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 $$f(t) = t^{s-1} + \Gamma(s) t^{-s}, \text{ where } s = \frac{1}{2} \pm \frac{i}{\pi} \log \left(\pi+\sqrt{\pi^2-1} \right) $$

  • $\begingroup$ here? $\endgroup$
    – K.defaoite
    Commented Nov 14, 2022 at 10:31

3 Answers 3


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]).

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


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


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.

  • 1
    $\begingroup$ I think you may have misunderstood the question being asked here. It is "Are there any useful functions that are their own Laplace transform, and can this property be exploited to give any interesting consequences?" $\endgroup$
    – 311411
    Commented Oct 5, 2021 at 3:01

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