# Thread: Holomorphic function defined as an integral.

1. ## Holomorphic function defined as an integral.

I do not fully understand the following statement:

"Let f(x) decay faster than any polynomial as x -> infinity, then int_0^infinity f(x) x^{s-1}dx defines a holomorphic function of s when Re(s) > 1.
This is because int_0^1 x^{s-1}ds converges and f(x) has a rapid decay."

When, in general, the function defined by an integral is holomorphic?

2. Originally Posted by Different
I do not fully understand the following statement:

"Let f(x) decay faster than any polynomial as x -> infinity, then int_0^infinity f(x) x^{s-1}dx defines a holomorphic function of s when Re(s) > 1.
This is because int_0^1 x^{s-1}ds converges and f(x) has a rapid decay."

When, in general, the function defined by an integral is holomorphic?
Do you know the theorem about continuity of functions defined by an integral? There is a very similar result for holomorphic functions:

Suppose $\displaystyle (x,z)\mapsto f(x,z)$ is
- holomorphic with respect to the $\displaystyle z$ variable and measurable with respect to the $\displaystyle x$ variable
- such that there is a function $\displaystyle \phi(x)$ with $\displaystyle |f(x,z)|\leq \phi(x)$ for any $\displaystyle x,z$ and $\displaystyle \int \phi(x) dx<\infty$. (this is a "domination" of $\displaystyle f$)
Then the function $\displaystyle F:z\mapsto \int f(x,z) dx$ is holomorphic, and $\displaystyle F'(z)=\int \partial_z f(x,z) dx$.

(I didn't specify the intervals or domains, this is whatever you want)

This theorem is quite powerful. It shows for instance that the "integral function" $\displaystyle F$ is indefinitely differentiable, using only a domination of the function (and not of its derivatives).

In your situation, you'll probably have to apply this theorem to compact subsets of the domain (otherwise you won't be able to get the domination).

3. 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.

4. Define, $\displaystyle g(s) = \int_0^{\infty} f(x) x^{s-1} dx$ where branch of $\displaystyle x^{s-1}$ is $\displaystyle (-\infty,0]$.

One way to show that functions defined by integrals are holomorphic is by using Morera's theorem.

Let $\displaystyle \Gamma$ be any rectangle in $\displaystyle \Re (s) > 1$ then:
$\displaystyle \oint_{\Gamma} g(s) ds = \oint_{\Gamma} \int_0^{\infty} f(x) x^{s-1}dx ~ ds = \int_0^{\infty} \oint_{\Gamma} f(x)x^{s-1} ds ~ dx = 0$

The result is zero because by Cauchy's theorem when we compute the integral over $\displaystyle \Gamma$ in $\displaystyle \Re (s)>1$ of an analytic function we get $\displaystyle 0$.
Thus, $\displaystyle g$ must be a holomorphic function.

The major questionable step here is whether or not changing integration order is justified here or not.
So we need to be careful.

5. 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 $\displaystyle \Gamma$.

6. Thanks. I understand that. You basically replicated a proof that Gamma function is analytic on Re(s)>0.

I was confused by the argument in the quoted text that said that the resulting integral defines a holomorphic function for Re(s)>1 since
a) integral from 0 to 1 of x^{s-1} converges
b) f(x) had a rapid decay.

...and, to be honest, I am still not sure how a) and b) imply that the function is holomorphic.

7. Originally Posted by Different
Thanks. I understand that. You basically replicated a proof that Gamma function is analytic on Re(s)>0.

I was confused by the argument in the quoted text that said that the resulting integral defines a holomorphic function for Re(s)>1 since
a) integral from 0 to 1 of x^{s-1} converges
b) f(x) had a rapid decay.

...and, to be honest, I am still not sure how a) and b) imply that the function is holomorphic.
I don't know what more you need. We gave you the statement of a theorem, the proof of it, and a quite precise proof of how it applies... In saying you already knew both the theorem and the proof for the $\displaystyle \Gamma$ function, you make me feel like my dedicated posts were somewhat unpredictably useless to you... I would have appreciated your being more specific.

Anyway, there's not much of another proof than the one I gave, and you can see that properties a) and b) provide the core of the proof of the domination. They respectively give both parts of the dominating function. I needed a simple additional ingredient that comes from the fact that the domination must hold for $\displaystyle s$ is a slab $\displaystyle a\leq\Re(s)\leq b$, so that I had to replace (simply by monotonicity) $\displaystyle x^{s-1}$ by $\displaystyle x^{a-1}$ or $\displaystyle x^{b-1}$ depending if $\displaystyle x<1$ or $\displaystyle x>1$. Since $\displaystyle a$ (and $\displaystyle b$) satisfy the same condition like $\displaystyle s$, namely $\displaystyle \Re(s)>0$, I could use property a) for $\displaystyle a$ instead of $\displaystyle s$. So a) and b) are almost the only crucial properties needed; they can help you write the proof but don't make one by themselves of course.

8. Originally Posted by Laurent
I don't know what more you need. We gave you the statement of a theorem, the proof of it, and a quite precise proof of how it applies... In saying you already knew both the theorem and the proof for the $\displaystyle \Gamma$ function, you make me feel like my dedicated posts were somewhat unpredictably useless to you... I would have appreciated your being more specific.

Anyway, there's not much of another proof than the one I gave, and you can see that properties a) and b) provide the core of the proof of the domination. They respectively give both parts of the dominating function. I needed a simple additional ingredient that comes from the fact that the domination must hold for $\displaystyle s$ is a slab $\displaystyle a\leq\Re(s)\leq b$, so that I had to replace (simply by monotonicity) $\displaystyle x^{s-1}$ by $\displaystyle x^{a-1}$ or $\displaystyle x^{b-1}$ depending if $\displaystyle x<1$ or $\displaystyle x>1$. Since $\displaystyle a$ (and $\displaystyle b$) satisfy the same condition like $\displaystyle s$, namely $\displaystyle \Re(s)>0$, I could use property a) for $\displaystyle a$ instead of $\displaystyle s$. So a) and b) are almost the only crucial properties needed; they can help you write the proof but don't make one by themselves of course.
thank you. your posts were useful (...didn't I click on "Thank You" button after your first post?)

I was probably just somewhat confused about the requirement of Re(s)>1 whereas it's true for Re(s)>0.

9. ## Re: Holomorphic function defined as an integral.

Originally Posted by Laurent
Do you know the theorem about continuity of functions defined by an integral? There is a very similar result for holomorphic functions:

Suppose $\displaystyle (x,z)\mapsto f(x,z)$ is
- holomorphic with respect to the $\displaystyle z$ variable and measurable with respect to the $\displaystyle x$ variable
- such that there is a function $\displaystyle \phi(x)$ with $\displaystyle |f(x,z)|\leq \phi(x)$ for any $\displaystyle x,z$ and $\displaystyle \int \phi(x) dx<\infty$. (this is a "domination" of $\displaystyle f$)
Then the function $\displaystyle F:z\mapsto \int f(x,z) dx$ is holomorphic, and $\displaystyle F'(z)=\int \partial_z f(x,z) dx$.

(I didn't specify the intervals or domains, this is whatever you want)

This theorem is quite powerful. It shows for instance that the "integral function" $\displaystyle F$ is indefinitely differentiable, using only a domination of the function (and not of its derivatives).

In your situation, you'll probably have to apply this theorem to compact subsets of the domain (otherwise you won't be able to get the domination).
Thanks Laurent. Could you please provide a reference to a general version of this result? It seems one could use this result to show that solutions $\displaystyle u(x,t)$ of initial-boundary problems for the heat equation $\displaystyle u_{xx} = 2 u_{t}$ are (real) analytic in $\displaystyle x$ even if boundary data is discontinuous. I have only seen such results with smooth boundary data. What am I missing?

Many thanks.

10. ## Re: Holomorphic function defined as an integral.

Originally Posted by Laurent
Do you know the theorem about continuity of functions defined by an integral? There is a very similar result for holomorphic functions:

Suppose $\displaystyle (x,z)\mapsto f(x,z)$ is
- holomorphic with respect to the $\displaystyle z$ variable and measurable with respect to the $\displaystyle x$ variable
- such that there is a function $\displaystyle \phi(x)$ with $\displaystyle |f(x,z)|\leq \phi(x)$ for any $\displaystyle x,z$ and $\displaystyle \int \phi(x) dx<\infty$. (this is a "domination" of $\displaystyle f$)
Then the function $\displaystyle F:z\mapsto \int f(x,z) dx$ is holomorphic, and $\displaystyle F'(z)=\int \partial_z f(x,z) dx$.

(I didn't specify the intervals or domains, this is whatever you want)

This theorem is quite powerful. It shows for instance that the "integral function" $\displaystyle F$ is indefinitely differentiable, using only a domination of the function (and not of its derivatives).

In your situation, you'll probably have to apply this theorem to compact subsets of the domain (otherwise you won't be able to get the domination).
Also, is there a name for results of this general type?