5 Integrals

5.1 Highly damped integral

$$I = \ensuremath{\int_{0}^{\infty} {\frac{1}{k^2 + z^2}} \dd{z}}$$ (66)

$$= \frac{1}{2} \ensuremath{\int_{-\infty}^{\infty} {\frac{1}{k^2 + z^2}} \dd{z}}$$ (67)

$$= \frac{1}{2k} \ensuremath{\int_{-\infty}^{\infty} {\frac{1}{u^2+1}} \dd{u}}$$ (68)

where $u \equiv z/k$, $du = dz/k$. There are simple poles at $u = \pm i$

$$I = \frac{1}{2k} \cdot 2 \pi i \operatorname{Res}\left({z=i},{f(u)}\right)$$ (69)

$$= \frac{1}{2k} \cdot \frac{2 \pi i}{i+i}$$ (70)

$$= \frac{1}{2k} \pi$$ (71)

$$= \frac{\pi}{2 k},$$ (72)

5.2 General case integral

We will show that for any $(a,b > 0) \in \ensuremath{\mathbb{R}}$

$$I = \ensuremath{\int_{-\infty}^{\infty} {\frac{1}{(a^2-z^2) + b^2 z^2}} \dd{z}} = \frac{\pi}{b a^2}.$$ (73)

First we note that $\vert f(z)\vert \rightarrow 0$ like $\vert z^{-4}\vert$ for $\vert z\vert \gg 1$, and that $f(z)$ is even, so

$$I = \ensuremath{\int_{\ensuremath{\mathcal{C}}}^{} {\frac{1}{(a^2-z^2)^2 + b^2 z^2}} \dd{z}},$$ (74)

where $\mathcal{C}$ is the contour shown in Figure 1.

Because the denominator is of the form $A^2 + B^2$, we can factor it into $(A+iB)(A-iB)$ like so

$$(a^2-z^2)^2 + b^2 z^2 = (a^2-z^2 \textcolor{Red}{+} ibz)(a^2-z^2 \textcolor{Red}{-} ibz)$$ (75)

And the roots of $z^2 \textcolor{Red}{\pm} ibz - a^2$

$$z_{r\textcolor{Blue}{\pm}} = \textcolor{Red}{\pm}\frac{ib}{2} \left( 1 \textcolor{Blue}{\pm} \sqrt{1-4\frac{-a^2}{(ib)^2}} \right)$$ (76)

$$= \pm\frac{ib}{2} \left( 1 \pm \sqrt{1-4\frac{a^2}{b^2}} \right)$$ (77)

$$= \pm\frac{ib}{2} \left( 1 \pm S \right)$$ (78)

Where $S \equiv \sqrt{1-4\frac{a^2}{b^2}}$.

To determine the nature and locations of the roots, consider the following cases (in order of increasing $a$).

• $a < b/2$, overdamped.
• $a = b/2$, critically damped.
• $a > b/2$, underdamped.

In the overdamped case $S \in \ensuremath{\mathbb{R}}$ and $S > 0$, so $z_{r\pm}$ is purely imaginary, and $z_{r+} != z_{r-}$. For any $a < b/2$, we have $0 < S < 1$, so $\ensuremath{\operatorname{Im}}(z_{r\pm}) > 0$. Thus, there are two single poles in the upper half plane ($z_{r\pm}$), and two single poles in the lower half plane ($-z_{r\pm}$).

In the critically damped case $S = 0$, so $z_{r+} = z_{r-}$, and we have double poles at $\pm z_{r+} = \frac{ib}{2}$.

In the underdamped case $S$ is purely imaginary, so $z_{r\pm}$ is complex, with $z_{r+}$ in the 2 $^{\mathrm{nd}}$quarter, and $z_{r-}$ in the 1 $^{\mathrm{st}}$quarter. The other two simple poles, $-z_{r-}$ and $-z_{r+}$, are in the 3 $^{\mathrm{rd}}$and 4 $^{\mathrm{th}}$quarters respectively.

Our contour $\mathcal{C}$ always encloses the poles $z_{r\pm}$. We will deal with the simple pole cases first, and then return to the critically damped case.

5.2.1 Over- and under-damped

Our factored function $f(z)$ is

$$f(z) = \frac{1}{(z-z_{r+})(z+z_{r+})(z+z_{r-})(z-z_{r-})}$$ (79)

Applying eq. 63 and 64 we have

$$I = 2\pi i \left( \operatorname{Res}\left({z=z_{r+}},{f(z)}\right) + \operatorname{Res}\left({z=z_{r-}},{f(z)}\right) \right)$$ (80)

$$= 2\pi i \left( \frac{1}{ (z_{r+}+z_{r+}) \textcolor{Red}{(z_{r+}... ...{1}{ \textcolor{Red}{(z_{r-}-z_{r+}) (z_{r-}+z_{r+})} (z_{r-}+z_{r-}) } \right)$$ (81)

$$= \frac{\pi i}{\textcolor{Red}{z_{r+}^2-z_{r-}^2}} \left( \frac{1}{z_{r+}} \textcolor{Red}{-} \frac{1}{z_{r-}} \right)$$ (82)

$$= \frac{\pi i}{ \left( \textcolor{Blue}{\frac{ib}{2}} (1+S) \righ... ...{Blue}{\frac{ib}{2}} (1-S) \right)^2 } \cdot \frac{z_{r-}-z_{r+}}{z_{r+}z_{r-}}$$ (83)

$$= \frac{\textcolor{Blue}{-4}\pi i / \textcolor{Blue}{b^2}}{ (1+2S... ...[(1-S) - (1+S)] } { \left(\frac{ib}{2}\right)^{\textcolor{Red}{2}} (1+S)(1-S) }$$ (84)

$$= \frac{-8\pi / b^3}{ 4S } \cdot \frac{-2S} {(1 - S^2)}$$ (85)

$$= \frac{ 4\pi }{ b^3 (1 - S^2)}$$ (86)

$$= \frac{ 4\pi }{ b^3 [1 - (1-4\frac{a^2}{b^2})]}$$ (87)

$$= \frac{ 4\pi }{ b^3 \cdot 4\frac{a^2}{b^2}}$$ (88)

$$= \frac{ \pi }{ b a^2 }$$ (89)

Hooray!

5.2.2 Critically damped

Our factored function $f(z)$ is

$$f(z) = \frac{1}{(z-z_{r+})^2(z-z_{r-})^2}$$ (90)

Applying eq. 63 and 65 we have

$$I = 2\pi i \operatorname{Res}\left({z=z_{r+}},{f(z)}\right)$$ (91)

$$= \textcolor{Red}{2}\pi i \left( \textcolor{Red}{\frac{1}{2!}} \lim_{z \rightarrow {z_{r+}}} \deriv{z}{} \frac{1}{(z + z_{r+})^2} \right)$$ (92)

$$= \pi i \lim_{z \rightarrow {z_{r+}}} -2 \cdot \frac{1}{(z_{r+} + z_{r+})^3}$$ (93)

$$= - 2 \pi i \frac{1}{z_{r+}^3}$$ (94)

$$= \textcolor{Red}{-} 2 \pi \textcolor{Red}{i} \frac{1}{(\frac{\textcolor{Red}{i}b}{2})^3}$$ (95)

$$= \frac{\pi}{b (\frac{b}{2})^2}$$ (96)

$$= \frac{\pi}{b a^2},$$ (97)

which matches eq. 90

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5 Integrals
Copyright © 2009-10-12, W. Trevor King (contact)
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Drexel Physics