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