I would like to compute the integral of the second modified Bessel function which has the following form $$ \int^{\infty}_{\epsilon}{dz \;z^a K_{\nu}(z)} $$
where I have some power of $z$ multiplied by the function and later I would like to take the limit of $\epsilon\rightarrow 0$. I am aware that this is some kind of hypergeometric function but I couldn't find an answer anywhere.
On
Provided $a \ge\nu\ge 0$, Maple says $$ \lim_{\epsilon\to 0^+}\int_{\epsilon}^{\infty }\!{z}^{a}{{\rm K}_{\nu}\left(z\right)} \,{\rm d}z = {\frac {{2}^{a-1}{\pi}^{2}}{\Gamma \left( 1/2-\nu/2-a/2 \right) \Gamma \left( \nu/2-a/2+1/2 \right) \left( \cos^2 \left( \pi\,a /2\right)+ \cos^2 \left( \pi\,\nu/2 \right) -1 \right) }} $$
In case you really need it, here it is without the limit:
(-sin(Pi*nu)*epsilon^(-nu+a+1)*cos(1/2*Pi*(-nu+a))*cos(1/2*Pi*(nu+a))*GAMMA(1/2-1/2 *nu-1/2*a)*GAMMA(1/2*nu-1/2*a+1/2)*nu*GAMMA(nu)^2*2^(-1+nu)*(nu+a+1)*hypergeom( [-1/2*nu+1/2*a+1/2],[1-nu, 3/2-1/2*nu+1/2*a],1/4*epsilon^2)+(epsilon^(nu+a+1)*cos(1/2* Pi*(-nu+a))*cos(1/2*Pi*(nu+a))*GAMMA(1/2-1/2*nu-1/2*a)*GAMMA(1/2*nu-1/2*a+1/2)* 2^(-nu-1)*(-nu+a+1)*hypergeom([1/2*nu+1/2*a+1/2],[nu+1, 3/2+1/2*nu+1/2*a],1/4* epsilon^2)+nu*GAMMA(nu)*sin(Pi*nu)*Pi*((a^2-nu^2+1)*2^(a-1)+a*2^a))*Pi)/cos(1/2*Pi* (nu+a))/cos(1/2*Pi*(-nu+a))/(-nu+a+1)/GAMMA(1/2-1/2*nu-1/2*a)/GAMMA(1/2*nu-1/2* a+1/2)/(nu+a+1)/sin(Pi*nu)/GAMMA(nu)/nu
On
Claim:
$$\int_{0}^{\infty}t^{s-1}K_{\nu}(t)dt=2^{s-2}\,\,\Gamma\left(\frac{s}{2}+\frac{\nu}{2} \right)\Gamma\left(\frac{s}{2}-\frac{\nu}{2} \right)$$
Recall the integral representation:
$$K_{\nu}(x)=\int_0^{\infty} e^{-x \cosh t} \cosh (\nu t) dt$$
Then
$$ \begin{aligned} I&=\int_{0}^{\infty}x^{s-1 }\int_0^{\infty} e^{-x \cosh t} \cosh (\nu t) dt\,dx\\ &=\int_{0}^{\infty}\cosh (\nu t) \int_0^{\infty} e^{-x \cosh t}x^{s-1 } dx\,dt\\ &=\int_{0}^{\infty}\cosh (\nu t)\cdot \frac{\Gamma\left(s\right)}{\left( \cosh t\right)^{s}} dt\\ &=\Gamma\left(s\right)\int_{0}^{\infty} \frac{\cosh (\nu t)}{ \cosh ^{s}(t)} dt\\ &=\frac{\Gamma\left(s\right)}{2}\int_{0}^{\infty} \frac{e^{\nu t}+e^{-\nu t}}{ \cosh ^{s}(t)} dt\\ &=2^{s-1}\Gamma\left(s\right)\int_{0}^{\infty} \frac{e^{\nu t}+e^{-\nu t}}{ \left( e^t+e^{-t} \right)^{s}} dt\\ &=2^{s-1}\Gamma\left(s\right)\int_{0}^{\infty}\frac{e^{-st}}{e^{-st}}\cdot \frac{e^{\nu t}+e^{-\nu t}}{ \left( e^t+e^{-t} \right)^{s}} dt\\ &=2^{s-1}\Gamma\left(s\right)\int_{0}^{\infty} \frac{e^{-(s-\nu) t}+e^{-(\nu+s) t}}{ \left( 1+e^{-2t} \right)^{s}} dt\\ &=2^{s-2}\Gamma\left(s\right)\int_{0}^1\frac{w^{\frac{(s-\nu)}{2}-1 }+w^{\frac{(\nu+s)}{2}-1 }}{ \left( 1+w \right)^{s}} \, dw &\left(e^{-2t}=w\right)\\ &=2^{s-2}\Gamma\left(s\right)\frac{\Gamma\left(\frac{s}{2}+\frac{\nu}{2} \right)\Gamma\left(\frac{s}{2}-\frac{\nu}{2} \right)}{\Gamma\left(s \right)} \qquad \blacksquare \end{aligned} $$
Where in the last line we used the Beta function (see below)
$$\int_0^{1}\frac{t^{x-1}+t^{y-1}}{\left(1+t \right)^{x+y}}dt=\frac{\Gamma\left(x \right)\Gamma\left(y \right)}{\Gamma\left(x+y \right)}$$
Proof:
Recall the integral representation for the Beta function
$$\int_0^{\infty}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt=\frac{\Gamma\left(x \right)\Gamma\left(y \right)}{\Gamma\left(x+y \right)}$$
Then,
$$ \begin{aligned} \int_0^{\infty}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt&=\int_0^{1}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt+\int_1^{\infty}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt\\ &=\int_0^{1}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt+\int_1^{0}\frac{w^{1-x}}{\left(1+\frac{1}{w} \right)^{x+y}}\frac{-dw}{w^2}\\ &=\int_0^{1}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt+\int_0^{1}\frac{w^{1-x}w^{x+y}w^{-2}}{\left(1+w \right)^{x+y}}dw\\ &=\int_0^{\infty}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt\\ &=\int_0^{1}\frac{t^{x-1}+t^{y-1}}{\left(1+t \right)^{x+y}}dt \end{aligned} $$
From the Digital Library of Mathematical Functions we know $$ \int_{0}^{\infty} t^{\alpha} K_\nu(t)\, \mathrm{d}t = 2^{\alpha -1} \Gamma\left( \frac{\alpha + \nu +1}{2}\right)\Gamma\left( \frac{\alpha - \nu +1}{2}\right) \qquad \text{for} \quad \lvert\Re(\nu)\rvert< \Re(\alpha)+1 $$