I'm interested in evaluating the integral $\int_{0}^{\infty} e^{ - a x } \sinh( \pi x ) K_{ix}(b) K_{ix}(c) \,dx$, where $0 < a < \pi$, and $b, c >0$. I've notice that if I evaluate the following integral then I should be in the clear: $$ I(A,B,C) \ = \ \int_{0}^{\infty} e^{ - A x } K_{ix}(B) K_{ix}(C)\,dx $$

Interestingly, in Gradshteyn and Ryzhik (Eq 3 in Ch 6.79) I have encountered the following integral: $$ \int_{-\infty}^{\infty} e^{(\pi - \gamma)x} K_{ix+iy}(a) K_{ix+iz}(b) = \pi e^{- \beta y - \alpha z} K_{iy-iz}(c)\,dx $$

where $0<\gamma<\pi$, and $\alpha, \beta,\gamma$ are the angles of the triangle with sides $a,b,c>0$. This reduces to the following integral more similar to mine: $$ \int_{-\infty}^{\infty} e^{(\pi - \gamma)x} K_{ix}(a) K_{ix}(b)\,dx = \pi K_{0}\left(\sqrt{ a^2 + b^2 - 2 a b \cos( \gamma ) } \right) $$

This is almost what I need! I just need the limits of integration to match up with mine - I have wasted many hours last night trying to transform this and other integrals in G&R to try and come up with something. Is there any way to do this?

P.S. $K_{ix}(\alpha)$ is of course the modified Bessel function of second kind, of order $ix$, evaluated at the point $\alpha$.

  • 1
    $\begingroup$ Have you tried the usual tricks, like replacing the function with its integral and/or infinite series expression, and then reverting the order of operations ? $\endgroup$ – Lucian Oct 4 '17 at 8:54
  • $\begingroup$ I have not tried the series idea yet, I will try that tonight. I have tried writing $K_{\nu}(x) K_{\nu}(y) \ = \ \frac{1}{2} \int_{0}^{\infty} \frac{1}{2a} \exp\left( - \frac{a}{2} - \frac{x^{2} + y^{2}}{2a}\right)\ K_{\nu}\left( \frac{xy}{a} \right) \ dp$ but so far no luck $\endgroup$ – Greg.Paul Oct 6 '17 at 2:14
  • 1
    $\begingroup$ Have you considered Lebedev's transform? $\endgroup$ – mathstackuser12 Oct 7 '17 at 7:01
  • $\begingroup$ Yes I have. I looked up a bunch of the lists of Kontorovich-Lebedev transforms, and nothing so far. I have found integrals over $\cosh(ax)K_{ix}(b)K_{ix}(b)$ as well as over $x\sinh(ax)K_{ix}(b)K_{ix}(b)$ for $x \in (0,\infty)$, but this has not brought me closer. $\endgroup$ – Greg.Paul Oct 9 '17 at 22:07

I’ll post what I have so far, as requested and because someone might pick up a way forward from the end result. It is also too long for a comment but should not be considered a solution to the problem. I feel that there might actually be a closed from worth chasing.

Firstly the function of interest has the integral representation of $$u\left( x,y,t \right)=\int\limits_{0}^{\infty }{{{e}^{-ts}}\sinh \left( \pi s \right){{K}_{is}}\left( x \right){{K}_{is}}\left( y \right)ds}.$$ It should be noted that this function appears to satisfy a diffusion like differential equation, so it might help to post some background information regarding where all this has come from. The reason I say that is that these sort of integrals – in terms of Lebedev’s transform / index transforms – have a very natural setting in eigenfunction expansions (see Titchmarsh - eigenfunction expansions vol 1). So there might be a natural interpretation of it, and it’s properties, within that framework. Note the identity $${{e}^{-ts}}=\frac{2}{\pi }\int\limits_{0}^{\infty }{\frac{w}{{{w}^{2}}+{{s}^{2}}}\sin \left( wt \right)du}\ \ \ \ \operatorname{Re}\left( t,s \right)>0$$ And so we have $$\frac{\partial }{\partial t}u\left( x,t \right)=-\int\limits_{0}^{\infty }{\frac{s}{{{w}^{2}}+{{s}^{2}}}\sinh \left( \pi s \right){{K}_{is}}\left( x \right){{K}_{is}}\left( y \right)ds}\frac{2}{\pi }\int\limits_{0}^{\infty }{w\sin \left( wt \right)dw}$$ Now from Bateman’s Integral Transforms vol. 2 pg 176 12.1.8 we have $$\int\limits_{0}^{\infty }{\frac{s}{{{n}^{2}}+{{s}^{2}}}\sinh \left( \pi s \right){{K}_{is}}\left( x \right){{K}_{is}}\left( y \right)ds}=\left\{ \begin{matrix} \frac{{{\pi }^{2}}}{2}{{I}_{n}}\left( y \right){{K}_{n}}\left( x \right) & 0<y<x \\ \frac{{{\pi }^{2}}}{2}{{I}_{n}}\left( x \right){{K}_{n}}\left( y \right) & x<y<\infty \\ \end{matrix} \right.$$ This is for n, a natural number. Suppose we can analytically continue it then we would have for example

$$\frac{\partial }{\partial t}u\left( x,y,t \right)=-\pi \int\limits_{0}^{\infty }{w{{I}_{w}}\left( y \right){{K}_{w}}\left( x \right)\sin \left( wt \right)dw}$$

for $0<y<x$, and a similar representation for $x<y$ etc. Integrating we find $$u\left( x,y,t \right)=\pi \int\limits_{0}^{\infty }{{{I}_{w}}\left( y \right){{K}_{w}}\left( x \right)\cos \left( wt \right)dw}+{{c}_{1}}\left( x,y \right)\ \ \ \ 0<y<x$$ $$u\left( x,y,t \right)=\pi \int\limits_{0}^{\infty }{{{I}_{w}}\left( x \right){{K}_{w}}\left( y \right)\cos \left( wt \right)dw}+{{c}_{2}}\left( x,y \right)\ \ \ \ 0<x<y$$

Note: ${{K}_{w}}\left( x \right)\sim \sqrt{\frac{\pi }{2w}}{{\left( \frac{ex}{2w} \right)}^{-w}}$ and ${{I}_{w}}\left( x \right)\sim \sqrt{\frac{\pi }{2w}}{{\left( \frac{ex}{2w} \right)}^{w}}$ for large w. Now observe that from the original representation that $\underset{t\to \infty }{\mathop{\lim }}\,u\left( x,y,t \right)=0$, by the Riemann-Lebesgue lemma. Therefore also by RL the first term in the above representations becomes zero in the limit, and hence we must have therefore ${{c}_{1}}={{c}_{2}}=0$. We have then $$u\left( x,y,t \right)=\pi \int\limits_{0}^{\infty }{{{\kappa }_{w}}\left( x,y \right)\cos \left( wt \right)dw}$$

where $\kappa$ takes on the different forms depending on $x<y$ or $y<x$. At the moment the reason why I think all this works and is worth pursuing is that a numerical integration comparison of this representation to that of the original representation is spot-on to many decimal places for a wide range of x,y that I have tested. Obviously not a proof, but I think one can ‘feel' that the justification to move n away from the naturals and to a real number wouldn’t be a stretch. Another reason is that if you now use Lebedev’s formula (pg 140 from special functions), you will arrive at the formula you have already derived (second last formula). Since Lebedev's formula is valid for arbitrary indices, then this might end up constituting a 'proof' of continuation.

Note from the form above we have $${{\hat{u}}_{C}}\left( x,y,w \right)=\pi {{K}_{w}}\left( x \right){{I}_{w}}\left( y \right)$$ where ${{\hat{u}}_{C}}$ is the cosine transform of u, which may be useful (?). I have various other representations but nothing that is heading to anything I’d call ‘closed’.


I have looked through various tables of Kontorovich-Lebedev transfroms and found nothing (any lists I have found are quite short).

My attempt at my integral of interest: $$ G(a,b,c) = \int_{0}^{\infty} dx\ e^{-ax} \sinh( \pi x ) K_{ix}(b) K_{ix}(c) $$

On page 111 of Yakubovich's ``the hypergeometric approach to integral transforms and convolutions'' I have found the following identity: $$ \sinh( \pi x ) K_{ix}(b) K_{ix}(c) = \frac{\pi}{2} \int_{\big| \ln\left( \frac{b}{c} \right) \big|}^{\infty} J_{0} \left( \sqrt{ 2 b c \cosh(u) - b^2 - c^2 } \right) \sin( u x ) du $$

Which allows me to write: $$ G(a,b,c) = \frac{\pi}{2} \int_{\big| \ln\left( \frac{b}{c} \right) \big|}^{\infty} du\ J_{0} \left( \sqrt{ 2 b c \cosh(u) - b^2 - c^2 } \right) \frac{u}{u^2 + a^2} $$

Doing a coordinate transformation, I was able to get this into the form: $$ G(a,b,c) = \frac{\pi}{2} \int_{0}^{\infty} d\lambda \ \frac{J_{0}(2\sqrt{\lambda})}{\sqrt{ \left( \lambda + \frac{1}{4}(b-c)^2 \right)\left( \lambda + \frac{1}{4}(b+c)^2 \right) }} \frac{\cosh^{-1}\left( \frac{4 \lambda + b^2 + c^2}{2 b c} \right)}{ \left[ \cosh^{-1}\left( \frac{4 \lambda + b^2 + c^2}{2 b c} \right) \right]^2 + a^2} $$

My hope in doing this was to take the series expansion of $J_{0}(2\sqrt{\lambda})$, but this has not been fruitful.

Does anyone have any ideas on how to complete this integral (or through a different avenue)?

  • 1
    $\begingroup$ In your second last formula, rather than performing a change of variable, you could, perhaps, use the summation theorem for the bessel funcction (see GR v7 section 8.53 for eg). A quick sketch seems to yeild tractable integrals (although you are left with a summation over G functions). However there may be issues wtih swapping summation and integration. $\endgroup$ – mathstackuser12 Oct 11 '17 at 10:03
  • $\begingroup$ Very interesting, there is this one in particular looks useful: $J_{0}\left( \sqrt{ 2 b c \cosh(u) - b^2 - c^2 } \right) = I_{0}(b) I_{0}(c) +2 \sum_{n=1}^{\infty} (-1)^n I_{n}(b) I_{n}(c) \cosh(u)$. The problem is though, when integrating over this series along with the factor $\frac{u}{u^2+a^2}$, even the first term in the series seems to diverge. $\endgroup$ – Greg.Paul Oct 11 '17 at 21:14
  • 1
    $\begingroup$ Yes as soon as I saw that I couldn’t edit the comment. The problems with convergence there forced me to go another way, so consider for example $$u\left( x,y,t \right)=\int\limits_{0}^{\infty }{{{e}^{-ts}}\sinh \left( \pi s \right){{K}_{is}}\left( x \right){{K}_{is}}\left( y \right)ds}$$ I have managed to get this to $$u\left( x,y,t \right)=\pi \int\limits_{0}^{\infty }{{{I}_{u}}\left( y \right){{K}_{u}}\left( x \right)\cos \left( ut \right)du},\,\,\,0<y<x$$ A similar result holds when x<y, (just swap them). More soon, hopefully. $\endgroup$ – mathstackuser12 Oct 15 '17 at 8:08
  • $\begingroup$ This seems like an interesting manipulation. Did you derive this using an identity for $I_u(x) K_u(y)$ from page 140 of Lebedev's book? I've been trying to use it with no success. If you post this manipulated form as an answer I can mark the question as answered, since I am starting to lose hope that this integral can actually be evaluated analytically. $\endgroup$ – Greg.Paul Oct 16 '17 at 4:35
  • $\begingroup$ I have been trying to reproduce your result with no luck - how did you do this? $\endgroup$ – Greg.Paul Oct 16 '17 at 19:02

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.