I need to do this integral:

$\int_0^{\infty} p^2 e^{-\beta\sqrt{p^2+1}} \, dp$

($\beta > 0$)

Any technique how to do this integral? (apparently not possible in closed form) or at least produce the result as a rapidly converging series.

A closely related formula (just a Gaussian integral):

$\int_0^{\infty} p^2 e^{-\beta p^2 } \, dp = \frac{\sqrt{\pi }}{4 \beta ^{3/2}}$

and on the other hand:

$\int_0^{\infty} p^2 e^{-\beta p } \, dp = \frac{2}{\beta^3}$

Progress 1: So, intuition tells that for $\beta\to 0$ the integral behaves like $\beta^{-3}$ and for $\beta\to\infty$ as $\beta^{-3/2}$.

Progress 2: From related texts in the field I am getting the hint that it maybe expressable in terms of a modified Bessel function $K_2$.

Progress 3 (solution): From experimenting with Bessel functions I get that

$\int_0^{\infty} p^2 e^{-\beta\sqrt{p^2+1}} \, dp = K_2(\beta)/\beta$

I'll still accept an answer with the derivation

  • $\begingroup$ My idea would be to attempt to expand the exponent into a series. $\endgroup$ Sep 24, 2013 at 10:16
  • $\begingroup$ @DepeHb, none of the integrals of the terms in the series will converge (I tried it). $\endgroup$
    – alfC
    Sep 24, 2013 at 10:22

1 Answer 1


Substitute $p=\sinh{t}$ and get the integral is equal to

$$\int_0^{\infty} dt \, \left ( \cosh^3{t}-\cosh{t}\right) e^{-\beta \cosh{t}}$$

Use the fact that

$$K_1(\beta) = \int_0^{\infty} dt \, \cosh{t}\, e^{-\beta \cosh{t}}$$

and the recurrence relations for modified Bessels; I get as the value of the integral

$$K_1''(\beta)-K_1(\beta) = \frac14 [K_3(\beta)- K_1(\beta)]$$

  • $\begingroup$ Perhaps, if you got $\frac{1}{4}[K_3(\beta) - K_1(\beta)]$ then it would be exactly equal to the answer I found $K_2(\beta)/\beta$. $\endgroup$
    – alfC
    Sep 24, 2013 at 10:04
  • $\begingroup$ @alfC: you were right; my mistake. $\endgroup$
    – Ron Gordon
    Sep 24, 2013 at 10:26

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