Problem:
For $ b^2 < 4ac $: $ \int\limits_{-\infty}^{+\infty}{\frac{dx}{(ax^2+bx+c)^2}} $
Solution:
Consider the contour concatenating $ -R $ to $ R $ and then go through a half circle in the upper half plane back to $ -R $, the half circle part goes to 0 (*), so all we need to do is to compute the residue in the upper half plane.
The integrand can be factorized as:
$ \begin{eqnarray*} & & \frac{1}{(ax^2 + bx + c)^2} \\ &=& \frac{1}{(\frac{1}{4a}(2ax + b + \sqrt{b^2 - 4ac})(2ax + b - \sqrt{b^2 - 4ac}))^2} \\ &=& \frac{16a^2}{(2ax + b + \sqrt{b^2 - 4ac})^2(2ax + b - \sqrt{b^2 - 4ac})^2} \\ \end{eqnarray*} $
So we have two poles with order 2: $ x = \frac{-b + \sqrt{b^2 - 4ac}}{2a} $ and $ x = \frac{-b - \sqrt{b^2 - 4ac}}{2a} $, only the first pole is in the upper half plane, so we compute the residue there.
$ \begin{eqnarray*} & & Res[\frac{16a^2}{(2ax + b + \sqrt{b^2 - 4ac})^2(2ax + b - \sqrt{b^2 - 4ac})^2}, \frac{-b + \sqrt{b^2 - 4ac}}{2a}] \\ &=& \lim\limits_{x \to \frac{-b + \sqrt{b^2 - 4ac}}{2a}}{((x - \frac{-b + \sqrt{b^2 - 4ac}}{2a})^2\frac{16a^2}{(2ax + b + \sqrt{b^2 - 4ac})^2(2ax + b - \sqrt{b^2 - 4ac})^2})'} \\ &=& \lim\limits_{x \to \frac{-b + \sqrt{b^2 - 4ac}}{2a}}{((\frac{2ax + b - \sqrt{b^2 - 4ac}}{2a})^2\frac{16a^2}{(2ax + b + \sqrt{b^2 - 4ac})^2(2ax + b - \sqrt{b^2 - 4ac})^2})'} \\ &=& \lim\limits_{x \to \frac{-b + \sqrt{b^2 - 4ac}}{2a}}{(\frac{4}{(2ax + b + \sqrt{b^2 - 4ac})^2})'} \\ &=& \lim\limits_{x \to \frac{-b + \sqrt{b^2 - 4ac}}{2a}}{(\frac{(-2)4}{(2ax + b + \sqrt{b^2 - 4ac})^3})(2a)} \\ &=& \lim\limits_{x \to \frac{-b + \sqrt{b^2 - 4ac}}{2a}}{\frac{-16a}{(2ax + b + \sqrt{b^2 - 4ac})^3}} \\ &=& \frac{-16a}{(2a(\frac{-b + \sqrt{b^2 - 4ac}}{2a}) + b + \sqrt{b^2 - 4ac})^3} \\ &=& \frac{-16a}{((-b + \sqrt{b^2 - 4ac}) + b + \sqrt{b^2 - 4ac})^3} \\ &=& \frac{-16a}{(2\sqrt{b^2 - 4ac})^3} \\ &=& \frac{-2a}{(\sqrt{b^2 - 4ac})^3} \\ \end{eqnarray*} $
Going back to the integral, we get:
$ \begin{eqnarray*} & & \int\limits_{-\infty}^{+\infty}{\frac{dx}{(ax^2+bx+c)^2}} \\ &=& 2\pi i \frac{-2a}{(\sqrt{b^2 - 4ac})^3} \\ &=& \frac{-4\pi a i}{(\sqrt{b^2 - 4ac})^3} \\ &=& \frac{4\pi a}{(\sqrt{4ac - b^2})^3} \\ \end{eqnarray*} $
For $ b^2 < 4ac $: $ \int\limits_{-\infty}^{+\infty}{\frac{dx}{(ax^2+bx+c)^2}} $
Solution:
Consider the contour concatenating $ -R $ to $ R $ and then go through a half circle in the upper half plane back to $ -R $, the half circle part goes to 0 (*), so all we need to do is to compute the residue in the upper half plane.
The integrand can be factorized as:
$ \begin{eqnarray*} & & \frac{1}{(ax^2 + bx + c)^2} \\ &=& \frac{1}{(\frac{1}{4a}(2ax + b + \sqrt{b^2 - 4ac})(2ax + b - \sqrt{b^2 - 4ac}))^2} \\ &=& \frac{16a^2}{(2ax + b + \sqrt{b^2 - 4ac})^2(2ax + b - \sqrt{b^2 - 4ac})^2} \\ \end{eqnarray*} $
So we have two poles with order 2: $ x = \frac{-b + \sqrt{b^2 - 4ac}}{2a} $ and $ x = \frac{-b - \sqrt{b^2 - 4ac}}{2a} $, only the first pole is in the upper half plane, so we compute the residue there.
$ \begin{eqnarray*} & & Res[\frac{16a^2}{(2ax + b + \sqrt{b^2 - 4ac})^2(2ax + b - \sqrt{b^2 - 4ac})^2}, \frac{-b + \sqrt{b^2 - 4ac}}{2a}] \\ &=& \lim\limits_{x \to \frac{-b + \sqrt{b^2 - 4ac}}{2a}}{((x - \frac{-b + \sqrt{b^2 - 4ac}}{2a})^2\frac{16a^2}{(2ax + b + \sqrt{b^2 - 4ac})^2(2ax + b - \sqrt{b^2 - 4ac})^2})'} \\ &=& \lim\limits_{x \to \frac{-b + \sqrt{b^2 - 4ac}}{2a}}{((\frac{2ax + b - \sqrt{b^2 - 4ac}}{2a})^2\frac{16a^2}{(2ax + b + \sqrt{b^2 - 4ac})^2(2ax + b - \sqrt{b^2 - 4ac})^2})'} \\ &=& \lim\limits_{x \to \frac{-b + \sqrt{b^2 - 4ac}}{2a}}{(\frac{4}{(2ax + b + \sqrt{b^2 - 4ac})^2})'} \\ &=& \lim\limits_{x \to \frac{-b + \sqrt{b^2 - 4ac}}{2a}}{(\frac{(-2)4}{(2ax + b + \sqrt{b^2 - 4ac})^3})(2a)} \\ &=& \lim\limits_{x \to \frac{-b + \sqrt{b^2 - 4ac}}{2a}}{\frac{-16a}{(2ax + b + \sqrt{b^2 - 4ac})^3}} \\ &=& \frac{-16a}{(2a(\frac{-b + \sqrt{b^2 - 4ac}}{2a}) + b + \sqrt{b^2 - 4ac})^3} \\ &=& \frac{-16a}{((-b + \sqrt{b^2 - 4ac}) + b + \sqrt{b^2 - 4ac})^3} \\ &=& \frac{-16a}{(2\sqrt{b^2 - 4ac})^3} \\ &=& \frac{-2a}{(\sqrt{b^2 - 4ac})^3} \\ \end{eqnarray*} $
Going back to the integral, we get:
$ \begin{eqnarray*} & & \int\limits_{-\infty}^{+\infty}{\frac{dx}{(ax^2+bx+c)^2}} \\ &=& 2\pi i \frac{-2a}{(\sqrt{b^2 - 4ac})^3} \\ &=& \frac{-4\pi a i}{(\sqrt{b^2 - 4ac})^3} \\ &=& \frac{4\pi a}{(\sqrt{4ac - b^2})^3} \\ \end{eqnarray*} $
No comments:
Post a Comment