Apunts de Matemàtiques i Física
PREAMBLE
Projecte
TRIGONOMETRY
Cordes en cercles
SPECIAL FUNCTIONS
Nu and related functions
ALGEBRA
Matrius i determinants
Racionalització

Apunts sobre la funció \(\nu\left(x\right)\)

Definició

Les funcions a considerar son la \(\nu\left(x\right)\) i les seves associades

\[ \begin{align} \nu\left(x\right) &= \int_{0}^{\infty}\!\! \frac{x^t}{\Gamma\left(t+1\right)}dt\\ \nu\left(x,\alpha\right) &= \int_{0}^{\infty}\!\! \frac{x^{\alpha+t}}{\Gamma\left(\alpha+t+1\right)}dt\\ \mu\left(x,\beta\right) &= \int_{0}^{\infty}\!\! \frac{x^{t}t^{\beta}}{\Gamma\left(\beta+1\right)\Gamma\left(t+1\right)}dt\\ \mu\left(x,\beta,\alpha\right) &= \int_{0}^{\infty}\!\! \frac{x^{\alpha+t}t^{\beta}}{\Gamma\left(\beta+1\right)\Gamma\left(\alpha+t+1\right)}dt \end{align} \]

Relacions i Recurrències

Entre les quatre funcions definides hi ha una sèrie de relacions:

\[ \begin{align} \nu\left(x\right) &= \nu\left(x,0\right) = \mu\left(x,0\right) = \mu\left(x,0,0\right) \\ \nu\left(x,\alpha\right)&= \mu\left(x,0,\alpha\right) \\ \mu\left(x,\beta\right) &= \mu\left(x,\beta,0\right) \end{align} \] \[ \beta\mu\left(x,\beta\right) = x\mu\left(x,\beta-1,-1\right) \] \[ x\nu\left(x,\alpha-1\right) - \alpha\nu\left(x,\alpha\right) = \mu\left(x,1,\alpha\right) \] \[ \left(\beta+1\right)\mu\left(x,\beta+1,\alpha\right) = x\mu\left(x,\beta,\alpha-1\right) - \alpha\mu\left(x,\beta,\alpha\right) \]

Demostracions de les relacions

Les tres primeres equacions son prou directes, només cal substituir els zeros en les definicions.

\(\beta\mu\left(x,\beta\right) = x\mu\left(x,\beta-1,-1\right)\)

La quarta equació (era incorrecta a [1] on apareix sense la \(\beta\) al denominador) requereix utilitzar la coneguda propietat \(\Gamma\left(z+1\right)=z\Gamma\left(z\right)\) i reescriure convenientment el resultat:

\[ \begin{align} \mu\left(x,\beta\right) &= \int_{0}^{\infty}\!\! \frac{x^{t}t^{\beta}}{\Gamma\left(\beta+1\right)\Gamma\left(t+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{t}t^{\beta}}{\beta\Gamma\left(\beta\right)t\Gamma\left(t\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{x x^{t-1}t^{\beta-1}}{\beta\Gamma\left(\beta\right)\Gamma\left(t\right)}dt\\ &= \frac{x}{\beta}\int_{0}^{\infty}\!\! \frac{x^{-1+t}t^{\beta-1}}{\Gamma\left(\beta-1+1\right)\Gamma\left(-1+t+1\right)}dt\\ &= \frac{x}{\beta}\mu\left(x,\beta-1,-1\right) \end{align} \]

\(x\nu\left(x,\alpha-1\right) - \alpha\nu\left(x,\alpha\right) = \mu\left(x,1,\alpha\right)\)

La cinquena equació també requereix utilitzar la propietat \(z\Gamma\left(z\right)=\Gamma\left(z+1\right)\) i reescriure convenientment el resultat:

\[ \begin{align} \nu\left(x,\alpha\right) &= \int_{0}^{\infty}\!\! \frac{x^{\alpha+t}}{\Gamma\left(\alpha+t+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{\left(\alpha+t+1\right)x^{\alpha+t}}{\left(\alpha+t+1\right)\Gamma\left(\alpha+t+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{\left(\alpha+t+1\right)x^{\alpha+t}}{\Gamma\left(\alpha+t+1+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{\left(\alpha+1+t\right)x^{\alpha+t}}{\Gamma\left(\alpha+1+t+1\right)}dt\\ &= \left(\alpha+1\right)\int_{0}^{\infty}\!\! \frac{x^{\alpha+t}}{\Gamma\left(\alpha+1+t+1\right)}dt + \int_{0}^{\infty}\!\! \frac{tx^{\alpha+t}}{\Gamma\left(\alpha+1+t+1\right)}dt\\ &= \frac{\alpha+1}{x}\int_{0}^{\infty}\!\! \frac{x^{(\alpha+1)+t}}{\Gamma\left((\alpha+1)+t+1\right)}dt + \frac{1}{x}\int_{0}^{\infty}\!\! \frac{x^{(\alpha+1)+t}t^1}{\Gamma\left((\alpha+1)+t+1\right)}dt\\ &= \frac{\alpha+1}{x}\nu\left(x,\alpha+1\right) + \frac{1}{x}\mu\left(x,1,\alpha+1\right) \end{align} \]

O sigui, el resultat obtingut és:

\[ x\nu\left(x,\alpha\right)= \left(\alpha+1\right)\nu\left(x,\alpha+1\right) + \mu\left(x,1,\alpha+1\right) \]

O també:

\[ x\nu\left(x,\alpha\right) - \left(\alpha+1\right)\nu\left(x,\alpha+1\right) = \mu\left(x,1,\alpha+1\right) \]

Si fem el canvi \(\alpha\rightarrow\alpha-1\) recuperem el resultat presentat a [1]:

\[ x\nu\left(x,\alpha-1\right) - \alpha\nu\left(x,\alpha\right) = \mu\left(x,1,\alpha\right) \]

\(\left(\beta+1\right)\mu\left(x,\beta+1,\alpha\right) = x\mu\left(x,\beta,\alpha-1\right) - \alpha\mu\left(x,\beta,\alpha\right)\)

Igual que en les anteriors cal utilitzar la propietat \(z\Gamma\left(z\right)=\Gamma\left(z+1\right)\) i reescriure convenientment el resultat (era incorrecta a [1] on apareix sense el factor \(\left(\beta+1\right)\)):

\[ \begin{align} \mu\left(x,\beta,\alpha\right) =&\int_{0}^{\infty}\!\! \frac{x^{\alpha+t}t^{\beta}}{\Gamma\left(\beta+1\right)\Gamma\left(\alpha+t+1\right)}dt\\ =&\int_{0}^{\infty}\!\! \frac{x^{\alpha+t}t^{\beta}\left(\alpha+t+1\right)}{\Gamma\left(\beta+1\right)\left(\alpha+t+1\right)\Gamma\left(\alpha+t+1\right)}dt\\ =&\int_{0}^{\infty}\!\! \frac{x^{\alpha+t}t^{\beta}\left(\alpha+1+t\right)}{\Gamma\left(\beta+1\right)\Gamma\left(\alpha+t+1+1\right)}dt\\ =&\frac{1}{x}\int_{0}^{\infty}\!\! \frac{x^{(\alpha+1)+t}t^{\beta}\left(\alpha+1+t\right)}{\Gamma\left(\beta+1\right)\Gamma\left((\alpha+1)+t+1\right)}dt\\ =&\frac{\alpha+1}{x}\int_{0}^{\infty}\!\! \frac{x^{(\alpha+1)+t}t^{\beta}}{\Gamma\left(\beta+1\right)\Gamma\left((\alpha+1)+t+1\right)}dt\\ &+\frac{1}{x}\int_{0}^{\infty}\!\! \frac{x^{(\alpha+1)+t}t^{\beta+1}}{\Gamma\left(\beta+1\right)\Gamma\left((\alpha+1)+t+1\right)}dt\\ =&\frac{\alpha+1}{x}\int_{0}^{\infty}\!\! \frac{x^{(\alpha+1)+t}t^{\beta}}{\Gamma\left(\beta+1\right)\Gamma\left((\alpha+1)+t+1\right)}dt\\ &+\frac{\beta+1}{x}\int_{0}^{\infty}\!\! \frac{x^{(\alpha+1)+t}t^{\beta+1}}{\Gamma\left((\beta+1)+1\right)\Gamma\left((\alpha+1)+t+1\right)}dt\\ =&\frac{\alpha+1}{x}\mu\left(x,\beta,\alpha+1\right)+\frac{\beta+1}{x}\mu\left(x,\beta+1,\alpha+1\right) \end{align} \]

O sigui, la primera forma del resultat obtingut és:

\[ \mu\left(x,\beta,\alpha\right) =\frac{\alpha+1}{x}\mu\left(x,\beta,\alpha+1\right) +\frac{\beta+1}{x}\mu\left(x,\beta+1,\alpha+1\right) \]

Petites variacions elementals serien:

\[ x\mu\left(x,\beta,\alpha\right) =\left(\alpha+1\right)\mu\left(x,\beta,\alpha+1\right) +\left(\beta+1\right)\mu\left(x,\beta+1,\alpha+1\right) \] \[ \left(\beta+1\right)\mu\left(x,\beta+1,\alpha+1\right) = x\mu\left(x,\beta,\alpha\right) -\left(\alpha+1\right)\mu\left(x,\beta,\alpha+1\right) \]

Com en el cas anterior, si fem el canvi \(\alpha\rightarrow\alpha-1\) recuperem el tipus de resultat presentat a [1]:

\[ \left(\beta+1\right)\mu\left(x,\beta+1,\alpha\right) = x\mu\left(x,\beta,\alpha-1\right) - \alpha\mu\left(x,\beta,\alpha\right) \]

on cal tornar a recordar que a [1] era incorrecta, apareix sense el factor \(\left(\beta+1\right)\).

Comentaris i presentacions alternatives

La funció \(\nu\left(x\right)\) es pot reescriure amb intervals d'integració finits. Una primera possibilitat podria ésser:

\[ \nu\left(x\right) = \int_{0}^{1} \frac{1}{ux^{\log u}\Gamma\left(1-\log u\right)}du \]

resultat de fer el canvi \(t=-\log u\) ò dit d'altra manera \(u=e^{-t}\):

\[ \left. \begin{array}{l} t=+\infty\\ t=0\\ \end{array} \right\} u=e^{-t} \left\{ \begin{array}{l} u=0\\ u=1\\ \end{array} \right. \;\;\;\;\;\; dt=\frac{-du}{u} \] \[ \begin{align} \int_{0}^{\infty} \frac{x^{t}}{\Gamma\left(t+1\right)}dt &=\int_{1}^{0} \frac{x^{-\log(u)}}{\Gamma\left(-\log u +1\right)}\left(\frac{-du}{u}\right)\\ &=\int_{0}^{1} \frac{1}{ux^{\log(u)}\Gamma\left(1-\log u\right)}du \end{align} \]

Una segona possibilitat podria ésser:

\[ \nu\left(x\right) = \int_{0}^{1} \frac{x^{\frac{u}{1-u}}}{\left(1-u\right)^2\Gamma\left(\frac{1}{1-u}\right)}du \]

resultat de fer el canvi \(t=\displaystyle\frac{u}{1-u}\) ò dit d'altra manera \(u=\displaystyle\frac{t}{t+1}\):

\[ \left. \begin{array}{l} t=+\infty\\ t=0\\ \end{array} \right\} u=\frac{t}{t+1} \left\{ \begin{array}{l} u=1\\ u=0\\ \end{array} \right. \;\;\;\;\;\; dt=\frac{du}{(1-u)^2} \] \[ \begin{align} \int_{0}^{\infty} \frac{x^{t}}{\Gamma\left(t+1\right)}dt &=\int_{0}^{1} \frac{x^{\frac{u}{1-u}}}{\Gamma\left(\frac{1}{1-u}\right)}\frac{du}{\left(1-u\right)^2}\\ &=\int_{0}^{1} \frac{x^{\frac{u}{1-u}}}{\left(1-u\right)^2\Gamma\left(\frac{1}{1-u}\right)}du \end{align} \]

La funció \(\nu\left(x,\alpha\right)\) es pot reescriure per \( \alpha\in \rm I\!R^{+}\) com

\[ \nu\left(x,\alpha\right) = \int_{0}^{\infty} \frac{x^{\alpha+t}}{\Gamma\left(\alpha+t+1\right)}dt = \int_{\alpha}^{\infty} \frac{x^{t}}{\Gamma\left(t+1\right)}dt \]

llavors es pot esperar, degut a la positivitat de l'integrand, que per \(x,\alpha\in\rm I\!R^{+}\) es satisfa que \(\nu\left(x,\alpha\right)\leq\nu\left(x\right)\).

És important veure que per les funcions \(\mu\left(x,\beta\right)\) i \(\mu\left(x,\beta,\alpha\right)\) el factor de normalització \(\Gamma\left(\beta+1\right)\) és extern a l'integral:

\[ \begin{align} \mu\left(x,\beta\right) &= \int_{0}^{\infty} \frac{x^{t}t^{\beta}dt}{\Gamma\left(\beta+1\right)\Gamma\left(t+1\right)} = \frac{1}{\Gamma\left(\beta+1\right)}\int_{0}^{\infty} \frac{x^{t}t^{\beta}}{\Gamma\left(t+1\right)}dt\\ \mu\left(x,\beta,\alpha\right) &= \int_{0}^{\infty} \frac{x^{\alpha+t}t^{\beta}dt}{\Gamma\left(\beta+1\right)\Gamma\left(\alpha+t+1\right)} = \frac{1}{\Gamma\left(\beta+1\right)}\int_{0}^{\infty} \frac{x^{\alpha+t}t^{\beta}}{\Gamma\left(\alpha+t+1\right)}dt \end{align} \]

Aquest factor \(\Gamma\left(\beta+1\right)\) és molt important per poder extendre la funció \(\mu\left(x,\beta,\alpha\right)\) per \(\beta\) en valors enters negatius. L'aparició d'aquest factor depen de l'autor consultat.

Representació de \(\nu\left(x\right)\)

Utilitzant que \(x=\left|x\right|e^{i\alpha_x}\)

\[ \begin{align} \nu\left(x\right) &= \int_{0}^{\infty}\!\! \frac{x^t}{\Gamma\left(t+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{\left|x\right|^{t}e^{i\alpha_x t}}{\Gamma\left(t+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{\left|x\right|^{t}\cos{\alpha_x t}}{\Gamma\left(t+1\right)}dt + i\int_{0}^{\infty}\!\! \frac{\left|x\right|^{t}\sin{\alpha_x t}}{\Gamma\left(t+1\right)}dt \end{align} \]

Per valors reals positius de \(x\) la funció \(\nu\left(x\right)\) és real. Per qualsevol altre es genera una part imaginaria. Si contemplem tota la recta real:

\[ \nu\left(x\right)= \left\{ \begin{array}{ll} \int_{0}^{\infty}\!\! \frac{x^t}{\Gamma\left(t+1\right)}dt & x\gt0\\ 0 & x=0\\ \int_{0}^{\infty}\!\! \frac{\left|x\right|^{t}\cos{\pi t}}{\Gamma\left(t+1\right)}dt + i\int_{0}^{\infty}\!\! \frac{\left|x\right|^{t}\sin{\pi t}}{\Gamma\left(t+1\right)}dt & x\lt0 \end{array} \right. \]


Part real de \(\nu(x)\) en vers de \(x\)


Part imaginària de \(\nu(x)\) en vers de \(x\)

Representació de \(\nu\left(x,\alpha\right)\)

Primer apropament al comportament asimptòtic de \(\nu\left(x\right)\) i \(\nu\left(x,\alpha\right)\)



\(\nu(x,-2)\)   \(\nu(x,-1)\)   \(\nu(x)=\nu(x,0)\)   \(\nu(x,+1)\)   \(\nu(x,+2)\)  

Representació de \(\displaystyle\frac{\nu(x,\alpha)}{e^x}\) en vers de \(x\) per estudiar a asimptocitat de les funcions

Sèries

\( \displaystyle\sum_{n=0}^{\infty} u^n\mu\left(x,n\right) = \nu\left(xe^u\right) \)

\[ \begin{align} \sum_{n=0}^{\infty} u^n\mu\left(x,n\right) &= \sum_{n=0}^{\infty} u^n\int_{0}^{\infty}\!\! \frac{x^{t}t^{n}}{\Gamma\left(n+1\right)\Gamma\left(t+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{t}}{\Gamma\left(t+1\right)}\left(\sum_{n=0}^{\infty}\frac{u^nt^{n}}{\Gamma\left(n+1\right)}\right)dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{t}}{\Gamma\left(t+1\right)}\left(\sum_{n=0}^{\infty}\frac{(ut)^{n}}{n!}\right)dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{t}}{\Gamma\left(t+1\right)}e^{ut}dt\\ &= \int_{0}^{\infty}\!\! \frac{(xe^u)^{t}}{\Gamma\left(t+1\right)}dt\\ &= \nu\left(xe^u\right) \end{align} \]

\( \displaystyle\sum_{n=0}^{\infty} u^n\mu\left(x,n,\alpha\right) = e^{-\alpha u}\nu\left(xe^u,\alpha\right) \)

\[ \begin{align} \sum_{n=0}^{\infty} u^n\mu\left(x,n,\alpha\right) &= \sum_{n=0}^{\infty} u^n\int_{0}^{\infty}\!\! \frac{x^{\alpha+t}t^{n}}{\Gamma\left(n+1\right)\Gamma\left(\alpha+t+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{\alpha+t}}{\Gamma\left(\alpha+t+1\right)}\left(\sum_{n=0}^{\infty}\frac{u^nt^{n}}{\Gamma\left(n+1\right)}\right)dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{\alpha+t}}{\Gamma\left(\alpha+1\right)}\left(\sum_{n=0}^{\infty}\frac{(ut)^{n}}{n!}\right)dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{\alpha+t}}{\Gamma\left(\alpha+t+1\right)}e^{ut}dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{\alpha}(xe^u)^{t}}{\Gamma\left(\alpha+t+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{e^{-\alpha u}e^{+\alpha u}x^{\alpha}(xe^u)^{t}}{\Gamma\left(\alpha+t+1\right)}dt\\ &= e^{-\alpha u}\int_{0}^{\infty}\!\! \frac{(xe^u)^{\alpha}(xe^u)^{t}}{\Gamma\left(\alpha+t+1\right)}dt\\ &= e^{-\alpha u}\int_{0}^{\infty}\!\! \frac{(xe^u)^{\alpha+t}}{\Gamma\left(\alpha+t+1\right)}dt\\ &= e^{-\alpha u}\nu\left(xe^u,\alpha\right) \end{align} \]

Diferenciació

\[\displaystyle\frac{d^n}{dx^n} \nu\left(x\right) = \nu\left(x,-n\right) \] \[\displaystyle\frac{d^n}{dx^n} \nu\left(x,\alpha\right) = \nu\left(x,\alpha-n\right) \] \[\displaystyle\frac{d^n}{dx^n} \mu\left(x,\beta\right) = \mu\left(x,\beta,-n\right) \] \[\displaystyle\frac{d^n}{dx^n} \mu\left(x,\beta,\alpha\right) = \mu\left(x,\beta,\alpha-n\right) \]

Demostracions

\(\displaystyle\frac{d^n}{dx^n} \nu\left(x,\alpha\right) = \nu\left(x,\alpha-n\right) \)

\[ \begin{align} \frac{d}{dx} \nu\left(x,\alpha\right) &= \frac{d}{dx} \int_{0}^{\infty}\!\! \frac{x^{\alpha+t}}{\Gamma\left(\alpha+t+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{(\alpha+t)x^{\alpha+t-1}}{(\alpha+t)\Gamma\left(\alpha+t\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{\alpha+t-1}}{\Gamma\left(\alpha+t\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{(\alpha-1)+t}}{\Gamma\left((\alpha-1)+t+1\right)}dt\\ &= \nu\left(x,\alpha-1\right) \end{align} \] \[\displaystyle\frac{d^n}{dx^n} \nu\left(x,\alpha\right) = \nu\left(x,\alpha-n\right)\]

\(\displaystyle\frac{d^n}{dx^n} \nu\left(x\right) = \nu\left(x,-n\right) \)

\[ \begin{align} \frac{d}{dx} \nu\left(x\right) &= \frac{d}{dx} \int_{0}^{\infty}\!\! \frac{x^{t}}{\Gamma\left(t+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{tx^{t-1}}{\Gamma\left(t+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{t-1}}{\Gamma\left(t\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{-1+t}}{\Gamma\left(-1+t+1\right)}dt\\ &= \nu\left(x,-1\right) \end{align} \] \[ \frac{d^n}{dx^n} \nu\left(x\right) = \nu\left(x,-n\right) \]

\(\displaystyle\frac{d^n}{dx^n} \mu\left(x,\beta,\alpha\right) = \mu\left(x,\beta,\alpha-n\right) \)

\[ \begin{align} \frac{d}{dx} \mu\left(x,\beta,\alpha\right) &= \frac{d}{dx} \int_{0}^{\infty}\!\! \frac{x^{\alpha+t}t^{\beta}}{\Gamma\left(\beta+1\right)\Gamma\left(\alpha+t+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{(\alpha+t)x^{\alpha+t-1}t^{\beta}}{\Gamma\left(\beta+1\right)(\alpha+t)\Gamma\left(\alpha+t\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{\alpha+t-1}t^{\beta}}{\Gamma\left(\beta+1\right)\Gamma\left(\alpha+t\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{(\alpha-1)+t}t^{\beta}}{\Gamma\left(\beta+1\right)\Gamma\left((\alpha-1)+t+1\right)}dt\\ &= \mu\left(x,\beta,\alpha-1\right) \end{align} \] \[ \frac{d^n}{dx^n} \mu\left(x,\beta,\alpha\right) = \mu\left(x,\beta,\alpha-n\right) \]

\(\displaystyle\frac{d^n}{dx^n} \mu\left(x,\beta\right) = \mu\left(x,\beta,-n\right) \)

\[ \begin{align} \frac{d}{dx} \mu\left(x,\beta\right) &= \frac{d}{dx} \int_{0}^{\infty}\!\! \frac{x^{t}t^{\beta}}{\Gamma\left(\beta+1\right)\Gamma\left(t+1\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{tx^{t-1}t^{\beta}}{\Gamma\left(\beta+1\right)t\Gamma\left(t\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{t-1}t^{\beta}}{\Gamma\left(\beta+1\right)\Gamma\left(t\right)}dt\\ &= \int_{0}^{\infty}\!\! \frac{x^{(-1)+t}t^{\beta}}{\Gamma\left(\beta+1\right)\Gamma\left((-1)+t+1\right)}dt\\ &= \mu\left(x,\beta,-1\right) \end{align} \] \[ \frac{d^n}{dx^n} \mu\left(x,\beta\right) = \mu\left(x,\beta,-n\right) \]



Referències

	Erdélyi, A.; Magnus, W.; Oberhettinger, F.; and Tricomi, F. G. 
	Higher Transcendental Functions, Vol. 1. 
	New York: Krieger, p. 388, 1981a.
	
	Erdélyi, A.; Magnus, W.; Oberhettinger, F.; and Tricomi, F. G. 
	Ch. 18 in Higher Transcendental Functions, Vol. 3. 
	New York: Krieger, p. 217, 1981b.
	
	Gradshteyn, I. S. and Ryzhik, I. M. 
	"The Functions nu(x), nu(x,a), mu(x,beta), mu(x,beta,alpha), lambda(x,y)." 
	§9.64 in Tables of Integrals, Series, and Products, 
	6th ed. San Diego, CA: Academic Press, p. 1109, 2000.
	
	Prudnikov, A. P.; Marichev, O. I.; and Brychkov, Yu. A. 
	Integrals and Series, Vol. 3: 
	More Special Functions. Newark, NJ: Gordon and Breach, 1990. 
	
	Jahnke, E. and Emde, F. 
	Tables of Functions with Formulae and Curves, 
	4th ed. New York: Dover, 1945.
	
	Weisstein, Eric W. "Nu Function." 
	From MathWorld--A Wolfram Web Resource. 
	http://mathworld.wolfram.com/NuFunction.html