Existence of self-Laplace transforms

Question

There are many functions that are self-Fourier transforms, such as $e^{-\pi x^2}$ or $\frac{1}{\cosh(\pi x)}$, and this property may be used to prove some interesting theorems such as the functional equation for the theta function or an integral relation like this.

I am wondering if the same can be said of self-Laplace transforms. Are there any useful functions that are their own Laplace transform, and can this property be exploited to give any interesting consequences? Here is one example of such a function that may be constructed, but it seems artificial and of no significance:

Suppose $f$ is a function of the form $f(t) = C_1t^{s-1} +C_2 t^{-s}$, where $0<\text{Re}(s)<1$ so that the Laplace transform exists and $C_1$ and $C_2$ are some constants. We may now assume $f$ is its own Laplace transform and solve for $s$ and the constants: $$ C_1 x^{s-1} + C_2 x^{-s} =\mathcal{L}(f(t))=\int_0^{\infty} f(t) e^{-xt} \, dt = C_1\int_0^{\infty} t^{s-1} e^{-xt} \, dt + C_2 \int_0^{\infty} t^{-s} e^{-xt} \, dt $$ $$ = C_1 \Gamma(s) x^{-s} + C_2 \Gamma(1-s) x^{s-1} $$ We therefore need $C_1 =C_2 \Gamma(1-s)$ and $C_2 = C_1 \Gamma(s)$, and so $ 1= \Gamma(s) \Gamma(1-s) = \pi \csc(\pi s)$, which has the unique solution $s=\frac{1}{2} \pm \frac{i}{\pi} \log(\pi + \sqrt{\pi^2-1}) $ in the strip $0<\text{Re}(s)<1$, and $\frac{C_2}{C_1}=\Gamma(s)$. Choosing $C_1=1$, our self-Laplace transform is $$f(t) = t^{s-1} + \Gamma(s) t^{-s}, \text{ where } s = \frac{1}{2} \pm \frac{i}{\pi} \log \left(\pi+\sqrt{\pi^2-1} \right) $$

Answer 1

I've run into a similar problem from Schaum's outlines, Murray Spiegel, Ch. 1 pb. 163.

Let $$a, b,\alpha, \beta, \;\text{and}\;\Lambda$$ be constants. Prove that $$\mathcal{L}[{at^{-\alpha} +bt^{-\beta}}]=\Lambda [{as^{-\alpha} +bs^{-\beta}}] $$

if and only if $$\alpha+\beta=1 $$ and $$\Lambda=\sqrt{\pi*\csc(\pi*\alpha)}.$$

What I'm having problems with is the Gamma factoring out of the s terms to produce the Lambda.

Answer 2

It depends on what you (somewhat subjectively) may consider as useful or not.
What you are asking for can be rephrased as:
"Are there eigenfunctions with unitary eigenvalue for the exponential kernel of the Laplace transform?"
or, more precisely:
"Is the exponential kernel of the Laplace transform a reproducing kernel?"
Also the answer depends on what you are looking for: an integrable function ($\in L_1(\mathbb{X})$), a square-integrable function ($\in L_2(\mathbb{X})$, a distribution?
For integrable functions, the answer is: $\frac{1}{\sqrt{x}}$ is an eigenfunction, but the eigenvalue is not unitary. Yet it is not in $L_2(\mathbb{]0,a]})$.
Your "self-Laplace" form is not a real Laplace transform as it is only defined at one point. You would need to define it for a given domain of $\mathbb{C}$.

However, there really is a highly useful close friend to the Laplace transform which is a self-reproducing kernel. It is the so-called exponential kernel, which is intensively used in machine leaning, PDEs and approximation methods. It is defined as:

$$ K(x, z) = \exp(-|x - z|)$$

It has cosine-type eigenfunctions and eigenvalues that may be unitary under certain conditions (see [1]).

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[1] Gregory Fasshauer & Michael McCourt, Kernel-based Approximation Methods using MATLAB, World Scientific.

Answer 3

Not directly an answer to your question, but this operator: $\mathcal{L}_t[t f(t)](x)$ has a fixed point at $f(x)=1/x$.