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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?

Community
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  • Perhaps you should cite the standard limit theorems from the Lebesgue theory (monotone convergence, dominated convergence, ...) and tell us whether they apply in your case. Or can we assume from your question that you do not know those theorems, so our answer should be a citation of one of them? – GEdgar Nov 23 '16 at 12:30

3 Answers3

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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.

cactus314
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  • this criterion is taken from the textbook Mathematical Analysis by Zorich Vol 1 and Vol II which is known for being very thorough. – cactus314 Nov 24 '16 at 15:52
  • Only a hint: It's right, that the uniform convergence has to be shown. The first term convergies uniformly only for $\alpha >0$ . The problem here is that you have no possibilty for an estimation because of the divergence. And therefore it's not possible to say, that in the infinity the first term is still uniformly (because it's not convergent and therefore also not uniformly convergent). It follows that the "lim" and integral are not exchangeable with your arguments. – user90369 Nov 25 '16 at 10:49
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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.

Dominik
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  • why do you have to introduce $g(x)$? there is no ln 2 in my question. –  Nov 19 '16 at 01:53
  • This is just to have a sequence of positive functions, because the most common version of the monotone convergence theorem needs a sequence of nonnegative monotone increasing functions. Thete are also other versions that you could apply directly here. – Dominik Nov 19 '16 at 08:06
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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.

user90369
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