Convergence of the integral $\int_0^{\pi/2}\ln(\cos(x))dx$ I want to Show that whether the integral
$$\int\limits_0^{\pi/2}\ln(\cos(x))dx$$
is convergent ot not.
My Approach: Let $y=cos(x)$, then the above integral reduces to $$\int\limits_0^{1}\frac{\ln(y)}{\sqrt{1-y^2}}dy.$$ At this step since $\ln(y)<<y^p$, $p=1,2,...$, I compare above integral from above with the integral$$\int\limits_0^{1}\frac{y}{\sqrt{1-y^2}}dy,$$ which is convergent. Hence by comprasion I obtain that the integral $\int\limits_0^{\pi/2}\ln(\cos(x))dx$ is convergent.
My Question: (1) Is my  approach true?
(2) Can you suggest any different aproach?
Thanks in advance...
 A: Quite surprisingly, that integral may be computed directly through Riemann sums. 
From the identity:
$$ \prod_{k=1}^{n-1}\sin\frac{\pi k}{n} = \frac{2n}{2^n}\tag{1} $$
it follows that:
$$ \frac{1}{n}\sum_{k=1}^{n-1}\log\sin\frac{\pi k}{n} = -\log 2+\frac{\log(2n)}{n} \tag{2} $$
hence by taking the limit as $n\to +\infty$ we get:

$$\int_{0}^{\pi/2}\log\cos\theta\,d\theta = \frac{1}{2} \int_{0}^{\pi}\log\sin\theta\,d\theta = -\frac{\pi}{2}\log 2. \tag{3} $$

By using differentiation under the integral sign, we may also notice that
$(3)$ just depends on the value of the digamma function $\psi(s)=\frac{d}{ds}\log\Gamma(s)$ in $s=\frac{1}{2}$ or $s=-\frac{1}{2}$.
A: First note that in the interval $(0,\pi/2)$, $0<\cos(x)<1$, so $\ln(\cos(x))<0$.  Therefore, we are interested in bounding the integral from below.
Since, in the interval $[0,\pi/2]$, $\frac{d}{dx}\cos(x)=-\sin(x)$, which is negative, you know that $\cos(x)$ is concave down.  Therefore, in this interval, $\cos(x)\geq1-\frac{2}{\pi}x$.  Therefore,
$$
\int_0^{\pi/2}\ln(\cos(x))dx\geq\int_0^{\pi/2}\ln(1-\frac{2}{\pi}x)dx.
$$
Using a $u$-substitution of $1-\frac{2}{\pi}x=u$, we have 
$$
\int_0^{\pi/2}\ln(1-\frac{2}{\pi}x)dx=\frac{\pi}{2}\int_0^1\ln(u)du=\frac{\pi}{2}[u\ln(u)-u]_0^1=-\frac{\pi}{2}.
$$
Therefore, the integral converges (and its value is between $0$ and $\pi/2$).
A: You have the right idea.  The issue is that you need to be a little more clear on the substitution of y for $ln (y) $.  For this, you need to use the power series expansion of ln near y=1 For the y=1 side Of the integral.  For the y=0 side, you can't use this, but you can use the power series expansion of the denominator to show convergence there. 
A: Your observation that $\ln (y) << y^p$ isn't really helpful near $0$ since $\ln (y) \to -\infty$ as $y \to 0^+$. Instead use the fact that the asymptotes of the function occur at the endpoints and that the integrals $$\int_0^1 \log(y) \, dy, \qquad \int_0^1 \frac{1}{\sqrt{1 - y}} \, dy$$ both converge.
A: Consider the Fourier series:
$$\ln(\cos(x))=-\ln(2)-\sum_{j=1}^{\infty} \frac{(-1)^j\cos(2jx)}{j},\,\,\,\,0\le x\lt \frac{\pi}{2}$$
Since the fourier series are uniformly convergence. So:
$$
\begin{align}
\int_0^{\frac{\pi}{2}} \log(\cos(x)) \, dx&=\lim_{b\to \frac{\pi}{2}}\int_0^{b} \log(\cos(x)) \, dx\\
&=\lim_{b\to \frac{\pi}{2}}\int_0^{b} (-\ln(2)-\sum_{j=1}^{\infty} \frac{(-1)^j\cos(2jx)}{j}) \, dx\\
&=\frac{-\pi}{2}\ln(2)+\lim_{b\to \frac{\pi}{2}}(\sum_{j=1}^{\infty} \frac{(-1)^j\sin(2jx)}{2j^2})]_0^{b}\\
&=\frac{-\pi}{2}\ln(2)
\end{align}
$$ 
