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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.

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  • $\begingroup$ didn't get the solution. what is $K_1$ and what are the other symbols? $\endgroup$ – MycrofD Dec 27 '14 at 11:59
  • $\begingroup$ Sorry, the first comment has some wrong symbols, fixed now. $K_1$ is the modified Bessel function $K_a(x)=\frac{\pi}2\frac{I_{-a}(x)-I_a(x)}{\sin(a\pi)}$ $\endgroup$ – Hippalectryon Dec 27 '14 at 12:02
  • $\begingroup$ Relevant : $$\int_{m}^\infty \dfrac{x e^{-\beta x}}{\sqrt{x^2 - m^2}} dx=m K_1(m\beta),m>0,\Re(\beta)>0$$ $\endgroup$ – Hippalectryon Dec 27 '14 at 12:08
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    $\begingroup$ The integrand function is not defined for $x\in[0,m)$, please fix your question. $\endgroup$ – Jack D'Aurizio Dec 27 '14 at 12:24
  • $\begingroup$ Thank u. question edited accordingly. $\endgroup$ – MycrofD Dec 27 '14 at 13:17
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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/

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  • $\begingroup$ Are you sure? it should be $\frac{m^2}{\beta}K_2(m\beta)$. $\endgroup$ – alexjo Dec 27 '14 at 13:35
  • $\begingroup$ @alexjo Yeah, it's just a different way of writing your result that looks simpler. $\endgroup$ – user 1357113 Dec 27 '14 at 13:43
  • $\begingroup$ plz elaborate the steps. and what is the final value? can K_2 be solved to a finite value? or do we have to leave it there? $\endgroup$ – MycrofD Dec 28 '14 at 12:27
  • $\begingroup$ @MycrofD Before posting this answer, I discussed with Hippalectryon in a chat the formula he posted it above. I used it and differentiated it properly with respect to $\beta$ such that I got the integral you wanted to evaluate. I don't think you can simplify further the last expression. $\endgroup$ – user 1357113 Dec 28 '14 at 13:10
  • $\begingroup$ ok. so the last exp cannot be simplified. but the steps.. i am sorry but i cannot get a hold of it. i am trying to use the definition (7) here at mathworld.wolfram.com/… K_1, K_2 and K_3 have different powers to [(x/m)^2 - 1]. So I dont see how they can be algebraically added or subtracted. $\endgroup$ – MycrofD Dec 28 '14 at 13:26
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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.

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  • $\begingroup$ what is the final answer? does it have to be in terms of K? can't it be some finite value? $\endgroup$ – MycrofD Dec 28 '14 at 12:44
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The obvious substitution is $x=m\cosh t$, followed by recognizing the integral expression for the modified Bessel function $K_\alpha(u)=\displaystyle\int_0^\infty\exp(-u\cosh t)\cosh(\alpha t)~dt.$

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