Question

Log(sinx ) is convergent
And limit is o to π/2

Answer 1

Let $x=\arcsin y$, so that $\sin x=y$ and $\mathrm dx=\dfrac{\mathrm dy}{\sqrt{1-y^2}}$. Then

$\displaystyle\int_0^{\pi/2}\log\sin x\,\mathrm dx=\int_0^1\dfrac{\log y}{\sqrt{1-y^2}}\,\mathrm dy$

Integrate by parts, setting

$u=\log y\implies \mathrm du=\dfrac{\mathrm dy}y$
$\mathrm dv=\dfrac{\mathrm dy}{\sqrt{1-y^2}}\implies v=\arcsin y$

so the integral is equivalent to

$\displaystyle\int_0^1\dfrac{\log y}{\sqrt{1-y^2}}\,\mathrm dy=uv-\int v\,\mathrm du$
$=\log y\arcsin y\bigg|_{y=0}^{y=1}-\displaystyle\int_0^1\frac{\arcsin y}y\,\mathrm dy$

The first term vanishes, since

$\displaystyle\lim_{y\to0^+}\log y\arcsin y=0$

Now for comparison, consider the function $f(y)=\dfrac1{\sqrt{1-y}}$. We have that the integral of this function over the same interval is convergent:

$\displaystyle\int_0^1\frac{\mathrm dy}{\sqrt{1-y}}=-2\sqrt{1-y}\bigg|_{y=0}^{y=1}=2$

so if we could show that $\dfrac{\arcsin y}y\le f(y)$ at least for $0<y<1$, then we're done. Replacing $y$ with $\sin x$ (which means $0<x<\dfrac\pi2$), the inequality can be written as

$\dfrac1{\sqrt{1-y}}\ge\dfrac{\arcsin y}y\implies \dfrac y{\sqrt{1-y}}\ge\arcsin y\implies\dfrac{\sin x}{\sqrt{1-\sin x}}\ge x$

This is certainly true, because as $x\to0^+$, both sides approach 0, while as $x\to\dfrac\pi2^-$, you have the right side approach $\dfrac\pi2$ while the left side approaches $+\infty$. You can compare these two functions' derivatives to see that the left side is monotonic over the interval.

So we've established that

$\displaystyle\frac{\arcsin y}y\le\frac1{\sqrt{1-y}}\implies\int_0^1\frac{\arcsin y}y\,\mathrm dy\le\int_0^1\frac{\mathrm dy}{\sqrt{1-y}}=2$

so the first integral must converge, and so must

$\displaystyle\int_0^{\pi/2}\log\sin x\,\mathrm dx$

to some value between -2 and 0.