TITLE: Show $\lim_{n\to \infty}n\int_{0}^{\frac{\pi}{2}}(1-\sqrt[n]{\sin(x)})\,\mathrm{d}x = \frac{\pi \ln(2)}{2}$
QUESTION [11 upvotes]: Here is an interesting limit of an integral I do not know how to begin. Any help is greatly appreciated.
$\lim_{n\to \infty}n\int_{0}^{\frac{\pi}{2}}(1-\sqrt[n]{\sin(x)})\,\mathrm{d}x$
I know it converges to $\frac{\pi \ln(2)}{2}$, but how?.
Thanks much

REPLY [3 votes]: Since
$$
\lim_{n\to\infty}n\left(x^{1/n}-1\right)=\log(x)\tag{1}
$$
converges monotonically,
$$
\begin{align}
\lim_{n\to\infty}n\int_0^{\pi/2}\left(1-\sqrt[\large n]{\sin(x)}\right)\,\mathrm{d}x
&=-\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x\\
&=-\frac12\int_0^\pi\log(\sin(x))\,\mathrm{d}x\\
&=-\int_0^{\pi/2}\log(\sin(2x))\,\mathrm{d}x\\
&=-\int_0^{\pi/2}\Big(\log(2)+\log(\sin(x))+\log(\cos(x))\Big)\,\mathrm{d}x\\
&=-\frac\pi2\log(2)-2\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x\tag{2}
\end{align}
$$
Solving $(2)$, we get
$$
-\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x=\frac\pi2\log(2)\tag{3}
$$