For a given $f$ when is there a unique $K$ s.t. $\int_{0}^{\infty}K(z,t)f(t)\mathrm{d}t=f(z)$? Recently I discovered two integral identities involving the sine integral and cosine integrals, $\mathrm{Si}(t)$ and $\mathrm{Ci}(t)$ respectively.
These are 
$$\int_{0}^{\infty} K_1(z,t)\frac{\gamma+\log(t)-\mathrm{Ci}(t)}{t}\mathrm{d}t = \frac{\gamma+\log(z)-\mathrm{Ci}(z)}{z}$$
$$\int_{0}^{\infty} K_2(z,t)\frac{\mathrm{Si}(t)}{t}\mathrm{d}t = \frac{\mathrm{Si}(z)}{z}$$
where
$$K_1(z,t) = \frac{t}{z^2-t^2}\left[tJ_1(z)J_0(t)-zJ_0(z)J_1(t)\right]$$
$$K_2(z,t) = \frac{t}{z^2-t^2}\left[zJ_1(z)J_0(t)-tJ_0(z)J_1(t)\right]$$
and $J_v(t)$ is the Bessel function of the first kind.
This got me thinking, are these $K$'s unique?  In general, for a given function $f(t)$ are there criteria for determining when there exists a unique $K(z,t)$ such that
$$\int_{0}^{\infty}K(z,t)f(t)\mathrm{d}t=f(z) \text{  ?}$$
Inasmuch, is there a name for this sort of integral identity, that is an integral where a function $f$ is sent to itself?
 A: They are not unique. E.g. For any function $g$ such that $\int_0^\infty f(t)g(t) dt \neq 0$ exists (i.e. $\int_0^\infty |f(t)g(t)| dt < \infty$), we can define $K(z,t) 
= \frac{f(z)g(t)}{\int_0^\infty f(t)g(t) dt}$, and so
$$\int_0^\infty K(z,t)f(t) dt = f(z)\frac{\int_0^\infty f(t)g(t) dt}{\int_0^\infty f(t)g(t) dt} = f(z).$$
We then see that there are a lot of these functions if $f \neq 0$ almost everywhere. And these are just "the trivial ones", because otherwise the operators are more interesting.
In general, one defines a function $K$ as a "kernel" and then we get the integral operator: $T[f](z) = \int K(z,t)f(z)$, where we restrict to the $f$ such that this integral makes sense. If $\lambda$, a complex number, is such that there is a nonzero function $f(z)$ such that $T[f](z) = \lambda f(z)$, then we call $\lambda$ an eigenvalue of $T$ and $f$ a $\lambda$-eigenvector of $T$. For your question, you are asking if every  (non-zero) function $f(z)$ can be expressed as a $1$-eigenvector of some integral operator. This is the language of linear algebra.
Here is a Wiki article on the subject:
https://en.wikipedia.org/wiki/Integral_transform
Note: To prove unqiueness, you would have to show that if $K_1$ and $K_2$ are kernels then $$\int_0^\infty K_1(z,t)f(t)dt = f(z) = \int_0^\infty K_2(z,t)f(t)dt$$ and hence $$\int_0^\infty (K_2(z,t)- K_1(z,t))f(t)dt = 0.$$ However this only requires that $f$ be in "the kernel" of the operator $T'$ with kernel $K_1 - K_2$. To show uniqueness, you would have to show that if $T'$ is an integral operator with kernel $K'$ and $T'[f] = 0$ then $K' = 0$ and hence $T' = 0$. However, this is not a restrictive statement at all. For example, we can find a non-zero function $h(t)$ such that $\int_0^\infty f(t)h(t)dt = 0$, in fact very many. (Try it yourself. Define $h$ first on $[0,1]$ to be anything such that $\int_0^1 f(t)h(t)dt$ makes sense. Then extend it to $(1,\infty)$ so that $\int_0^\infty f(t)h(t)dt$ makes sense and equals zero.) Then if $T'$ has kernel $h(t)$, $T[f](z) = \int_0^\infty f(t)h(t)dt = 0$.
These ideas are all in the area where advanced calculus/analysis interact with linear algebra. Namely, function spaces and their linear operators.
