Quadratic Integral - Negative-discriminant Case

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

\begin{align} & {} \qquad \frac{1}{c} \int \frac{ du} {u^2 + A^2} \\ & = \frac{1}{cA} \int \frac{du/A}{(u/A)^2 + 1 } \\ & = \frac{1}{cA} \int \frac{dw}{w^2 + 1} \\ & = \frac{1}{cA} \arctan(w) + \mathrm{constant} \\ & = \frac{1}{cA} \arctan\left(\frac{u}{A}\right) + \text{constant} \\ & = \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} \\ & = \frac{2}{\sqrt{4ac - b^2\, }} \arctan\left(\frac{2cx + b}{\sqrt{4ac - b^2}}\right) + \text{constant}. \end{align}

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