Originally Posted by

**Different** Thanks.

I know this theorem....it looks like we can break this integral (from 0 to infinity) to two from 0 to 1 and from 1 to infinity.

Then I can see how the theorem applies to the second integral. Not 100% sure about the first one though...I think if we assume that f(x) is absolutely integrable on [0,1] then we can apply this theorem to the 1st integral.

It looked like you were asking for a general theorem, that's why I didn't give the details of how it applies to your situation. Here they are:

First, a remark: your function is actually holomorphic on $\displaystyle \{s|\Re (s) > 0\}$. Like the $\displaystyle \Gamma$ function, which is the usual example.

As I said, you can't prove the domination on the whole $\displaystyle \{s|\Re (s) > 0\}$ at once. So let $\displaystyle 0<a<b$, and we shall prove that the theorem applies on $\displaystyle \{s|a\leq\Re (s)\leq b\}$. Since $\displaystyle a,b$ are arbitrary, this gives the analyticity on $\displaystyle \{s|\Re(s)>0\}$.

Using the hypotheses, there exist $\displaystyle C,M>0$ such that $\displaystyle |f(x)|\leq \frac{C}{x^{b+1}}$ if $\displaystyle x\geq 1$, and $\displaystyle |f(x)|\leq M$ if $\displaystyle 0\leq x\leq 1$.

Now, let's prove the domination. Let $\displaystyle s\in\mathbb{C}$ such that $\displaystyle a\leq\Re(s)\leq b$, and let $\displaystyle x>0$. Recalling that $\displaystyle |x^{s-1}|=x^{\Re(s)-1}$, we have:

$\displaystyle |f(x)x^{s-1}|\leq \left\{\begin{array}{cl} Mx^{a-1} & \mbox{ if }0< x\leq 1\\ \frac{C}{x^{b+1}}x^{b-1}=\frac{C}{x^2}&\mbox{ if }x>1.\end{array}\right.$

and the right-hand side function (which does not depend on $\displaystyle s$) is integrable on $\displaystyle (0,+\infty)$. The previously mentioned theorem may thus be applied. This solves your problem.

---

To TPH: Thanks for the remark about Morera's theorem, it provides a proof for the one I quoted. Indeed, Fubini's theorem applies under the domination hypothesis, and thus justifies the change of integration order: we have, for any rectangle $\displaystyle \Gamma$ in the domain where we apply the theorem,

$\displaystyle \int_0^{\infty} \oint_{\Gamma} |f(x,z)| dz dx \leq \int_0^{\infty} \oint_{\Gamma} \phi(x) dz dx $ $\displaystyle = |\Gamma| \int_0^\infty \phi(x) dx <\infty$,

where $\displaystyle |\Gamma|$ is the length of the rectangle $\displaystyle \Gamma$.