2

Let's say that for certain restrictions on the complex parameters $\alpha$ and $\beta$ , you determine that $$\int_{a}^{b} f(x,\alpha, \beta) \, dx = g(\alpha,\beta).$$

But you can show that $ \int_{a}^{b} f(x,\alpha,\beta) \, dx$ converges for values of $\beta$ in a larger region.

Can you conclude that $\int_{a}^{b} f(x,\alpha,\beta) \, dx = g(\alpha,\beta)$ in this larger region if $g(\alpha,\beta)$ is analytic with respect to $\beta$ in the larger region?

For example, if $\text{Re}(\beta) > \text{Re}(\alpha) >-1$, one can use contour integration to show that $$\int_{0}^{\frac{\pi}{2}} \cos^{\alpha}(x) \cos (\beta x) \, dx = \frac{\pi}{2^{\alpha+1}} \frac{\Gamma(\alpha+1)}{\Gamma(\frac{\alpha}{2} + \frac{\beta}{2} +1)\Gamma(\frac{\alpha}{2} - \frac{\beta}{2} +1)}. $$

I want to argue that this equation holds for all complex values of $\beta$.

Random Variable
  • 42,026

1 Answers1

2

The Gamma function has simple poles on $-\mathbb{N}$, and is holomorphic and nonzero everywhere else. So your right hand side is an entire function with respect to $\beta$, for $\alpha$ fixed.

On the left-hand side, still for a fixed $\alpha$, the integrand is clearly entire with respect to $\beta$ for each $t$. Integrating with respect to $x$ yields an entire function here. Indeed, the partial derivative with respect to $\beta$ is locally uniformly (in $\beta$) dominated by an integrable function (no longer depending on $\beta$) over $x\in[0,\pi/2]$. So holomorphy follows by dominated convergence. We could avoid DCT, but it does not make the argument shorter. Note also that, unlike what you said in the comments, we need to be careful: if $g(z)=\int f(t,z)dt$ converges on some open set $\Omega$ in $\mathbb{C}$, and if $z\longmapsto f(t,z)$ is holomorphic on $\omega$ for each $t$, it does not necessarily imply that $g$ is holomorphic. See the example below.

So you have two entire functions which coincide on $(\alpha,+\infty)$. By the principle of isolated zeroes, they must coincide on the whole of $\mathbb{C}$. So your formula holds for all $\alpha>-1$ and all $\beta\in\mathbb{C}$.

Note: I couldn't find a wikipedia reference for the principle of isolated zeroes. So I suspect it is not the most popular terminology. What I mean is: the zeroes of a nonconstant holomorphic function on a domain are isolated. Conversely: if the zero set of a holomorphic function on a domain has a limit point in this domain, then the function is constant equal to zero on the whole domain.

Example: Here is why we need to be careful before claiming the left hand side is holomorphic. Consider $g(z):=\int_0^1\sin(z/t)dt$. This is well-defined on $\mathbb{C}$, and $g(0)=0$. Now consider $$\frac{g(1/n)}{1/n}=\int_0^1n\sin (1/(nt))dt.$$ If $g$ was holomorphic at $0$, this would converge when $n$ tends to $+\infty$. Now recall $\sin x\geq 2x/\pi$ on $[0,\pi/2]$. Hence $n\sin (1/(nt))\geq 2/(\pi t)$ on $[2/(\pi n),1]$. Hence $$ \frac{g(1/n)}{1/n}\geq \frac{2}{\pi}\int_{\frac{2}{\pi n}}^1\frac{1}{t}dt=\frac{2}{\pi}\log\left(\frac{\pi n}{2}\right)\longrightarrow +\infty. $$ So $g$ is not holomorphic at $0$.

Julien
  • 44,791
  • What value is $a$ being fixed at? Any real value less than $-1$? And is a function defined by an integral not necessarily analytic/holomorphic where the integral converges? – Random Variable Mar 22 '13 at 22:55
  • 1
    Yes, you must fix $a>-1$ where you say the formula holds. No, there are theorems for functions defined by an integral to be analytic. I don't have counterexamples on top of my head, though...And I actually was not careful enough, I'll edit. – Julien Mar 22 '13 at 22:58
  • I meant greater than $-1$. I know you can use Morera's theorem. But I was thinking that in the region where the integral converges, the function is analytic. That's completely wrong? – Random Variable Mar 22 '13 at 23:03
  • Sorry I have to go. Maybe I'm over careful. I'll think about it. – Julien Mar 22 '13 at 23:04
  • @RandomVariable Consider $z\longmapsto \int_0^1\sin(z/t^2)dt$. – Julien Mar 23 '13 at 00:51
  • I'll consider it, but I don't know what I'm supposed to notice about it. The integrand is continuous on $(0,1]$ and bounded as $t \to 0$, so the integral converges for all real $z$. But I still don't know if that fact has anything to do with where the function is analytic. I'm sort of lost. – Random Variable Mar 23 '13 at 01:15
  • The derivative of the function above with respect ot $z$ is not integrable. – Julien Mar 23 '13 at 01:50
  • Regarding your second question, now please read carefully. – Julien Mar 23 '13 at 01:51
  • @RandomVariable I don't know if the example I gave you is not entire. I'll look for a better one. For the rest, the main argument is Lebesgue DCT. It proves that the function is differentiable with respect to $\beta$, i.e holomorphic. On $\mathbb{C}$. Sorry I can't come up with an obvious example showing why one needs to be careful. I haven't touched these things in long while. – Julien Mar 23 '13 at 02:22
  • @RandomVariable I did my homework and I added a detailed example showing that we must be careful in these matters. Integrable and partially holomorphic does not imply that the integral is holomorphic. – Julien Mar 23 '13 at 11:58