moving limit through the integral sign Let $F(\alpha) = \int\limits_0^{\pi/2} \ln( \alpha^2 - \sin^2 x) \mathrm{d} x $ where $\alpha>1$
Im tempting to argue that $F(1) = \int\limits_0^{\pi/2} \ln (1 - \sin^2 x) dx = \int\limits_0^{\pi/2} \ln \cos^2 x dx $
But, $\alpha > 1$. Thus, the only way we can do  this is if we can do the following
$$ \lim_{\alpha \to 1^+ } \int\limits_0^{\pi/2} \ln( \alpha^2 - \sin^2 x) \mathrm{d} x = \int\limits_0^{\pi/2} \lim_{\alpha \to 1^+} \ln( \alpha^2 - \sin^2 x) \mathrm{d} x $$
Qs: are we allowed to move the limit in such a way?
 A: Let's examine the function being integrated.  If $\alpha = 1$ the function is just:
$$ \lim_{x \to \pi/2}\;\bigg[ \log (1 - \sin^2 x) \bigg] = -\infty $$
We have something to be concerned about.  Our integral is improper but really it looks kind of OK
$$  \int_0^{\pi/2} \bigg[ \log (1 - \sin^2 x) \bigg] \, dx $$
Near the value $x = 1$ can we find a cuttoff where the error is uniform with respect to $\alpha$?
$$ \int_0^{\frac{\pi}{2}(1-\epsilon)}  \bigg[ \log ( \alpha^2 - \sin^2 x) \bigg] \, dx 
+ \int_{\frac{\pi}{2}(1-\epsilon)}^{\frac{\pi}{2}}  \bigg[ \log ( \alpha^2 - \sin^2 x) \bigg] \, dx$$
The first term now coverges unifromly since we have removed the controversial art.  But now the other part:
$$  \frac{\pi \epsilon}{2} \bigg[ \log ( \alpha^2 - 0) \bigg]  > \int_{\frac{\pi}{2}(1-\epsilon)}^{\frac{\pi}{2}}  \bigg[ \log ( \alpha^2 - \sin^2 x) \bigg] \, dx > 
 \frac{\pi \epsilon}{2} \bigg[ \log \Big( 1^2 - (1-\frac{\pi \epsilon}{2})^2 \Big) \bigg] $$
where $\alpha > 1$ (and really $\alpha = 1 + \epsilon'$).  We should also say $\alpha < \sqrt{2}$.  
Very important RHS does not depend on $\alpha$.  And I am using that $0 < \sin x < x$ whenever $x > 0$.

Statements like this can be found in chapters on uniform convergence  in analysis textbooks.  Here's an exercise from Rudin: 

Suppose $g,f$ are defined on $(0, \infty)$ are Riemann-Integrable functions on $[a,b]$ with $0 < a < b < \infty$  and $|f_n| \leq g$ and $f_n \to f$ uniformly on every compact subset of $(0, \infty)$ and that 
  $\int g(x) \, dx < \infty$ then  prove
  $$ \lim_{n \to \infty} \int_0^\infty f_n(x) \, dx = \int_0^\infty f(x) \, dx$$
  Rubin Mathematical Principles of Real Analysis Chapter 7 Ex 12

I have my doubts here.  Certainly the statement is true and the book remarks this is a weak case of dominated convergence.   However,


*

*your domain of integration is $[0, \frac{\pi}{2}$ and your integrand is infinite whenever $\sin x = 1$. 

*we are told to compare $\log(1 - \sin^2 x)$ with something larger but still provably finite.  

*The goal is to consider interval $[a,b] \subseteq (0, \infty)$ and ultimately let $a \to 0$.  In our case we'd like $b \to \frac{\pi}{2}$ 


and various other small things such that I'd rather do the estimates myself than call on a theorem which might be incorrect.
A: You can interchange the limit and the integral.
The function $g(x) = \ln 2 - \ln(\alpha^2 -\sin^2(x))$ is nonnegative for $1 < \alpha < 2$ and increases for $\alpha \searrow 1$. By the monotone  onvergence theorem you can interchange the limit and the integral for $g$ and consequently also in your problem.
A: It’s $\enspace\lim\limits_{\alpha\to 1^+} \ln(\alpha^2-\sin^2 x) =\ln\cos^2 x\enspace$ and we have to proof with
$$\lim\limits_{\alpha\to 1^+}((  \int\limits_0^{t\pi/2}\ln(\alpha^2-\sin^2 x)dx - \int\limits_0^{t\pi/2}\ln(\cos^2 x) dx )|_{t\to 1^-}) =\frac{\pi}{2} \lim\limits_{\alpha\to 1^+} \int\limits_0^1 \ln(1+\frac{\alpha^2-1}{\cos^2(\frac{\pi}{2}x)}) dx$$ if 
$$\lim\limits_{\alpha\to 1^+} \int\limits_0^1 \ln(1+\frac{\alpha^2-1}{\cos^2(\frac{\pi}{2}x)}) dx = \int\limits_0^1 \lim\limits_{\alpha\to 1^+} \ln(1+\frac{\alpha^2-1}{\cos^2(\frac{\pi}{2}x)}) dx =0$$ holds. 
It's $\enspace\cos(\frac{\pi}{2}x)>1-x\enspace$ for $\enspace0<x<1\enspace$ and therefore $$0< \int\limits_0^t \ln(1+\frac{\alpha^2-1}{\cos^2(\frac{\pi}{2}x)}) dx <\int\limits_0^t \ln(1+\frac{\alpha^2-1}{(1-x)^2}) dx $$ gives us an upper bound for $\enspace 0<t\leq 1 $ .  
With $\enspace a:=\sqrt{\alpha^2-1}\to 0\enspace $ for $\enspace \alpha\to 1^+\enspace $ it's
$$\int\limits_0^1 \ln(1+\frac{a^2}{(1-x)^2}) dx =$$ $$=[2a\arctan\frac{a}{1-x}-(1-x)(\ln((1-x)^2+a^2)-2)+(1-x)(\ln((1-x)^2)-2)]_0^1$$ $$\enspace \enspace = a\pi-2a\arctan a+\ln(1+a^2)\leq a\pi$$ for all $\enspace a\geq 0\enspace $ 
and it follows $$\lim\limits_{a\to 0} \int\limits_0^1 \ln(1+\frac{a^2}{(1-x)^2}) dx=0 $$ and therefore the answer to your question is Yes.
