In mathematical analysis, Haar's Tauberian theorem^[1] named after Alfréd Haar, relates the asymptotic behaviour of a continuous function to properties of its Laplace transform. It is related to the integral formulation of the Hardy–Littlewood Tauberian theorem.

William Feller gives the following simplified form for this theorem:^[2]

Suppose that ${\displaystyle f(t)}$ is a non-negative and continuous function for ${\displaystyle t\geq 0}$, having finite Laplace transform

${\displaystyle F(s)=\int _{0}^{\infty }e^{-st}f(t)\,dt}$

for ${\displaystyle s>0}$. Then ${\displaystyle F(s)}$ is well defined for any complex value of ${\displaystyle s=x+iy}$ with ${\displaystyle x>0}$. Suppose that ${\displaystyle F}$ verifies the following conditions:

1. For ${\displaystyle y\neq 0}$ the function ${\displaystyle F(x+iy)}$ (which is regular on the right half-plane ${\displaystyle x>0}$) has continuous boundary values ${\displaystyle F(iy)}$ as ${\displaystyle x\to +0}$, for ${\displaystyle x\geq 0}$ and ${\displaystyle y\neq 0}$, furthermore for ${\displaystyle s=iy}$ it may be written as

${\displaystyle F(s)={\frac {C}{s}}+\psi (s),}$

where ${\displaystyle \psi (iy)}$ has finite derivatives ${\displaystyle \psi '(iy),\ldots ,\psi ^{(r)}(iy)}$ and ${\displaystyle \psi ^{(r)}(iy)}$ is bounded in every finite interval;

2. The integral

${\displaystyle \int _{0}^{\infty }e^{ity}F(x+iy)\,dy}$

converges uniformly with respect to ${\displaystyle t\geq T}$ for fixed ${\displaystyle x>0}$ and ${\displaystyle T>0}$;

3. ${\displaystyle F(x+iy)\to 0}$ as ${\displaystyle y\to \pm \infty }$, uniformly with respect to ${\displaystyle x\geq 0}$;

4. ${\displaystyle F'(iy),\ldots ,F^{(r)}(iy)}$ tend to zero as ${\displaystyle y\to \pm \infty }$;

5. The integrals

${\displaystyle \int _{-\infty }^{y_{1}}e^{ity}F^{(r)}(iy)\,dy}$ and ${\displaystyle \int _{y_{2}}^{\infty }e^{ity}F^{(r)}(iy)\,dy}$

converge uniformly with respect to ${\displaystyle t\geq T}$ for fixed ${\displaystyle y_{1}<0}$, ${\displaystyle y_{2}>0}$ and ${\displaystyle T>0}$.

Under these conditions

${\displaystyle \lim _{t\to \infty }t^{r}[f(t)-C]=0.}$

A more detailed version is given in.^[3]

Suppose that ${\displaystyle f(t)}$ is a continuous function for ${\displaystyle t\geq 0}$, having Laplace transform

${\displaystyle F(s)=\int _{0}^{\infty }e^{-st}f(t)\,dt}$

with the following properties

1. For all values ${\displaystyle s=x+iy}$ with ${\displaystyle x>a}$ the function ${\displaystyle F(s)=F(x+iy)}$ is regular;

2. For all ${\displaystyle x>a}$, the function ${\displaystyle F(x+iy)}$, considered as a function of the variable ${\displaystyle y}$, has the Fourier property ("Fourierschen Charakter besitzt") defined by Haar as for any ${\displaystyle \delta >0}$ there is a value ${\displaystyle \omega }$ such that for all ${\displaystyle t\geq T}$

${\displaystyle {\Big |}\,\int _{\alpha }^{\beta }e^{iyt}F(x+iy)\,dy\;{\Big |}<\delta }$

whenever ${\displaystyle \alpha ,\beta \geq \omega }$ or ${\displaystyle \alpha ,\beta \leq -\omega }$.

3. The function ${\displaystyle F(s)}$ has a boundary value for ${\displaystyle \Re s=a}$ of the form

${\displaystyle F(s)=\sum _{j=1}^{N}{\frac {c_{j}}{(s-s_{j})^{\rho _{j}}}}+\psi (s)}$

where ${\displaystyle s_{j}=a+iy_{j}}$ and ${\displaystyle \psi (a+iy)}$ is an ${\displaystyle n}$ times differentiable function of ${\displaystyle y}$ and such that the derivative

${\displaystyle \left|{\frac {d^{n}\psi (a+iy)}{dy^{n}}}\right|}$

is bounded on any finite interval (for the variable ${\displaystyle y}$)

4. The derivatives

${\displaystyle {\frac {d^{k}F(a+iy)}{dy^{k}}}}$

for ${\displaystyle k=0,\ldots ,n-1}$ have zero limit for ${\displaystyle y\to \pm \infty }$ and for ${\displaystyle k=n}$ has the Fourier property as defined above.

5. For sufficiently large ${\displaystyle t}$ the following hold

${\displaystyle \lim _{y\to \pm \infty }\int _{a+iy}^{x+iy}e^{st}F(s)\,ds=0}$

Under the above hypotheses we have the asymptotic formula

${\displaystyle \lim _{t\to \infty }t^{n}e^{-at}{\Big [}f(t)-\sum _{j=1}^{N}{\frac {c_{j}}{\Gamma (\rho _{j})}}e^{s_{j}t}t^{\rho _{j}-1}{\Big ]}=0.}$