Question: I have a two-sided Laplace-Stieltjes transform,
$$
\int\_{-\infty}^{+\infty} e^{-xt}d\mu(t)
$$
that converges absolutely in $(a,b)$.

If the measure $\mu$ is finite,then 
$$
\int\_{-\infty}^{+\infty}d\mu(t)=\mu(\mathbb{R})<\infty
$$
can I conclude that $(a,b)$ MUST contain the origin?

In general, how changes the interval of convergence of a two-sided Laplace-Stieltjes transform with respect to $\mu$? 

Thank you

Answer: 
$$x^{2/3}+y^{2/3}=a^{2/3}\longleftrightarrow\;x=a\cos^3t\;,\;\;y=a\sin^3t\;,\;\;0\le t\le 2\pi$$

So

$$\frac12\int\limits\_0^{2\pi}\left[(a\cos^3t\cdot3a\sin^2t\cos t)-(a\sin^3t\cdot(-3a\cos^2t\sin t)\right]dt =$$

$$=\frac{3a^2}2\int\limits\_0^{2\pi}(\cos^4t\sin^2t+\cos^2t\sin^4t)dt=\frac{3a^2}2\int\limits\_0^{2\pi}\cos^2t\sin^2t\,dt=$$

$$=\left.\frac{3a^2}{64}\left(4x-\sin 4x\right)\right|\_0^{2\pi}=\frac{3a^2}{64}8\pi=\frac38a^2\pi$$
**Hint:** You need to parametrize $x^{2/3}+y^{2/3}=a^{2/3}$ to compute the integral on the boundary.

Consider that the circle
$$
x^2+y^2=a^2
$$
is parametrized by
$$
x=a\cos(t)\qquad y=a\sin(t)
$$
Modifying slightly, we get that
$$
x=a\cos^3(t)\qquad y=a\sin^3(t)
$$
parametrizes $x^{2/3}+y^{2/3}=a^{2/3}$.