Opening the book…
Gaussian Distribution
The Gaussian or Normal distribution function in one dimension is
\[P(x)=\frac{1}{\sqrt{2 \pi \sigma^{2}}} \, \exp \frac{\left(x-x_{0}\right)^{2}}{2 \sigma^{2}}\]
where \(x_{0}\) is the mean value of the distribution, \(\sigma\) is the standard deviation, and \(\sigma^{2}=\langle x^2 \rangle -\langle x\rangle ^2 \) is the variance of the distribution.
Gaussian integrals
The general strategy with solving Gaussian definite integrals is to complete the square in exponential argument to recast it as a simpler integral. To illustrate, let's integrate the product of a Gaussian and a complex exponential of the form
\[ I=\int_{-∞}^{+∞} dt \, e^{-\alpha t^2} e^{\mathrm{i}\omega t-\gamma t} \]
where \(\alpha\) is a real positive number. Assigning \(\beta=\mathrm{i}\omega-\gamma\), we can add \(-\beta^2/4\alpha\) to the exponential argument, we can write it as
\[ I=e^{\beta^2/4\alpha} \int_{-∞}^{+∞} dt \, \exp\left(-\alpha\left(t-\frac{\beta}{2\alpha}\right)^2\right) \]
The simpler Gaussian integral here is \(\sqrt{\pi∕\alpha}\), so
\[ I=\sqrt{\frac{\pi}{\alpha}} e^{\beta^2/4\alpha} \]
Table of Gaussian integrals
\[\begin{align} & \int_{-\infty}^{+\infty} dt\, \mathrm{e}^{-\alpha t^{2}}=\sqrt{\frac{\pi}{\alpha}} \\
& \int_{0}^{+\infty} dt\, \mathrm{e}^{-\alpha t^{2}}=\frac{1}{2} \sqrt{\frac{\pi}{\alpha}} \\
& \int_{0}^{\infty} dt\, t\, \mathrm{e}^{-\alpha t^{2}}=\frac{1}{2 \alpha} \\
& \int_{-\infty}^{+\infty} dt\, t^{2} \mathrm{e}^{-\alpha t^{2}}=\sqrt{\frac{\pi}{4 \alpha^{3}}} d \int_{-\infty} dt\, t^{2} \mathrm{e}^{-\alpha t^{2}}=-\frac{d}{d \alpha} \int_{-\infty}^{+\infty} dt\, \mathrm{e}^{-\alpha t^{2}} \\
& \int_{-\infty}^{+\infty} d y \int_{-\infty}^{+\infty} d x \mathrm{e}^{-\left(a x^{2}+2 b x y+c y^{2}\right)}=\frac{\pi}{\sqrt{a c}-b^{2}} \\
& \int_{0}^{\infty} dt\, t^{2 n} \mathrm{e}^{-\alpha t^{2}}=\frac{1 \cdot 3 \cdot \ldots(2 n-1)}{2^{n+1} a^{n}} \sqrt{\frac{\pi}{\alpha}} \quad & (n=0,1,2 \ldots)\\
& \int_{0}^{+\infty} dt\, t^{2 n+1} \mathrm{e}^{-\alpha t^{2}}=\frac{n !}{2 a^{n+1}} \quad & (n=0,1,2 \ldots) \\
& \int_{-\infty}^{\infty} dt\, t^{2 n+1} \mathrm{e}^{-\alpha t^{2}}=0 \quad & (n=0,1,2 \ldots) \\
& \int_{-\infty}^{+\infty} \exp^{-\left(a t^{2}+b t+c\right)}=\sqrt{\frac{\pi}{a}} \mathrm{exp}\left[\frac{b^{2}}{4 a}-c\right] \\
& \int_{-\infty}^{+\infty} dt\, \exp \left(-\alpha\left(t-\frac{\mathrm{i} x}{2 \alpha}\right)^{2}\right)=\int_{-\infty}^{+\infty} d t \mathrm{e}^{-\alpha t^{2}+\mathrm{i} x t+x^{2} / 4 \alpha}=\sqrt{\frac{\pi}{\alpha}} \\
& \int_{-\infty}^{+\infty} \rm{exp} \left( { - \beta x^2} \right)\cos \left( {bx} \right)\,dx = \sqrt{\frac{\pi}{\beta}} \exp \left( - \frac{b^2}4\beta \right) \\
& \int_{ - \infty }^{ + \infty } dy\, \int_{ - \infty }^{ + \infty } dx\, {\rm{e}}^{ - \left( a{x^2} + 2bxy + c{y^2} \right)} = \frac{\pi}{\sqrt{ac-b^2}} \end{align}\]
Error Function
The error function, \(\operatorname{erf}(z)\), is a complex sigmoidal step function that appears in integrals over Gaussian distributions and Gaussian convolutions. If is defined through a definite integral over a Gaussian function as
\[\operatorname{erf}(z)=\frac{2}{\sqrt{\pi}} \int_{0}^{z} e^{-t^{2}} d t\]
The complementary error function, \(\operatorname{erfc}(z)\), is
\[\operatorname{erfc}(z) \equiv \frac{2}{\sqrt{\pi}} \int_{z}^{\infty} e^{-t^{2}} d t\]
and \(\operatorname{erfc}(x)=1-\operatorname{erf}(x)\).
The imaginary error function is defined as
\[\operatorname{erfi}(z) \equiv-\mathrm{i} \operatorname{erf}(\mathrm{i} z)=\frac{2}{\sqrt{\pi}} \int_{0}^{z} e^{t^{2}} d t\]
Error Function Identities
\[\begin{align} & \operatorname{erf}(-z)=-\operatorname{erf}(z) \\ & \frac{2}{\sqrt{\pi}} \int_{a}^{b} dz\, \mathrm{e}^{-\alpha z^{2}}=\operatorname{erf}(b)-\operatorname{erf}(a) \\ & \operatorname{erf}\left(z^{*}\right)=(\operatorname{erf}(z))^{*} \\ & \frac{d}{d x} \operatorname{erf}(z)=\sqrt{\frac{4}{\pi}} \mathrm{e}^{-z^{2}} \\ & \operatorname{erf}(z)=\sqrt{\frac{4}{\pi}} \sum_{n=0}^{\infty} \frac{(-1)^{n} z^{2 n+1}}{n !(2 n+1)} \end{align}\]
Gaussian integrals involving the error function
\[\begin{align} & \int e^{-a^{2} z^{2}+b z}\, dz=\frac{\sqrt{\pi}}{2 a} \exp \left(\frac{b^{2}}{4 a^{2}}\right) \operatorname{erf}\left(a z-\frac{b}{2 a}\right) \\ & \int_{0}^{\infty} e^{-a^{2} x^{2}+b x}\, dx=\frac{\sqrt{\pi}}{2 a} \exp \left(\frac{b^{2}}{4 a^{2}}\right)\left[1+\operatorname{erf}\left(\frac{b}{2 a}\right)\right] \end{align}\]
\[\begin{align} & \int dt\, \mathrm{e}^{-\left(a t^{2}+2 b t+c\right)}=\frac{1}{2} \sqrt{\frac{\pi}{a}} \exp \left(\frac{b^{2}-a c}{a}\right) \operatorname{erf}\left(\sqrt{a} t+\frac{b}{\sqrt{a}}\right) \\ & \int_{0}^{\infty} dt\, \mathrm{e}^{-\left(a t^{2}+2 b t+c\right)}=\frac{1}{2} \sqrt{\frac{\pi}{a}} \exp \left(\frac{b^{2}-a c}{a}\right) \operatorname{erfc}\left(\frac{b}{\sqrt{a}}\right) \\ & \int_{-\infty}^{\infty} dt\, \mathrm{e}^{-\left(a t^{2}+2 b t\right)}=\sqrt{\frac{\pi}{a}} \mathrm{e}^{\frac{-b^{2}}{a}} \\ & \int_{0}^{\infty} dt\, \mathrm{e}^{-\left(a t^{2}+2 b t\right)}=\frac{1}{2} \sqrt{\frac{\pi}{a}} \mathrm{e}^{b^{2} / a} \operatorname{erfc}\left(\frac{b}{\sqrt{a}}\right) \\ & \int_{0}^{x} dt\, \mathrm{e}^{-\alpha t^{2}}=\sqrt{\frac{\pi}{4 \alpha}} \operatorname{erf}\left(\frac{x}{\sqrt{\alpha}}\right) \\ & \int_{0}^{x} dt\, t^{2} \mathrm{e}^{-\alpha t^{2}}=\frac{2}{\sqrt{\pi}} x \mathrm{e}^{-\alpha x^{2}}+\sqrt{\frac{\pi}{4 \alpha^{3}}} \operatorname{erf}\left(\frac{x}{\sqrt{\alpha}}\right) \end{align}\]
Error function integrals
\[\begin{align} & \int \operatorname{erf}(a z)\, dz=z \operatorname{erf}(a z)+\frac{1}{a \sqrt{\pi}} \exp \left(-a^{2} z^{2}\right) \\ & \int \operatorname{erfc}(a z)\, dz=z \operatorname{erfc}(a z)-\frac{1}{a \sqrt{\pi}} \exp \left(-a^{2} z^{2}\right) \\ & \int \operatorname{erf}(a z) e^{b z}\, dz=\frac{1}{b} e^{b z} \operatorname{erf}(a z)-\frac{1}{b} \exp \left(\frac{b^{2}}{4 a^{2}}\right) \operatorname{erf}\left(a z-\frac{b}{2 a}\right) \\ & \int_{0}^{\infty} \operatorname{erf}(a x) e^{-b x} \, d x=\frac{1}{b} \exp \left(\frac{b^{2}}{4 a^{2}}\right) \operatorname{erfc}\left(\frac{b}{2 a}\right) \quad && \left(\operatorname{Re}(b)>0,|\arg a|<\frac{\pi}{4}\right) \\ & \int_{0}^{\infty} e^{-a t} \operatorname{erf} \sqrt{b t}\, dt=\frac{1}{a} \sqrt{\frac{b}{a+b}} \quad && (\mathrm{Re}(a+b)>0) \\ & \int_{0}^{\infty} e^{-a t} \operatorname{erfc} \sqrt{\frac{b}{t}}\, dt=\frac{1}{a} e^{-2 \sqrt{a b}} \quad && (\mathrm{Re}(a)>0, \mathrm{Re}(b)>0) \\ & \int_{0}^{\infty} dt\, \mathrm{e}^{-a t^{2}} \operatorname{erf}(b x)=\sqrt{\frac{1}{a \pi}} \operatorname{atan} \frac{1}{a} \\ & \int_{0}^{\infty} dt\, \frac{\mathrm{e}^{-a t^{2}}}{t^{2}+b^{2}}=\frac{1}{2} \frac{\pi}{b} \mathrm{e}^{a b^{2}} \operatorname{erfc}(\sqrt{a} b) \end{align}\]
References
| \[\tag {1} \label {Cite1}\] |
Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 7th ed.; Abramowitz, M., Stegun, I. A., Eds.; National Bureau of Standards, 1968. |
This page titled 21.1.3: Gaussian Integrals and Error Function was last modified on Fri, 05 Jun 2026 02:20:26 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Andrei Tokmakoff via source content that was edited to the style and standards of the LibreTexts platform.
The book could not be loaded just now.