# Laplace transform identity

Is there a function equal to its Laplace transform?

I mean

$$\int_{0}^{\infty}dt\exp(-st)f(t)= f(s).$$

Of course I know $f(t)=0$ satisfy the equation.

For the case of the Fourier transform, I know the Hermite Polynomials are eigenfunction of the Fourier transform, perhaps it's enough with a shift or rotation into the complex plane ($s \rightarrow i\omega$)?

For $l$ with real part greater than $1$ a standard computation yields $$\mathcal{L}(t^l) = s^{-l-1} \Gamma(l+1).$$ Pick $z$ with real part between $0$ and $1$ and let $f(t)=\sqrt{\Gamma(z)} t^{-z} + \sqrt{\Gamma(1-z)} t^{z-1}.$ Then we have $$\mathcal{L}\left( f \right) = \sqrt{ \Gamma(z) \Gamma(1-z) } \cdot f(s).$$
Since $\displaystyle \Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$ it suffices to find a $z$ with real part between $0$ and $1$ such that $\sin(\pi z)=\pi.$ A solution is $z= \displaystyle \frac{\sin^{-1} \pi}{\pi}$ where $\sin^{-1}(\pi) = \dfrac{\pi}{2} - i \log(\pi + \sqrt{\pi^2-1}) .$ Therefore the Laplace transform operator has a fixed point.