Prove $\ln[\sin(x)] \in L_1 [0,1].$
Since the problem does not require actually solving for the value, my strategy is to bound the integral somehow. I thought I was out of this one free since for $\epsilon > 0$ small enough, $$\lim_{\epsilon \to 0}\int_\epsilon^1 e^{\left|\ln(\sin(x))\right|}dx=\cos(\epsilon)-\cos(1) \to 1-\cos(1)<\infty$$
and so by Jensen's Inequality, $$e^{\int_0^1 \left| \ln(\sin(x))\right|\,dx}\le \int_0^1e^{\left|\ln(\sin(x))\right|}\,dx\le1-\cos(1)<\infty$$ so that $\int_0^1 \left|\ln(\sin(x))\right|\,dx<\infty$.
The problem, of course, is that the argument begs the question, since Jensen's assumes the function in question is integrable to begin with, and that's what I'm trying to show.
Any way to save my proof, or do I have to use a different method? I attempted integration by parts to no avail, so I am assuming there is some "trick" calculation I do not know that I should use here.
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$\begingroup$We can show this using the fact that $\sin x \sim x$ for small values of $x$; precisely, we have the inequality
$$\frac 1 2 x \le \sin x$$
for all $x \in [0,1]$; this leads to
$$\ln\left(\frac{x}{2}\right) \le \ln \sin x$$
almost everywhere on $[0,1]$. We'll actually use that
$$-\ln \left(\frac x 2\right) \ge - \ln \sin x$$Noting that $\ln(x/2) = \ln x - \ln 2$, and that our measure space is finite, it is sufficient to show that $\ln x \in L^1[0,1]$. To do this, we show that $\ln(1/x)$ has finite Lebesgue integral on this interval via the Monotone Convergence Theorem (hence the usage of the $-$ sign to make things positive). Since $\ln x$ is continuous and bounded on every interval $[\epsilon, 1]$, the Lebesgue integral coincides with the Riemann integral, and applying the MCT to the functions $-\chi_{[1/n,1]} \ln x$ gives
\begin{align*} \int_0^1 - \ln x dx &= \lim_{n \to \infty} \int_{1/n}^1 - \ln x dx \\ &= - \lim_{n \to \infty} x (\ln x - 1) \Big|_{1/n}^1 \\ &= - \lim_{n \to \infty} \Big(1 (\ln 1 - 1)\Big) - \Big(\frac 1 n \left(\ln \frac 1 n - 1\right)\Big) \\ &= 1 - \lim_{n \to \infty} \left(\frac 1 n + \frac{\ln n}{n}\right) \\ &= 1 \end{align*}
Now by comparison, the original function is integrable.
$\endgroup$ $\begingroup$First observe that $\ln \sin x = \ln {\sin x \over x} + \ln x$.
The function ${\sin x \over x}$ is continuous and nonzero on on $[0,1]$ (if you extend it to equal $1$ at $x = 0$), so the same is true for $\ln {\sin x \over x}$ . Thus $\ln {\sin x \over x}$ is in $L^1[0,1]$.
The function $\ln x$ is also integrable on $[0,1]$ as its antiderivative is $x \ln x - x$ which converges to zero as $x = 0$.
So their sum $\ln \sin x$ is in $L^1[0,1]$ too.
$\endgroup$ $\begingroup$A simpler approach would be to observe that the function $x^{1/2}\ln \sin x$ is bounded on $(0,1]$, because it has a finite limit as $x\to 0$ -- by L'Hôpital's rule applied to $\dfrac{\ln \sin x}{x^{-1/2}}$. This gives $|\ln \sin x|\le Mx^{-1/2}$.
As Byron Schmuland noted, $e^{|\ln \sin x|} = 1/\sin x$, which is nonintegrable; this is fatal for your approach.
$\endgroup$ $\begingroup$Here is a proof that hides behind a theorem on swapping order of integration:
We have $0 \le {x \over 2} \le \sin x$, and $-\log$ is decreasing on $(0,1]$.
Then $\int_0^1 | \log(\sin x)| dx = \int_0^1 - \log( \sin x) dx \le \int_0^1 - \log( { x \over 2}) dx = \log 2 + \int_0^1 - \log( { x}) dx$.
Tonelli gives $\int_0^1 -\log(x) dx = \int_{x=0}^1 \int_{t=x}^1 {dt \over t} dx = \int_{t=0}^1 \int_{x=0}^t {dx \over t} dt = 1$.
Hence the upper bound $\int_0^1 | \log(\sin x)| dx \le 1+ \log 2$.
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