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Stephen Dando-Collins

In the expressions in this article,

$\varphi (x)={\frac {1}{\sqrt {2\pi }}}e^{-{\frac {1}{2}}x^{2}}$

is the standard normal probability density function,

$\Phi (x)=\int _{-\infty }^{x}\varphi (t)\,dt={\frac {1}{2}}\left[1+\operatorname {erf} \left({\frac {x}{\sqrt {2}}}\right)\right]$

is the corresponding cumulative distribution function (where erf is the error function), and

$T(h,a)=\varphi (h)\int _{0}^{a}{\frac {\varphi (hx)}{1+x^{2}}}\,dx$

is Owen's T function.

Owen^[1] has an extensive list of Gaussian-type integrals; only a subset is given below.

Indefinite integrals

$\int \varphi (x)\,dx=\Phi (x)+C$
$\int x\varphi (x)\,dx=-\varphi (x)+C$
$\int x^{2}\varphi (x)\,dx=\Phi (x)-x\varphi (x)+C$
$\int x^{2k+1}\varphi (x)\,dx=-\varphi (x)\sum _{j=0}^{k}{\frac {(2k)!!}{(2j)!!}}x^{2j}+C$ ^[2]
$\int x^{2k+2}\varphi (x)\,dx=-\varphi (x)\sum _{j=0}^{k}{\frac {(2k+1)!!}{(2j+1)!!}}x^{2j+1}+(2k+1)!!\,\Phi (x)+C$

In the previous two integrals, $n!!$ is the double factorial: for even $n$ it is equal to the product of all even numbers from 2 to $n$ , and for odd $n$ it is the product of all odd numbers from 1 to $n$ ; additionally it is assumed that $0!! = (-1)!! = 1$ .

$\int \varphi (x)^{2}\,dx={\frac {1}{2{\sqrt {\pi }}}}\Phi {\left(x{\sqrt {2}}\right)}+C$
$\int \varphi (x)\varphi (a+bx)\,dx={\frac {1}{t}}\varphi {\left({\frac {a}{t}}\right)}\Phi {\left(tx+{\frac {ab}{t}}\right)}+C,\qquad t={\sqrt {1+b^{2}}}$ ^[3]
$\int x\varphi (a+bx)\,dx=-{\frac {1}{b^{2}}}\left[\varphi (a+bx)+a\Phi (a+bx)\right]+C$
$\int x^{2}\varphi (a+bx)\,dx={\frac {1}{b^{3}}}\left[\left(a^{2}+1\right)\Phi (a+bx)+\left(a-bx\right)\varphi (a+bx)\right]+C$
$\int \varphi (a+bx)^{n}\,dx={\frac {1}{b{\sqrt {n{\left(2\pi \right)}^{n-1}}}}}\Phi {\left({\sqrt {n}}(a+bx)\right)}+C$
$\int \Phi (a+bx)\,dx={\frac {1}{b}}\left[\left(a+bx\right)\Phi (a+bx)+\varphi (a+bx)\right]+C$
$\int x\Phi (a+bx)\,dx={\frac {1}{2b^{2}}}\left[\left(b^{2}x^{2}-a^{2}-1\right)\Phi (a+bx)+\left(bx-a\right)\varphi (a+bx)\right]+C$
$\int x^{2}\Phi (a+bx)\,dx={\frac {1}{3b^{3}}}\left[\left(b^{3}x^{3}+a^{3}+3a\right)\Phi (a+bx)+\left(b^{2}x^{2}-abx+a^{2}+2\right)\varphi (a+bx)\right]+C$
$\int x^{n}\Phi (x)\,dx={\frac {1}{n+1}}\left[\left(x^{n+1}-nx^{n-1}\right)\Phi (x)+x^{n}\varphi (x)+n(n-1)\int x^{n-2}\Phi (x)\,dx\right]+C$
$\int x\varphi (x)\Phi (a+bx)\,dx={\frac {b}{t}}\varphi {\left({\frac {a}{t}}\right)}\Phi {\left(xt+{\frac {ab}{t}}\right)}-\varphi (x)\Phi (a+bx)+C,\qquad t={\sqrt {1+b^{2}}}$
$\int \Phi (x)^{2}\,dx=x\Phi (x)^{2}+2\Phi (x)\varphi (x)-{\frac {1}{\sqrt {\pi }}}\Phi {\left(x{\sqrt {2}}\right)}+C$
$\int e^{cx}\varphi (bx)^{n}\,dx={\frac {e^{\frac {c^{2}}{2nb^{2}}}}{b{\sqrt {n{\left(2\pi \right)}^{n-1}}}}}\Phi {\left({\frac {b^{2}xn-c}{b{\sqrt {n}}}}\right)}+C,\qquad b\neq 0,n>0$

Definite integrals

$\int _{-\infty }^{\infty }x^{2}\varphi (x)^{n}\,dx={\frac {1}{\sqrt {n^{3}{\left(2\pi \right)}^{n-1}}}}$
$\int _{-\infty }^{\infty }\varphi (x)\varphi (a+bx)\,dx={\frac {1}{\sqrt {1+b^{2}}}}\varphi {\left({\frac {a}{\sqrt {1+b^{2}}}}\right)}$
$\int _{-\infty }^{0}\varphi (ax)\Phi (bx)\,dx={\frac {1}{2\pi |a|}}\left({\frac {\pi }{2}}-\arctan \left({\frac {b}{|a|}}\right)\right)$
$\int _{0}^{\infty }\varphi (ax)\Phi (bx)\,dx={\frac {1}{2\pi |a|}}\left({\frac {\pi }{2}}+\arctan \left({\frac {b}{|a|}}\right)\right)$
$\int _{0}^{\infty }x\varphi (x)\Phi (bx)\,dx={\frac {1}{2{\sqrt {2\pi }}}}\left(1+{\frac {b}{\sqrt {1+b^{2}}}}\right)$
$\int _{0}^{\infty }x^{2}\varphi (x)\Phi (bx)\,dx={\frac {1}{4}}+{\frac {1}{2\pi }}\left({\frac {b}{1+b^{2}}}+\arctan(b)\right)$
$\int _{-\infty }^{\infty }x\varphi (x)^{2}\Phi (x)\,dx={\frac {1}{4\pi {\sqrt {3}}}}$
$\int _{0}^{\infty }\Phi (bx)^{2}\varphi (x)\,dx={\frac {1}{2\pi }}\left(\arctan(b)+\arctan {\sqrt {1+2b^{2}}}\right)$
$\int _{-\infty }^{\infty }\Phi (a+bx)^{2}\varphi (x)\,dx=\Phi {\left({\frac {a}{\sqrt {1+b^{2}}}}\right)}-2T{\left({\frac {a}{\sqrt {1+b^{2}}}},{\frac {1}{\sqrt {1+2b^{2}}}}\right)}$
$\int _{-\infty }^{\infty }x\Phi (a+bx)^{2}\varphi (x)\,dx={\frac {2b}{\sqrt {1+b^{2}}}}\varphi {\left({\frac {a}{t}}\right)}\Phi {\left({\frac {a}{{\sqrt {1+b^{2}}}{\sqrt {1+2b^{2}}}}}\right)}$ ^[4]
$\int _{-\infty }^{\infty }\Phi (bx)^{2}\varphi (x)\,dx={\frac {1}{\pi }}\arctan {\sqrt {1+2b^{2}}}$
$\int _{-\infty }^{\infty }x\varphi (x)\Phi (bx)\,dx=\int _{-\infty }^{\infty }x\varphi (x)\Phi (bx)^{2}\,dx={\frac {b}{\sqrt {2\pi (1+b^{2})}}}$
$\int _{-\infty }^{\infty }\Phi (a+bx)\varphi (x)\,dx=\Phi {\left({\frac {a}{\sqrt {1+b^{2}}}}\right)}$
$\int _{-\infty }^{\infty }x\Phi (a+bx)\varphi (x)\,dx={\frac {b}{\sqrt {1+b^{2}}}}\varphi {\left({\frac {a}{\sqrt {1+b^{2}}}}\right)},$
$\int _{0}^{\infty }x\Phi (a+bx)\varphi (x)\,dx={\frac {b}{t}}\varphi {\left({\frac {a}{t}}\right)}\Phi {\left(-{\frac {ab}{t}}\right)}+{\frac {1}{\sqrt {2\pi }}}\Phi (a),\qquad t={\sqrt {1+b^{2}}}$
$\int _{-\infty }^{\infty }\ln(x^{2}){\frac {1}{\sigma }}\varphi {\left({\frac {x}{\sigma }}\right)}\,dx=\ln(\sigma ^{2})-\gamma -\ln 2\approx \ln(\sigma ^{2})-1.27036$

References

^ Owen 1980.
^ Patel & Read (1996) lists this integral without the minus sign, which is an error. See calculation by WolframAlpha.
^ Patel & Read (1996) report this integral with error, see WolframAlpha.
^ Patel & Read (1996) report this integral incorrectly by omitting x from the integrand.

Owen, D. (1980). "A table of normal integrals". Communications in Statistics: Simulation and Computation. B9 (4): 389–419. doi:10.1080/03610918008812164.
Patel, Jagdish K.; Read, Campbell B. (1996). Handbook of the normal distribution (2nd ed.). CRC Press. ISBN 0-8247-9342-0.

Lists of integrals
Rational functions Irrational functions Trigonometric functions Inverse trigonometric functions Hyperbolic functions Inverse hyperbolic functions Exponential functions Logarithmic functions Gaussian functions Definite integrals

Credit: Wikipedia https://en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions

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