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Let $K_{\alpha}(z)$ be the modified Bessel function of the second kind of order $\alpha$.

I need to compute the following integral:

$$\int_0^\infty\;\;K_0\left(\sqrt{a(k^2+b)}\right)dk$$

where $a>0$ and $b>0$.

I have tried several substitutions and played around a lot in Mathematica, and can't seem to solve this. Perhaps an integral representation of $K_{0}(z)$ would be helpful here.

Even if this can't be done exactly, a sensible approximation strategy would also be useful.

Any advice would be greatly appreciated. Thanks in advance for your time!

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    $\begingroup$ substituting $\sqrt{a}k=x$ it is enough to consider the case $a=1$. $\endgroup$ – vesszabo Jul 11 '13 at 16:58
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Using some of the ideas mentioned in Ron Gordon's answer we can evaluate this integral. Change the variable to $x=\sqrt{a} k$, so that the integral becomes $$ \int_0^\infty K_0(\sqrt{ak^2+ab})\,dk = \frac1{\sqrt{a}} \int_0^\infty K_0(\sqrt{x^2+ab})\,dx, $$ and introduce the function $$ I(b) = \int_0^\infty K_0(\sqrt{x^2+b^2})\,dx, $$ so that the integral is $I(\sqrt{ab})/\sqrt{a}$.

First, making the substitution $x=\sqrt{s^2-b^2}$, we get that $$ I(b) = \int_b^\infty K_0(s)\frac{s\,ds}{\sqrt{s^2-b^2}}, $$ and we can use $K_0'(s) = -K_1(s)$ to write $K_0(s)=\int_s^\infty K_1(u)\,du$, so that $$ I(b) = \int_b^\infty \frac{s\,ds}{\sqrt{s^2-b^2}}\int_s^\infty K_1(u)\,du. $$ The integral has the range $b<s<u<\infty$, and we can do the integral over $s$ explicitly, giving $$ I(b) = \int_b^\infty\sqrt{u^2-b^2}K_1(u)\,du. $$

Second, using the formula $$ K_0(u) = \int_0^\infty e^{-u\cosh t}\,dt, $$ we can write $$ I(b) = \int_b^\infty \frac{s\,ds}{\sqrt{s^2-b^2}}\int_0^\infty e^{-s\cosh t}\,dt = \int_0^\infty b K_1(b\cosh t)\,dt, $$ where the integral over $s$ can be done in closed form in Mathematica. Substituting $t = \text{arccosh}(u/b)$ we get $$ I(b) = \int_b^\infty \frac{b\,du}{\sqrt{u^2-b^2}}\,K_1(u). $$

From the two expressions above we can easily see that $$ \frac{dI}{db} = -I(b), $$ and Mathematica will tell us that $$ I(0) = \int_0^\infty K_0(x)\,dx = \frac\pi2. $$ Therefore, $$I(b) = \frac\pi2 e^{-b}, $$ and $$ \int_0^\infty K_0(\sqrt{a x^2+a b})\,dx = \frac{\pi}{2\sqrt{a}}e^{-\sqrt{a b}}. $$ This matches Ron Gordon's answer, even though his was only an asymptotic calculation.

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  • $\begingroup$ Whoa...no need for me to do any error analysis...or maybe, I should figure out whether all my errors cancel out. Still, good stuff. $\endgroup$ – Ron Gordon Jul 12 '13 at 1:53
  • $\begingroup$ @RonGordon Definitely an amusing integral. Wouldn't even have tried this without your asymptotic being correct to $O(10^{-16})$. $\endgroup$ – Kirill Jul 12 '13 at 1:59
  • $\begingroup$ Next time when I get suspiciously amazing accuracy, I think I'll have reason to find an exact solution. Good work, man! $\endgroup$ – Ron Gordon Jul 12 '13 at 2:02
  • $\begingroup$ Amazing! Thanks so much to @RonGordon and Kirill for helping me with this. Very impressive :) $\endgroup$ – Can't integrate Jul 12 '13 at 7:50
  • $\begingroup$ Minor edit: the "dx" in the first integral should be a "dk" :) $\endgroup$ – Can't integrate Jul 12 '13 at 7:59
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OK, I think I have a way to make an approximation that holds up pretty well by using Laplace's method twice.

Begin by using the representation

$$K_0(u) = \int_0^{\infty} dt \, e^{-u\cosh{t}}$$

Then the integral you seek may be written, when we reverse the order of integration, as

$$\int_0^{\infty} dk \, K_0\left(\sqrt{a (k^2+b)}\right) = \int_0^{\infty} dt \, \int_0^{\infty} dk \, e^{-\sqrt{a} \cosh{t} \sqrt{k^2+b}}$$

This is justified in that both integrals clearly converge absolutely. Now, to apply Laplace's method on the integral over $k$, we may assume that $a \cosh{t}$ is sufficiently large, and we use the approximation that

$$\sqrt{k^2+b} \sim \sqrt{b} + \frac{k^2}{2 \sqrt{b}}$$

We end up with a familiar integral that may be evaluated immediately, and we are left with a single integral:

$$\int_0^{\infty} dk \, K_0\left(\sqrt{a (k^2+b)}\right) \sim \sqrt{\frac{\pi \sqrt{b}}{2 \sqrt{a}}} \int_0^{\infty} \frac{dt}{\sqrt{\cosh{t}}} e^{-\sqrt{a b} \cosh{t}} \quad (a \to \infty)$$

At this point I should say that I am not providing an error estimate, although deriving one should be fairly straightforward. (This involves providing an additional term of the Taylor expansion of the square root above, and then Taylor expanding the exponential in the integrand.)

Now we are left with the problem of evaluating the above integral, which looks only a little less problematic than the original one. Nevertheless, we may apply Laplace's method again, as we are claiming that $a$ is sufficiently large, and $b \gt 0$. In this case, we use the Taylor expansion of $\cosh{t} \sim 1+t^2/2$ and we again end up with a familiar integral. The final result is

$$\int_0^{\infty} dk \, K_0\left(\sqrt{a (k^2+b)}\right) \sim \frac{\pi}{2 \sqrt{a}} e^{-\sqrt{a b}} \quad (a \to \infty)$$

Some random numerical samples - even with moderate values of $a=1$ and $b=1$ in WA have verified the correctness of this approximation.

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    $\begingroup$ In the internal integral over $k$ you can do integration explicitly to find that for $a=1$ the main integral is $\int_{\sqrt{b}}^{\infty}\frac{\sqrt{b}K_1(u)\,du}{\sqrt{u^2-b}}. $ (This makes it easier to apply the method of steepest descent to get the full asymptotic expansion.) $\endgroup$ – Kirill Jul 11 '13 at 21:49
  • $\begingroup$ @Kirill: that is an interesting result, no doubt. But does it make life any easier? The advantage of the method I outline is that a) I stick with very simple, elementary functions, b) I use Laplace's method, from which full asymptotic results are indeed obtainable, and c) the end result is unbelievably accurate for even moderate values of $a$ and $b$. But, tell you what: let's see. Maybe you are right that corrections are more easily done with your integral. I don't know. But I claim that it's a tough world that requires those corrections. $\endgroup$ – Ron Gordon Jul 11 '13 at 21:54
  • $\begingroup$ @Kirill: BTW why $a=1$? I don't see how the value of $a$ determines the difficulty of the integral over $k$. $\endgroup$ – Ron Gordon Jul 11 '13 at 21:58
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    $\begingroup$ It's just a nice special form that I wanted to point out. The integral is equal to $\frac{1}{\sqrt{a}}\int_0^\infty K_0(\sqrt{x^2+ab})\,dx$, so setting $a=1$ just simplifies expressions. $\endgroup$ – Kirill Jul 11 '13 at 22:03
  • $\begingroup$ @Kirill: after knocking myself in the head, I totally see what you mean. When I get home tonight, I will do an error analysis in Mathematica and figure out how far we can push these approximations. I will also see if your suggestion leads anywhere. Thanks! $\endgroup$ – Ron Gordon Jul 11 '13 at 22:05
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$\int_0^\infty K_0\left(\sqrt{a(k^2+b)}\right)dk$

$=\int_0^\infty\int_0^\infty e^{-\sqrt{a(k^2+b)}\cosh t}~dt~dk$

$=\int_0^\infty\int_0^\infty e^{-(\sqrt{a}\cosh t)\sqrt{k^2+b}}~dk~dt$

$=\int_0^\infty\int_0^\infty e^{-(\sqrt{a}\cosh t)\sqrt{(\sqrt{b}\sinh u)^2+b}}~d(\sqrt{b}\sinh u)~dt$

$=\sqrt{b}\int_0^\infty\int_0^\infty e^{-\sqrt{a}\sqrt{b}\cosh t\cosh u}\cosh u~du~dt$

$=\sqrt{b}\int_0^\infty K_1(\sqrt{a}\sqrt{b}\cosh t)~dt$

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The evaluation is a bit easier if we use the integral representation $$K_{0}(x) = \frac{1}{2} \int_{0}^{\infty} \frac{1}{t} \, \exp \left(-t - \frac{x^{2}}{4t} \right) \, dt , \quad x>0, $$ which can be derived from the integral representation $$K_{0}(x) = \int_{0}^{\infty} \exp(-x \cosh t) \, dt, \quad x>0, $$ by extending the interval of integration to the entire real line and then making the substitution $ e^{t} = \frac{2}{z} u$.

Using this representation of the modified Bessel function of the second kind of order zero, we get

$$\begin{align}\int_{0}^{\infty} K_{o}\sqrt{a^{2}\left(x^{2}+b^{2} \right)} &= \int_{0}^{\infty} K_{0} \left(a\sqrt{x^{2}+b^{2}} \right) \, dx \\ &= \frac{1}{2} \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{t} \, \exp \left(-t - \frac{a^{2}(x^{2}+b^{2})}{4t} \right) \, dt \, dx \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{1}{t} \, \exp \left(-t-\frac{a^{2}b^{2}}{4t} \right) \int_{0}^{\infty} \exp \left(-\frac{a^{2}x^{2}}{4t} \right) \, dx \, dt \tag{1}\\ &= \frac{\sqrt{\pi}}{2a} \int_{0}^{\infty} \frac{1}{\sqrt{t}} \, \exp \left(-t-\frac{a^{2}b^{2}}{4t}\right) \, dt \\ &= \frac{\sqrt{\pi}}{a} \int_{0}^{\infty} \exp \left(-u^{2}-\frac{a^{2}b^{2}}{4u^{2}} \, \right) \, du \\ &= \frac{\sqrt{\pi}}{a} \, \sqrt{\frac{\pi}{4}} \exp \left(-2 \, \sqrt{\frac{a^{2}b^{2}}{4}} \right) \tag{2}\\ &= \frac{\pi}{2a} \, \exp(-ab). \end{align}$$


$(1)$ Since the integrand is nonnegative, we're allowed to switch the order of the integration.

$(2)$ How to evaluate $\int_{0}^{+\infty}\exp(-ax^2-\frac b{x^2})\,dx$ for $a,b>0$


This particular integral is a limiting case of the more general integral

$$\small \frac{1}{\beta^{\mu}} \int_{0}^{\infty} J_{\mu}(\beta t) \, \frac{K_{\nu}(a\sqrt{t^{2}+b^{2}}{})}{(t^{2}+b^{2})^{\nu/2}} \, t^{\mu+1} \, dt = \frac{1}{a^{\nu}} \left(\frac{\sqrt{a^{2}+\beta^{2}}}{b} \right)^{\nu-\mu-1}K_{\nu-\mu-1} \left(b \sqrt{a^{2}+\beta^{2}} \right),\tag{3}$$ where $a, \beta >0$, $\Re(b) >0$, and $\Re(\mu) >-1$.

(As $\beta\downarrow 0$, $J_{\mu}(\beta t)\sim \left(\frac{\beta t}{2} \right)^{\mu} \frac{1}{\Gamma(\mu+1)}$.)

Integral $(3)$ is entry (2) in section 47 of chapter 13 of the textbook A Treatise on the Theory of Bessel Functions.

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