Analytically continuing a function defined by an integral to a larger region 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 $\beta$.
 A: 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 $0$.
