Consider $$ g(y)=\int_{-\infty}^{+\infty} e^{-yt}d\mu(t) $$ convergent for $y\in(a,b)$ for some $a,b>0$; and with $\mu(t)$ a $\sigma$-finite and non-negative Borel measure on $\mathbb{R}$. I'm trying to understand if such $g(y)$ is an analytic function in $(a,b)$.
this is what I'm doing:
$$ g(y)=\int_{-\infty}^{+\infty} \lim_k \sum_n^k \frac{(-t)^n e^{-y_0 t}}{n!} (y-y_0)^n d\mu(t) $$ for every fixed $y_0\in(a,b)$.
Now, since $\sum_n^k \frac{(-t)^n e^{-y_0 t}}{n!} (y-y_0)^n$ are continuous and so measurable function such that $\lim_k |\sum_n^k \frac{(-t)^n e^{-y_0 t}}{n!} (y-y_0)^n|=|e^{-yt}|=e^{-yt}$ and $$\int_{-\infty}^{+\infty}e^{-yt}d\mu(t)$$ is convergent in $(a,b)$ by hypothesis. By dominated convergence's theorem, I can pass the limit and the sum out of the integral and say that $$ g(y)=\lim_k \sum_n^k \frac{\int_{-\infty}^{+\infty}(-t)^n e^{-y_0 t}d\mu(t)}{n!} (y-y_0)^n $$ But how can I prove that $\int_{-\infty}^{+\infty}(-t)^n e^{-y_0 t}d\mu(t)$ is convergente for every $y_0\in(a,b)$? There is a way to conclude that these integrals are convergent in $(a,b)$ without further assumption on the measure $\mu$? Thank you
If you take any $\alpha$ such that $a<\alpha<y_0$, you can write $$\vert (-t)^ne^{-y_0t}\vert =\vert t^n e^{-(y_0-\alpha)t}\vert e^{-\alpha t}=g_n(t)e^{-\alpha t} $$ Since $y_0-\alpha >0$, the function $g_n$ is bounded on $\mathbb R_+$, this gives the integrability for $(-t)^ne^{-y_0t}$ on $\mathbb R_+$ because $e^{-\alpha t}$ is integrable wrt $\mu$.
To show that $(-t)^ne^{-y_0t}$ is integrable on $\mathbb R_-$, do the same thing starting with $\beta$ such that $y_0<\beta<b$.