Day: October 18, 2023

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202310181051 Exercise 7.9.21

Evaluate the integral

\displaystyle{\int \frac{\mathrm{d}x}{\sqrt{x^2-4x}}}.

Extracted from James Stewart. (2012). Calculus (8e)

Roughwork.

There are as many ways to evaluate definite/indefinite integrals as there are to evaluate derivatives/differentials, for example, by i. basic integration formulae, ii. substitution by rational polynomials, trigonometric functions, or hyperbolic functions, iii. integration by parts, iv. reduction formulae, iv. simplification by partial fractions, v. parity argument, and vi. parametrization.

Make use of none above but the following

Residue Theorem. Suppose that the function f is analytic within and on a positively oriented simple closed contour C except for a finite number of isolated singular points \{ z_j, j=1,2,\dots ,N\} interior to C, then

\displaystyle{\int_{C}f(z)\,\mathrm{d}z=2\pi i\sum_{j=1}^{N}\mathrm{Res}_{z=z_j}f(z)}.

Residue Formula. If f(z) has a pole of order k at z=z_0, then

\displaystyle{\mathrm{Res}\, (f,z_0)=\frac{1}{(k-1)!}\frac{\mathrm{d}^{k-1}}{\mathrm{d}z^{k-1}}\Big( (z-z_0)^kf(z)\Big)\bigg|_{z=z_0}}

Write for some real number x

\begin{aligned} & \quad\enspace \int \frac{\mathrm{d}x}{\sqrt{x^2-4x}} \\ & \hookrightarrow \textrm{let }R(x)=\frac{1}{\sqrt{x^2-4x}} \\ \end{aligned}

and also for some complex number z=x+iy,

\begin{aligned} & \quad\enspace \int \frac{\mathrm{d}z}{\sqrt{z^2-4z}} \\ & \hookrightarrow \textrm{let }R(z)=\frac{1}{\sqrt{z^2-4z}} \\ \end{aligned}

Then there are two singular points (/singularities) (0,0) and (4,0) on the Argand plane, because, for instance

\displaystyle{R\big( (0,0)\big)=\frac{1}{\sqrt{(0)^2-4(0)}}}

is undefined.

From the denominator of R(z) we see there

\sqrt{z(z-4)}

2 poles of order \frac{1}{2} (*fractional order seems disputable, but no worries as you read further along), by the fact that

\begin{aligned} & \lim_{z\to 0} (z)^{(\frac{1}{2})}\frac{1}{\sqrt{z^2-4z}} = -\frac{i}{2} \neq 0 \\ & \lim_{z\to 4} (z-4)^{(\frac{1}{2})}\frac{1}{\sqrt{z^2-4z}} = \frac{1}{2} \neq 0 \\ \end{aligned}

and their residues being:

z(x,y)=\{ (0,0)\} \cup \{ (4,0)\}.

We introduce the below

Lemma 1. (positive fractional derivatives)

\displaystyle{\frac{\mathrm{d}^nx^m}{\mathrm{d}x^n}=\frac{\Gamma (m+1)}{\Gamma (m-n+1)}x^{m-n}}

In particular, if letting m=1 and n=1/2, then

\displaystyle{\frac{\mathrm{d}^{1/2}x}{\mathrm{d}x^{1/2}}=\frac{2\sqrt{x}}{\sqrt{\pi}}}.

Lemma 2. (negative fractional derivatives)

\displaystyle{D^{-\nu}x^{\mu}=\frac{\Gamma (\mu +1)}{\Gamma (\mu +\nu +1)}x^{\mu +\nu}}

where \nu >0, \mu >-1, and x>0. In particular, if \nu =\frac{1}{2}, then

\begin{aligned} D^{-1/2}x^0 & = \frac{\Gamma (1)}{\Gamma (3/2)}x^{1/2}=2\sqrt{\frac{x}{\pi}} \\ D^{-1/2}x^1 & = \frac{\Gamma (2)}{\Gamma (5/2)}x^{3/2}=\frac{4}{3}\sqrt{\frac{x^3}{\pi}} \\ D^{-1/2}x^2 & = \frac{\Gamma (3)}{\Gamma (7/2)}x^{5/2}=\frac{16}{15}\sqrt{\frac{x^5}{\pi}} \\ \end{aligned}

where the Legendre’s Gamma function is given by

\Gamma (n+1)=n!.

For brevity we state without proof the definition of factorial of fractions:

\displaystyle{N!=\prod (N)}

where N\in\mathbb{Q} and Gauss product Pi is

\displaystyle{\prod (x) = \int_{0}^{\infty}t^xe^{-t}\,\mathrm{d}t}

Cited from Kimeu, Joseph M., “Fractional Calculus: Definitions and Applications” (2009). Masters Theses & Specialist Projects. Paper 115.

Therefore

\begin{aligned} &\quad\enspace \int R(x)\,\mathrm{d}x \\ & = \pi i\bigg\{ \mathrm{Res}\, [R(z),0] + \mathrm{Res}\, [R(z),4] \bigg\} \\ \end{aligned}

\begin{aligned} \hookrightarrow\quad\mathrm{Res}(R(z),0) & = \frac{1}{(-1/2)!}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z^{-1/2}}(z-0)^{1/2}R(z)\bigg|_{z=0} \\ & = \frac{1}{\sqrt{\pi}}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z^{-1/2}}\frac{\sqrt{z}}{\sqrt{z^2-4z}}\bigg|_{z=0} \\ & = \frac{1}{\sqrt{\pi}}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z^{-1/2}}\frac{1}{\sqrt{z-4}}\bigg|_{z=0} \\ & \stackrel{\dagger}{=}\frac{1}{\sqrt{\pi}}\sqrt{\frac{-2}{z'^3}}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z'^{-1/2}}\big( z'\big) \\ & = \frac{1}{\sqrt{\pi}}\sqrt{\frac{-2}{z'^3}}\bigg(\frac{4}{3}\sqrt{\frac{z'^3}{\pi}}\bigg) \\ & = \frac{4\sqrt{2}i}{3\pi} \\ & \\ \textrm{\dag} :\qquad\qquad & \\ \quad \textrm{let }z' & = \frac{1}{\sqrt{z-4}} \\ \textrm{then }\frac{\mathrm{d}z'}{\mathrm{d}z} & = -\bigg(\frac{1}{2}\bigg)(z-4)^{-3/2}(1) \\ & = -\frac{z'^3}{2} \\ \mathrm{d}z & = -\frac{2}{z'^3}\,\mathrm{d}z' \\ \end{aligned}

\begin{aligned} \hookrightarrow\quad\mathrm{Res}(R(z),4) & = \frac{1}{(-1/2)!}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z^{-1/2}}(z-4)^{1/2}R(z)\bigg|_{z=0} \\ & = \frac{1}{\sqrt{\pi}}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z^{-1/2}}\frac{\sqrt{z-4}}{\sqrt{z^2-4z}}\bigg|_{z=0} \\ & = \frac{1}{\sqrt{\pi}}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z^{-1/2}}\frac{1}{\sqrt{z}}\bigg|_{z=0} \\ & \stackrel{\textrm{\ddag}}{=}\frac{1}{\sqrt{\pi}}\sqrt{\frac{-2}{z''^3}}\frac{\mathrm{d}^{-1/2}}{\mathrm{d}z''^{-1/2}}\big( z''\big) \\ & = \frac{1}{\sqrt{\pi}}\sqrt{\frac{-2}{z''^3}}\bigg(\frac{4}{3}\sqrt{\frac{z''^3}{\pi}}\bigg) \\ & = \frac{4\sqrt{2}i}{3\pi} \\ & \\ \textrm{\ddag} :\qquad\qquad & \\ \quad \textrm{let }z'' & = \frac{1}{\sqrt{z-4}} \\ \textrm{then }\frac{\mathrm{d}z''}{\mathrm{d}z} & = -\bigg(\frac{1}{2}\bigg)(z-4)^{-3/2}(1) \\ & = -\frac{z''^3}{2} \\ \mathrm{d}z & = -\frac{2}{z''^3}\,\mathrm{d}z'' \\ \end{aligned}

\begin{aligned} &\quad\enspace \int R(x)\,\mathrm{d}x \\ & = \pi i\bigg\{ \mathrm{Res}\, [R(z),0] + \mathrm{Res}\, [R(z),4] \bigg\} \\ & = \pi i \bigg\{\frac{4\sqrt{2}i}{3\pi}+\frac{4\sqrt{2}i}{3\pi}\bigg\} \\ & = -\frac{8\sqrt{2}}{3} \\ \end{aligned}

Of course this \textrm{\scriptsize{MUST}} be a joke; how can an indefinite integral give a definite value, let alone be independent of any variable?

Countercheck.

The correct answer is

2\ln (\sqrt{x}+\sqrt{x-4})+C\enspace\textrm{for some constant}.

(to be continued)