6
$\begingroup$

How does one find the following representation of the bessel function $K_v(y)$: $$K_v(y) = \frac{1}{2} \int_{0}^{\infty} t^{v-1} \exp \left(-\frac{1}{2}y\left(t+t^{-1}\right) \right)\,\mathrm{d}t.$$ I have seen many different integral presentations in different sources but couldn't find a proof for this one.

$\endgroup$
6
  • $\begingroup$ where does this beauty come from? $\endgroup$
    – tired
    Mar 1 '16 at 13:40
  • $\begingroup$ @tired If I remember right I think I first found this in a book by Y. Motohashi. $\endgroup$ Mar 1 '16 at 14:14
  • $\begingroup$ For me this is the definition of $K_{\nu}$. What is your preferred definition? $\endgroup$ Mar 1 '16 at 17:44
  • 1
    $\begingroup$ @Startwearingpurple Well none of the books I'm reading gives this one as the definition. Im mostly using the book by Lebedev which defines $K_v(y)$ using $I_v(z)$ which is given as a series. But I've seen proofs for some of the integral representations but not for this one. $\endgroup$ Mar 1 '16 at 19:15
  • $\begingroup$ I don't think expressing $K_v$ in terms of $I_{\pm v}$ is the right way of looking at this function. It is useful as the solution of Bessel equation having nice canonical behavior at infinity (irregular singular point) instead of zero (regular one). The integral representation comes from the fact that Bessel equation appears when one separates the variables in 2D Helmholtz equation: namely, it expresses the corresponding solution as a superposition of plane waves. $\endgroup$ Mar 1 '16 at 22:30
1
$\begingroup$

I have seen many different integral presentations in different sources

Hint: In that case, let $t=e^u,$ maybe it will jog your memory$\ldots$ ;-$)$

$\endgroup$
1
  • $\begingroup$ That actually might work.. there is one step however where I'm stuck at: I can proof for example that $K_v(y)=\int_{0}^{\infty} e^{-y(\text{cosh}(t))} \text{cosh}(vt) \text{d}t$. After letting $z=e^{t}$ I get $\frac{1}{2} \int_{-\infty}^{\infty} e^{-\frac{y}{2} (z+z^{-1}) }(z^{v-1}+z^{-v-1})\text{d}z$. Then I tried to separate the integral to parts from $(-\infty, 0]$ and $[0, \infty)$ but couldn't get to the form I'm looking for (it's close though). Any tips for that? $\endgroup$ Mar 1 '16 at 19:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.