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I am looking for book recommendations or hints on numerical integration over infinite intervals. I am particularly interested in integrals of the form

$\int\limits_{-\infty}^{+\infty} g(x) \exp(p_d(x)) \mathrm{d} x$,

where $g(x)$ is an arbitrary continuous function (but not necessarily continuously differentiable) and $p_d(x)$ is some polynomial of even degree $d > 2$ with negative leading coefficient.

Moreover, I know the first few weights and abscissas of the corresponding Gaussian quadrature but have no rule to compute more for higher accuracy as there is no known family of orthogonal polynomials with respect to $\exp(p_d(x)), \;d>2$, that I can make use of.

I'd be grateful for any hints or literature recommendations because I haven't found a nice summary of suitable methods yet. Thank you in advance.

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    $\begingroup$ It won't be pleasant, but can you not define your own families of orthogonal polynomials based on $exp(-x^4)$, $exp(-x^6)$, and so on? Alternatively, and this is extremely half-baked, do a change-of-variables $y=x^d$ to switch to the half-line and do Gauss-Laguerre? $\endgroup$
    – JCK
    Commented Nov 18, 2020 at 21:44
  • $\begingroup$ @JCK Thank you for your ideas. I thought about that, too. But the exponent is a general polynomial including terms of odd degree. So the integrand would include a product of weight functions. Your second suggestion doesn't work for a general polynomial because it is not clear how express $g(x)$ in terms of $t$, is it? $\endgroup$
    – BernieD
    Commented Nov 19, 2020 at 14:00
  • $\begingroup$ I was thinking you'd write $f(x) * exp(-a_0x^d + a_1 x^{d+1} + ...)$ as $exp(-a_0 x^d) * f(x)*exp(whatever)$, then do a change-of-variables to get $exp(-y^d) g(y)$ and use the $d-$specific quadrature rule. $\endgroup$
    – JCK
    Commented Nov 19, 2020 at 16:12
  • $\begingroup$ OK, I could try that. Then the quality of the approximation should primarily depend on how well (in terms of the integral) the last term $\exp(\text{whatever})$ is approximated by the truncated power series, right? $\endgroup$
    – BernieD
    Commented Nov 19, 2020 at 18:25
  • $\begingroup$ Or in other words, if the power series is sufficiently accurate (whatever that means) if truncated after let's say $N$ terms the required number of quadrature nodes would be $(N - N \text{mod} 2)/2 + 1$. $\endgroup$
    – BernieD
    Commented Nov 19, 2020 at 18:35

1 Answer 1

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I would convert the integration range to a finite interval, $$\int_{-\infty}^\infty f(x)dx=\int_0^1\left[f(1/t-1)+f(-1/t+1)\right]t^{-2}dt,$$ and then use an adaptive Gauss-Kronrod routine. Many computational libraries have code for that, for example, Matlab or Mathematica.

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  • $\begingroup$ Thank you. I am going to try that. But somehow I will have to quantify the degree of accuracy in the original domain, which should be possible somehow, I suppose. $\endgroup$
    – BernieD
    Commented Nov 19, 2020 at 12:54
  • $\begingroup$ I'm not quite sure what you mean by "accuracy in the original domain", the transformation to the finite domain is exact, it does not introduce an error; for a pointer to the error estimates of the Gauss-Kronrod integration, you might take a look at the references in nag.com/numeric/fl/nagdoc_latest/html/d01/d01amf.html $\endgroup$
    – Carlo Beenakker
    Commented Nov 19, 2020 at 13:12
  • $\begingroup$ Maybe I wasn't clear about what I meant: A Gaussian quadrature for example has an accuracy of $2N - 1$ which means that an $N$-node quadrature of the $\int g(x)w(x) \mathrm{d} x$ is exact if $g(x)$ is a polynomial of degree up to $2N -1$. If I recall correctly the idea of a Gauss-Kronrod quadrature consists in adding quadrature nodes to raise the degree to $3N-1$ to get an error estimate for the integral. As I am looking at different forms of $g(x)$ I was wondering how to relate the accuracy in terms of $g(t(x))$ after the variable transformation $x \rightarrow t$ to that of the original form. $\endgroup$
    – BernieD
    Commented Nov 19, 2020 at 14:17
  • $\begingroup$ I am going to have a look at your linked website first. Perhaps that will answer my question already. $\endgroup$
    – BernieD
    Commented Nov 19, 2020 at 14:20

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