Quadratic integral
In mathematics, a quadratic integral is an integral of the form
It can be evaluated by completing the square in the denominator.
Positive-discriminant case
Assume that the discriminant q = b2 − 4ac is positive. In that case, define u and A by
,
and
The quadratic integral can now be written as
The partial fraction decomposition
allows us to evaluate the integral:
The final result for the original integral, under the assumption that q > 0, is
Negative-discriminant case
- This (hastily written) section may need attention.
In case the discriminant q = b2 − 4ac is negative, the second term in the denominator in
is positive. Then the integral becomes
References
- Weisstein, Eric W. "Quadratic Integral." From MathWorld--A Wolfram Web Resource, wherein the following is referenced:
- Gradshteyn, Izrail Solomonovich; Ryzhik, Iosif Moiseevich; Geronimus, Yuri Veniaminovich; Tseytlin, Michail Yulyevich; Jeffrey, Alan (2015) [October 2014]. Zwillinger, Daniel; Moll, Victor Hugo, eds. Table of Integrals, Series, and Products. Translated by Scripta Technica, Inc. (8 ed.). Academic Press, Inc. ISBN 0-12-384933-0. LCCN 2014010276. ISBN 978-0-12-384933-5.
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![\begin{align}
& {} \qquad \frac{1}{c} \int \frac{ du} {u^2 + A^2} \\[9pt]
& = \frac{1}{cA} \int \frac{du/A}{(u/A)^2 + 1 } \\[9pt]
& = \frac{1}{cA} \int \frac{dw}{w^2 + 1} \\[9pt]
& = \frac{1}{cA} \arctan(w) + \mathrm{constant} \\[9pt]
& = \frac{1}{cA} \arctan\left(\frac{u}{A}\right) + \text{constant} \\[9pt]
& = \frac{1}{c\sqrt{\frac{a}{c} - \frac{b^2}{4c^2}}} \arctan
\left(\frac{x + \frac{b}{2c}}{\sqrt{\frac{a}{c} - \frac{b^2}{4c^2}}}\right) + \text{constant} \\[9pt]
& = \frac{2}{\sqrt{4ac - b^2\, }}
\arctan\left(\frac{2cx + b}{\sqrt{4ac - b^2}}\right) + \text{constant}.
\end{align}](../I/m/4c0d85ba4dbafa5b97ff01b24a1a6274.png)