High Accurate Simple Approximation of Normal Distribution Integral

Table 1

Gaussian function-related integrals.

Integral of Gaussian function	Approximate and exact solution	Relative error figure
(T.1) ∫𝜙(𝑥)𝑑𝑥; 𝐶=0	≈exp−35823𝑥+111arctan37𝑥294+1^−1+𝐶=Φ(𝑥)+𝐶
(T.2) ∫𝑥^2𝜙(𝑥)𝑑𝑥; 𝐶=0	≈exp−35823𝑥+111arctan37𝑥294+1^−1−𝑥𝜙(𝑥)+𝐶=Φ(𝑥)−𝑥𝜙(𝑥)+𝐶
(T.3) ∫𝑥^2𝑘+2𝜙(𝑥)𝑑𝑥; 𝑘=1,𝐶=0	≈−𝜙(𝑥)∑^𝑘𝑗=0(2𝑘+1)!!𝑥(2𝑗+1)!!^2𝑗+1+(2𝑘+1)!!exp−35823𝑥+111arctan37𝑥294+1^−1+𝐶=−𝜙(𝑥)∑^𝑘𝑗=0(2𝑘+1)!!𝑥(2𝑗+1)!!^2𝑗+1+(2𝑘+1)!!Φ(𝑥)+𝐶
(T.4) ∫𝜙(𝑥)^2𝑑𝑥; 𝐶=0	≈12√𝜋−√exp358𝑥2√23+111arctan37𝑥2294+1^−1=1+𝐶2√𝜋Φ(𝑥√2)+𝐶
(T.5) ∫𝜙(𝑥)𝜙(𝑎+𝑏𝑥)𝑑𝑥; 𝑎=1,𝑏=2,𝐶=0	≈1𝑡𝜙𝑎𝑡exp−35823𝑡𝑥+𝑎𝑏𝑡+111arctan37294𝑡𝑥+𝑎𝑏𝑡+1^−1−12+𝐶,𝑡=√1+𝑏^2=1𝑡𝜙𝑎𝑡Φ𝑡𝑥+𝑎𝑏𝑡−12+𝐶,𝑡=√1+𝑏^2
(T.6) ∫𝑥𝜙(𝑎+𝑏𝑥)𝑑𝑥; 𝑎=1,𝑏=2,𝐶=0	1≈−𝑏^2−𝑎𝜙(𝑎+𝑏𝑥)𝑏^2exp−35823(𝑎+𝑏𝑥)+111arctan37294(𝑎+𝑏𝑥)+1^−11+𝐶=−𝑏^2𝑎𝜙(𝑎+𝑏𝑥)−𝑏^2Φ(𝑎+𝑏𝑥)+𝐶
(T.7) ∫𝑥^2𝜙(𝑎+𝑏𝑥)𝑑𝑥; 𝑎=2,𝑏=2,𝐶=0	≈𝑎^2+1𝑏^3exp−35823(𝑎+𝑏𝑥)+111arctan37294(𝑎+𝑏𝑥)+1^−1+𝑎−𝑏𝑥𝑏^3=𝑎𝜙(𝑎+𝑏𝑥)+𝐶2+1𝑏^3Φ(𝑎+𝑏𝑥)+𝑎−𝑏𝑥𝑏^3𝜙(𝑎+𝑏𝑥)+𝐶
(T.8) ∫𝜙(𝑎+𝑏𝑥)^𝑛𝑑𝑥; 𝑎=1,𝑏=2,𝑛=2,𝐶=0	≈(2𝜋)^−(𝑛−1)/2𝑏√𝑛×exp−358(√23𝑛(𝑎+𝑏𝑥))+111arctan37(√294𝑛(𝑎+𝑏𝑥))+1^−1=+𝐶(2𝜋)^−(𝑛−1)/2𝑏√𝑛Φ(√𝑛(𝑎+𝑏𝑥))+𝐶
(T.9) ∫Φ(𝑎+𝑏𝑥)𝑑𝑥; 𝑎=1,𝑏=2,𝐶=0	≈1𝑏(𝑎+𝑏𝑥)exp−35823(𝑎+𝑏𝑥)+111arctan37294(𝑎+𝑏𝑥)+1^−1+1𝑏=1𝜙(𝑎+𝑏𝑥)+𝐶𝑏1(𝑎+𝑏𝑥)Φ(𝑎+𝑏𝑥)+𝑏𝜙(𝑎+𝑏𝑥)+𝐶
(T.10) ∫𝑥Φ(𝑎+𝑏𝑥)𝑑𝑥; 𝑎=1,𝑏=2,𝐶=0	≈12𝑏^2((𝑏^2𝑥^2−𝑎^2×−−1)exp35823(𝑎+𝑏𝑥)+111arctan37294(𝑎+𝑏𝑥)+1^−1=1+(𝑏𝑥−𝑎)𝜙(𝑎+𝑏𝑥))+𝐶2𝑏^2((𝑏^2𝑥^2−𝑎^2−1)Φ(𝑎+𝑏𝑥)+(𝑏𝑥−𝑎)𝜙(𝑎+𝑏𝑥))+𝐶
(T.11) ∫𝑥^2Φ(𝑎+𝑏𝑥)𝑑𝑥; 𝑎=1,𝑏=2,𝐶=0	≈13𝑏^3((𝑏^3𝑥^3+𝑎^3×−+3𝑎)exp35823(𝑎+𝑏𝑥)+111arctan37294(𝑎+𝑏𝑥)+1^−1+(𝑏^2𝑥^2−𝑎𝑏𝑥+𝑎^2=1+2)𝜙(𝑎+𝑏𝑥))+𝐶3𝑏^3((𝑏^3𝑥^3+𝑎^3+3𝑎)Φ(𝑎+𝑏𝑥)+(𝑏^2𝑥^2−𝑎𝑏𝑥+𝑎^2+2)𝜙(𝑎+𝑏𝑥))+𝐶
(T.12) ∫𝑥𝜙(𝑥)Φ(𝑎+𝑏𝑥)𝑑𝑥; 𝑎=1,𝑏=2,𝐶=0	≈𝑏𝑡𝜙𝑎𝑡exp−35823𝑥𝑡+𝑎𝑏𝑡+111arctan37294𝑥𝑡+𝑎𝑏𝑡+1^−1−𝜙(𝑥)exp−35823(𝑎+𝑏𝑥)+111arctan37294(𝑎+𝑏𝑥)+1^−1+𝐶,𝑡=√1+𝑏^2=𝑏𝑡𝜙𝑎𝑡Φ𝑥𝑡+𝑎𝑏𝑡−𝜙(𝑥)Φ(𝑎+𝑏𝑥)+𝐶,𝑡=√1+𝑏^2
(T.13) ∫Φ(𝑥)^2𝑑𝑥; 𝐶=0	≈𝑥exp−35823𝑥+111arctan37𝑥294+1^−2+2𝜙(𝑥)exp−35823𝑥+111arctan37𝑥294+1^−1−1√𝜋exp−358√23(𝑥2)+111arctan37√294(𝑥2)+1^−1+𝐶=𝑥Φ(𝑥)^21+2Φ(𝑥)𝜙(𝑥)−√𝜋Φ(𝑥√2)+𝐶
(T.14) ∫𝑒^𝑐𝑥𝜙(𝑏𝑥)^𝑛𝑑𝑥; 𝑐=3,𝑛=2,𝑏=2,𝐶=0	≈1𝑏𝑛(2𝜋)^𝑛−1𝑐exp^22𝑛𝑏^2×exp−358√23𝑏𝑥𝑐𝑛−𝑏√𝑛+111arctan37√294𝑏𝑥𝑐𝑛−𝑏√𝑛+1^−1=1+𝐶,𝑏≠0,𝑛>0𝑏𝑛(2𝜋)^𝑛−1𝑐exp^22𝑛𝑏^2Φ𝑏𝑥√𝑐𝑛−𝑏√𝑛+𝐶,𝑏≠0,𝑛>0
(T.15) ∫^∞−∞Φ(𝑎+𝑏𝑥)𝜙(𝑥)𝑑𝑥; 𝑎=1,𝑏=2,𝐶=0	≈exp−358𝑎23√1+𝑏^2+111arctan37𝑎294√1+𝑏^2+1^−1𝑎=Φ√1+𝑏^2	−0.3691843425𝐸−4
(T.16) ∫^∞0𝑥Φ(𝑎+𝑏𝑥)𝜙(𝑥)𝑑𝑥; 𝑎=1,𝑏=2,𝐶=0	≈𝑏𝑡𝜙𝑎𝑡exp358𝑎𝑏−23𝑡+111arctan37𝑎𝑏294𝑡+1^−1+(2𝜋)^−1/2exp−358𝑎23+111arctan37𝑎294+1^−1,𝑡=√1+𝑏^2=𝑏𝑡𝜙𝑎𝑡Φ−𝑎𝑏𝑡+(2𝜋)^−1/2Φ(𝑎),𝑡=√1+𝑏^2	−0.1961451412𝐸−4