The Gaussian integral, also called the Probability Integral, is the integral of the 1-D Gaussian over
. It can be computed using the trick of combining two 1-D Gaussians
and switching to Polar Coordinates,
However, a simple proof can also be given which does not require transformation to Polar Coordinates (Nicholas and
Yates 1950).
The integral from 0 to a finite upper limit
can be given by the
Continued Fraction
![\begin{displaymath}
\int_0^a e^{-x^2}\,dx = {\sqrt{\pi}\over 2} {1\over a+} {2\over 2a+} {3\over a+} {4\over 2a+\ldots}.
\end{displaymath}](g_918.gif) |
(3) |
The general class of integrals of the form
![\begin{displaymath}
I_n(a)\equiv \int_0^\infty e^{-ax^2}x^n\,dx
\end{displaymath}](g_919.gif) |
(4) |
can be solved analytically by setting
Then
For
, this is just the usual Gaussian integral, so
![\begin{displaymath}
I_0(a)= {\sqrt{\pi}\over 2} a^{-1/2} = {1\over 2}\sqrt{\pi\over a}.
\end{displaymath}](g_928.gif) |
(9) |
For
, the integrand is integrable by quadrature,
![\begin{displaymath}
I_1(a)= a^{-1} \int_0^\infty e^{-y^2}y\,dy = a^{-1}[-{\textstyle{1\over 2}}e^{-y^2}]^\infty_0 = {\textstyle{1\over 2}}a^{-1}.
\end{displaymath}](g_929.gif) |
(10) |
To compute
for
, use the identity
For
Even,
so
![\begin{displaymath}
\int^\infty_0 x^{2s}e^{-ax^2}\,dx = {(s-{\textstyle{1\over 2...
...ver 2a^{s+1/2}}= {(2s-1)!!\over 2^{s+1}a^s} \sqrt{\pi\over a}.
\end{displaymath}](g_940.gif) |
(13) |
If
is Odd, then
so
![\begin{displaymath}
\int^\infty_0 x^{2s+1}e^{-ax^2}\,dx = { s!\over 2a^{s+1}}.
\end{displaymath}](g_945.gif) |
(15) |
The solution is therefore
![\begin{displaymath}
\int_0^\infty e^{-ax^2}x^n\,dx =\cases{
{(n-1)!!\over 2^{n/...
...$\ even\cr
{[(n+1)/2]!\over 2a^{(n+1)/2}} & for $n$\ odd.\cr}
\end{displaymath}](g_946.gif) |
(16) |
The first few values are therefore
![$\displaystyle I_0(a)$](g_947.gif) |
![$\textstyle =$](g_15.gif) |
![$\displaystyle {1\over 2}\sqrt{\pi\over a}$](g_948.gif) |
(17) |
![$\displaystyle I_1(a)$](g_949.gif) |
![$\textstyle =$](g_15.gif) |
![$\displaystyle {1\over 2a}$](g_950.gif) |
(18) |
![$\displaystyle I_2(a)$](g_951.gif) |
![$\textstyle =$](g_15.gif) |
![$\displaystyle {1\over 4a}\sqrt{\pi\over a}$](g_952.gif) |
(19) |
![$\displaystyle I_3(a)$](g_953.gif) |
![$\textstyle =$](g_15.gif) |
![$\displaystyle {1\over 2a^2}$](g_954.gif) |
(20) |
![$\displaystyle I_4(a)$](g_955.gif) |
![$\textstyle =$](g_15.gif) |
![$\displaystyle {3\over 8a^2}\sqrt{\pi\over a}$](g_956.gif) |
(21) |
![$\displaystyle I_5(a)$](g_957.gif) |
![$\textstyle =$](g_15.gif) |
![$\displaystyle {1\over a^3}$](g_958.gif) |
(22) |
![$\displaystyle I_6(a)$](g_959.gif) |
![$\textstyle =$](g_15.gif) |
![$\displaystyle {15\over 16a^3}\sqrt{\pi\over a}.$](g_960.gif) |
(23) |
A related, often useful integral is
![\begin{displaymath}
H_n(a)\equiv {1\over\sqrt{\pi}}\int_{-\infty}^\infty e^{-ax^2}x^n\,dx,
\end{displaymath}](g_961.gif) |
(24) |
which is simply given by
![\begin{displaymath}
H_n=\cases{
{2I_n(a)\over\sqrt{\pi}} & for $n$\ even\cr
0 & for $n$\ odd.\cr}
\end{displaymath}](g_962.gif) |
(25) |
References
Nicholas, C. B. and Yates, R. C. ``The Probability Integral.'' Amer. Math. Monthly 57, 412-413, 1950.
© 1996-9 Eric W. Weisstein
1999-05-25