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Moment recurrence for the scaled complementary error function

Statement

For $x > 0$ and integer $k \geq 0$ , the moments in Scaled complementary error function and its moments are finite and positive. They satisfy

$2 x M_{0} + 2 M_{1} = 1, k M_{k - 1} = 2 x M_{k} + 2 M_{k + 1} (k \geq 1) .$

With $r_{k} = M_{k + 1} / M_{k}$ , it follows that

$r_{k} = \frac{k + 1}{2 (x + r_{k + 1})}, 0 < r_{k} < \frac{k + 1}{2 x}, erfcx (x) = \frac{1}{\sqrt{π} (x + r_{0})} .$

Facts & Assumptions

Given: Real $x > 0$ , integer $k \geq 0$ , and the moment and scaled-error definitions in Scaled complementary error function and its moments.

[F1]

Scaled complementary error function and its moments: $M_{k} (x) = \int_{0}^{\infty} t^{k} e^{- t^{2} - 2 x t} d t$ and $erfcx (x) = e^{x^{2}} erfc (x)$ .

[F2]

Fresnel, error, and Gauss hypergeometric series: $erfc (x) = 1 - \erf (x)$ .

[F3]

Error-function series geometric tail bound: for real $x$ , $\erf (x) = 2 π^{- 1 / 2} \int_{0}^{x} e^{- u^{2}} d u$ .

[F4]

The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ : $\int_{- \infty}^{\infty} e^{- u^{2}} d u = \sqrt{π}$ .

[F5]

If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ : on compact intervals, if its hypotheses hold, $\int u v^{'} = u v ∣_{\partial} - \int u^{'} v$ . In particular the derivatives used below are integrable on each compact interval because they are continuous.

[F6]

Change of variable in an improper integral: an admissible increasing $C^{1}$ substitution preserves convergence and value of an improper integral.

[F7]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion: each continuous real derivative on a compact truncation interval is Riemann integrable; apply to real and imaginary parts for the complex-valued integration-by-parts formulas.

[F8]

Euler's Gamma function on the right half-plane and Euler's Gamma integral converges locally uniformly on the right half-plane: for $a > 0$ , $\int_{0}^{\infty} s^{a - 1} e^{- s} d s = Γ (a) < \infty$ .

[L9]

The exponential function is smooth and $(\exp)^{'} = \exp$ , Sum, product, quotient, and chain rules from difference quotients: the displayed real functions are smooth on their stated domains, and their derivatives follow by the product, power and chain rules; iterate these rules for the finite higher derivatives used below.

Proof

1.1givenF1F6F7F8algebra

The integrands in [F1] are positive. Near $0$ they are bounded by $t^{k}$ . For $t \geq 1$ , $e^{- 2 x t} \leq 1$ ; the substitution $s = t^{2}$ and [F6] bound the tail by $\frac{1}{2} \int_{1}^{\infty} s^{(k - 1) / 2} e^{- s} d s$ , which is finite by the Gamma integral convergence in [F8]. The integrands are continuous on every compact truncation interval, so their proper integrals there exist by [F7]. Thus each $M_{k}$ is finite; it is strictly positive because its integrand is positive on $[1, 2]$ .

2.1F1F5F7L9step 1.1algebra

Apply [F5] on $[ε, R]$ to $w (t) = e^{- t^{2} - 2 x t}$ and to $t^{k} w (t)$ . These functions and their derivatives are continuous on the compact interval, so all derivatives required by [F5] are integrable by [F7]; the complex-valued integration-by-parts statement is applied componentwise. The boundary terms tend to $0$ at $R \to \infty$ ; for $k \geq 1$ they also tend to $0$ at $ε ↓ 0$ , while for $k = 0$ the lower boundary tends to $1$ . For the infinite endpoint, choose an integer $m > k / 2$ ; the exponential series gives $e^{t^{2} / 2} \geq (t^{2} / 2)^{m} / m!$ , so $t^{k} e^{- t^{2} - 2 x t} \leq 2^{m} m! t^{k - 2 m} e^{- t^{2} / 2} \to 0$ . Passing to the limits in the integration-by-parts identities gives $2 x M_{0} + 2 M_{1} = 1$ and, for $k \geq 1$ , $k M_{k - 1} = 2 x M_{k} + 2 M_{k + 1}$ .

3.1step 2.1algebra

Divide the recurrence with index $k + 1$ by $2 M_{k} > 0$ . Since $r_{k} = M_{k + 1} / M_{k}$ and $r_{k + 1} = M_{k + 2} / M_{k + 1}$ , this gives $(k + 1) / 2 = x r_{k} + r_{k} r_{k + 1}$ and hence $r_{k} = (k + 1) / (2 (x + r_{k + 1}))$ . Positivity gives $0 < r_{k} < (k + 1) / (2 x)$ .

4.1F1F2F3F4F6step 2.1algebra∎

The integrand in [F4] is even, so $\int_{0}^{\infty} e^{- u^{2}} d u = \sqrt{π} / 2$ . Subtract the integral identity in [F3] and use the definition in [F2] to get $erfc (x) = \frac{2}{\sqrt{π}} \int_{x}^{\infty} e^{- u^{2}} d u .$ The increasing substitution $u = t + x$ from [F6] then gives $erfcx (x) = \frac{2}{\sqrt{π}} \int_{0}^{\infty} e^{- t^{2} - 2 x t} d t = \frac{2 M_{0}}{\sqrt{π}} .$ The first identity in step 2.1 gives $2 M_{0} (x + r_{0}) = 1$ ; substitution proves $erfcx (x) = 1 / (\sqrt{π} (x + r_{0}))$ .

Depends on

Used by

Cited to discharge well-definedness by Scaled complementary error function and its moments.

Dependency tree · two levels

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Sources

DLMF, §7.2, definitions and elementary properties (standard reference, not scraped)
DLMF, §§7.2 and 7.12(i) (standard reference, not scraped)
N. M. Temme, “Numerical aspects of special functions,” Acta Numerica 17 (2008), §§3 and 5 (standard reference, not scraped)