# Modified Bessel Function and Dirac Delta function

Is there a representation of the Dirac delta function of the form $$\int_0^\infty \frac{d z}{z} K_{i \nu} (z) K_{i \nu'} (z) = f(\nu) \delta ( \nu - \nu' ) , \qquad \nu,\nu'>0.$$ for some function $$f(\nu)$$ which I would like to fix.

I believe that there should be an identity of this form, but I'm struggling to find a reference to this anywhere.

• $K_\nu(z)$ is the modified Bessel function. I can't just choose it arbitrarily. Commented Jun 5 at 16:35
• Oh, I thought it was some kernel you where trying to solve for. I see the (bessel-functions) tag now, but you could've also mentioned that in the question itself. Commented Jun 5 at 16:59
• @coiso - it’s literally in the title of the question. Commented Jun 5 at 17:02
• Welp. That's embarassing. Commented Jun 5 at 17:06
• @coiso hahah :) Commented Jun 5 at 17:07

EDIT: After much search, I finally found that the result I am looking for is a consequence of the Kontorovich–Lebedev transform (Wiki). A relevant math paper can be found here. The statement is the following: $$\boxed{ \int_0^\infty \frac{d z}{z} K_{ia} (z) K_{ib} (z) = \frac{\pi}{2} |\Gamma(ia) |^2 \delta ( a-b ) , \qquad a,b>0}$$ This holds in an appropriate space of test functions, as discussed in the linked paper above.

EDIT - Apparently, it was not clear how this is derived from the formulas in the paper so I'm adding the details here. Equations (1.1) and (1.2) of the paper are $$G(\tau) = \int_0^\infty \frac{K_{i\tau}(x)}{x} g(x) dx \tag{1.1}$$ $$g(x) = \frac{2}{\pi^2} \int_0^\infty \tau \sinh (\pi \tau) K_{i\tau}(x) G(\tau) d \tau \tag{1.2}$$ Substituting (1.2) into (1.1), we find \begin{align} G(\tau) &= \int_0^\infty \frac{K_{i\tau}(x)}{x} \left( \frac{2}{\pi^2} \int_0^\infty \tau' \sinh (\pi \tau') K_{i\tau'}(x) G(\tau') d \tau' \right) dx \\ &= \int_0^\infty G(\tau') \left( \frac{2}{\pi^2} \tau' \sinh (\pi \tau') \int_0^\infty \frac{K_{i\tau}(x)}{x} K_{i\tau'}(x) dx \right) d \tau' \end{align} This holds for any function $$G(\tau)$$. From this, we can read off $$\frac{2}{\pi^2} \tau' \sinh (\pi \tau') \int_0^\infty \frac{K_{i\tau}(x)}{x} K_{i\tau'}(x) dx = \delta(\tau-\tau')$$ This holds for all $$\tau,\tau' > 0$$. Rearranging and using the property $$|\Gamma(i\tau)|^2 = \frac{\pi}{\tau \sinh(\pi \tau)}$$, we find $$\int_0^\infty \frac{1}{x} K_{i\tau}(x) K_{i\tau'}(x) dx = \frac{\pi}{2}|\Gamma(i\tau)|^2 \delta(\tau-\tau')$$

• I cannot find the equation you are mentioning in the paper you linked, could you indicate where precisely it is in the paper? Commented Jun 7 at 22:32
• @LL3.14 - The relevant equations are the first two equations of the paper (1.1) and (1.2). You have to substitute g(x) given by (1.2) into (1.1) and the identity mentioned here follows. Commented Jun 8 at 5:46
• @StevenClark - Please see the comment above. I've also clearly mentioned that the identity holds for $a,b>0$, so you cannot set $-b$ in the formulas. Of course, given that the LHS is an even function, you can extend the result to include negative values of $b$, if you wish. Commented Jun 8 at 5:48
• @StevenClark - To be extremely precise, what I mean by the boxed equation (as is always true with Dirac delta functions), is that for any function $f(b)$, we have $\int_0^\infty d b f(b) \left[ \int_0^\infty \frac{d z}{z} K_{ia} (z) K_{ib} (z) \right] = \int_0^\infty d b f(b) \left[ \frac{\pi}{2} |\Gamma(ia) |^2 \delta ( a-b ) \right] = \frac{\pi}{2} |\Gamma(ia) |^2 f(a)$. That is the correct way to interpret the boxed formula in my answer. This is the version of the formula you will find in the paper I linked. Commented Jun 8 at 6:02
• @LL3.14 - $|\Gamma(ia)|^2 = \frac{\pi}{a \sinh(\pi a)}$. I'll add details. Commented Jun 8 at 7:56

I originally suspected you were confusing poles (complex infinities) with Dirac delta functions.

Consider the Mellin transform result

$$F_{a,b}(s)=\mathcal{M}_z[K_{i a}(z)\, K_{i b}(z)](s)=\int_0^{\infty } K_{i a}(z)\, K_{i b}(z)\, z^{s-1} \, dz=\frac{2^{s-3} \Gamma \left(\frac{1}{2} (s+i (a+b))\right) \Gamma \left(\frac{1}{2} (s-i (a+b))\right) \Gamma \left(\frac{1}{2} (s+i (a-b))\right) \Gamma \left(\frac{1}{2} (s-i (a-b))\right)}{\Gamma (s)}\tag{1}$$

provided by Mathematica and WolframAlpha.

Note that when $$b\ne\pm a$$, $$F_{a,b}(s)$$ evaluates to zero at $$s=0$$ because the $$\Gamma(s)$$ term in the denominator has a pole at $$s=0$$, but when $$b=\pm a$$ two of the $$\Gamma$$ terms in the numerator of $$F_{a,b}(s)$$ have poles at $$s=0$$ as well as the $$\Gamma(s)$$ term in the denominator having a pole at $$s=0$$.

Assuming the possibility

$$\underset{s\to 0+}{\text{lim}}\, F_{a,b}(s)=f(a, b)\, (\delta(a-b)+\delta(a+b))\tag{2}$$

consider the table of Mathematica evaluations below using $$b=1$$ where the integral is evaluated using numeric integration.

$$\begin{array}{cc} s & \int\limits_{-\infty}^{\infty} F_{a,1}(s) \, da \\ \frac{1}{10} & 0.962155 \\ \frac{1}{100} & 0.866912 \\ \frac{1}{1000} & 0.855844 \\ \frac{1}{10000} & 0.233828 \\ \frac{1}{100000} & 0.0644596 \\ \frac{1}{1000000} & 0.00659243 \\ \end{array}$$

The table of Mathematica evaluations above seems to suggest that

$$\lim\limits_{s\to 0} \left(\int\limits_{-\infty}^{\infty} F_{a,1}(s) \, da\right)=0\tag{3}$$

whereas

$$\int\limits_{-\infty}^{\infty} f(a, 1)\, (\delta(a-1)+\delta(a+1)) \, da=f(1, 1)+f(-1, 1)\tag{4}.$$

If formula (3) above were true the only way formula (2) above could also be true is if $$f(1, 1)=f(-1, 1)=0$$ which implies $$\underset{s\to 0+}{\text{lim}}\, F_{a,1}(s)=0$$ or if $$f(1, 1)=-f(-1, 1)$$ which is inconsistent with $$\underset{s\to 0+}{\text{lim}}\, F_{a,1}(s)$$ and $$\delta(a-1)+\delta(a+1)$$ both being even functions of $$a$$.

I originally evaluated $$\int\limits_{-\infty}^{\infty} F_{a,1}(s) \, da$$ for various values of s in a table context and Mathematica didn't complain, but I more recently noticed Mathematica issues warnings about precision and accuracy when evaluating the integral for various values of s in standalone contexts. So I now suspect the results of the numerical integration in the table above are incorrect.

I see now the Kontorovich–Lebedev transform (defined in formula 1.1 here)

$$\mathcal{KL}_x[K_{i b}(x)](a)=\int\limits_0^\infty K_{i b}(x) \frac{K_{i a}(x)}{x} \, dx=\frac{\pi}{2}\, |\Gamma(i a)|^2\, \delta(a-b)\,,\quad a,b>0$$

is consistent with the inverse Kontorovich–Lebedev transform (defined in formula 1.2 here)

$$\mathcal{KL}^{-1}_a[\frac{\pi}{2}\, |\Gamma(i a)|^2\, \delta(a-b)](x)=\frac{2}{\pi^2} \int\limits_0^{\infty} \frac{\pi}{2}\, |\Gamma(i a)|^2\, \delta(a-b)\, a\, \sinh(\pi a)\, K_{i a}(x) \, da\\=\frac{1}{\pi}\, |\Gamma(i b)|^2\, b \sinh(\pi b)\, K_{i b}(x)=K_{i b}(x)\,,\quad b>0$$

• Poles can be related to Dirac delta functions and I think the function you've written down has exactly the right structure for it. Let me think about and I'll get back to you. Commented Jun 6 at 11:36
• For instance, consider the function $\frac{s}{x^2+s^2}$. If $x \neq 0$, then it vanishes at $s=0$. On the other hand, if $x=0$, then it has a simple pole at $s=0$. This function precisely represents the Dirac delta function and we can write $\lim_{s \to 0} \frac{s}{x^2+s^2} = \pi \delta(x)$. I believe the complicated products of Gamma functions you have written down have precisely this structure and limits to the Dirac delta function I conjectured in my answer. I am checking it now. Commented Jun 6 at 11:39
• Inspired by your idea, I believe I have a derivation of the required result. Can you check if you agree with my answer? I disagree with the second part of your answer. Commented Jun 7 at 10:14
• After much effort, I have found a reference that gives the result I derived (up to a factor of 2 that I have been unable to fix). Something is incorrect with your derivation. The numerical integration, in particular, is certainly incorrect. Commented Jun 7 at 19:16