|cite|improve this answer

The factor $e^{-x^2/2}$ "zeroes out" the $\cos(x)$ term.

share|cite|improve this answer

Another way to look at why this is by recalling that $A \cos x$ has amplitude $|A|$. Here the amplitude is not constant but a decreasing function of $x$: $A = e^{-x^2}$. The graph of $y=e^{-x^2} \cos x$ is bound between $y = \pm e^{-x^2}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.