Integration $\int_{m}^\infty {x}{\sqrt{x^2 - m^2}}e^{-\beta x} dx$ How can the following integration be performed? Does it involve Bessel functions?$$\int_{m}^\infty {x}{\sqrt{x^2 - m^2}}e^{-\beta x} dx$$
EDIT: Actually, the original question is:
$$\int_{0}^\infty \frac{y^2}{e^{-\beta \sqrt{y^2 + m^2}}} dy$$
Which means the lower limit is $m$ and not $0$ when $y$ is changed to $x$ as above. Accordingly the question has been edited. Thanks to @Jack D'Aurizio
Please give the final answer in some finite number. Elaboration of steps is encouraged.
 A: Yes, it involves the modified Bessel function of the second kind.
$$\int_{m}^\infty {x}{\sqrt{x^2 - m^2}}e^{-\beta x} dx$$
$$=\frac{1}{2} m^2 \left(m K_1(\beta m)+\frac{1}{2} m (K_1(\beta m)+K_3(\beta m))\right)-m^3 K_1(\beta m)$$
$$=\frac{m^2}{β}K_2(mβ)$$
For getting the last equality, use the recurrence relation $$K_v(z)=K_{v-2}(z)+2\frac{v-1}{z}K_{v-1}(z)$$
http://functions.wolfram.com/Bessel-TypeFunctions/BesselK/introductions/Bessels/05/
A: With the substitution $x=mt$, the integral is:
$$J=m^3\int_1^{\infty} t\sqrt{t^2-1}\,e^{-m\beta t}\,dt$$
Consider,
$$I(a)=\int_1^{\infty} \sqrt{t^2-1}\,e^{-at}\,dt=\frac{K_1(a)}{a}$$
where I used identity 7 from here. 
To get the original integral, differentiate $I(a)$ with respect to $a$ and substitute $a=m\beta$, hence
$$I'(a)=-\frac{K_2(a)}{a} \Rightarrow J=\frac{m^2K_2(m\beta)}{\beta}$$
For the derivative, I had to use Wolfram Alpha.
A: The obvious substitution is $x=m\cosh t$, followed by recognizing the integral expression for the modified Bessel function $K_\alpha(u)=\displaystyle\int_0^\infty\exp(-u\cosh t)\cosh(\alpha t)~dt.$
