Show $\lim_{n\to \infty}n\int_{0}^{\frac{\pi}{2}}(1-\sqrt[n]{\sin(x)})\,\mathrm{d}x = \frac{\pi \ln(2)}{2}$ 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
 A: A trick (and it's only a trick) is to set $$f(x) = n(1-\sqrt[n]{\sin(x)})$$ so
$$ \sin(x) = \left(1-\frac{f(x)}{n}\right)^n$$ which in the limit $n\to\infty$ is
$$ \sin(x) = e^{-f_{\infty}(x)}$$ or $$f_{\infty}(x) = -\ln (\sin(x))$$ which you integrate to the result: $$-\int_0^{\frac{\pi}{2}}\ln(\sin(x))dx = \frac{\pi\ln 2}{2}$$
A: You can integrate by parts in your integral, differentiating $1 - (\sin(x))^{1 \over n}$ and integrating 1. You get
$$n\int_0^{\pi \over 2}(1 -  \sin(x)^{1 \over n})\,dx = \int_0^{\pi \over 2}x \sin(x)^{{1 \over n} - 1}\cos(x)\,dx$$
(Both of the boundary terms evaluate to zero). In the right-hand integral above, the integrand increases monotonically to $x(\sin(x))^{-1}\cos(x) = x\cot(x)$ as 
$n$ goes to infinity. So your limit is exactly
$$\int_0^{\pi \over 2}x\cot(x)\,dx$$
This is a pretty famous integral in itself, but you can integrate by parts again in the reverse direction (making sure the endpoints go to zero in the improper integral) to convert it into the integral
$$-\int_0^{\pi \over 2}\ln(\sin(x))\,dx$$
As you indicated this is well known and evaluates to ${\displaystyle{\pi \ln2 \over 2}}$. 
Effectively what is going on here is that you are using that the limit as $\epsilon \rightarrow 0$ of ${\displaystyle{1 - u^{\epsilon} \over \epsilon}}$ is negative the derivative of $u^x$ at $x = 0$, or $-\ln(u)$. Letting $\epsilon = {1 \over n}$ and $u = \sin(x)$ gives that your integrand (including the $n$ factor) converges to $-\ln(\sin(x))$. So your answer should be the integral of that. But making this rigorous seems a little tricky (to me) so I integrated by parts and then took limits, then integrated by parts back again.
A: 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
$$
-\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x=\frac\pi2\log(2)\tag{3}
$$
A: Suggestion: substitute $u=\sin x$, and consider the Beta function.
EDIT: A better suggestion. To arrive to
$$
 - \int_0^{\pi /2} {\ln (\sin x)dx} \, \bigg( = \frac{{\pi \ln 2}}{2}\bigg),
$$
consider
$$
\mathop {\lim }\limits_{r \to \infty } \frac{{\int_0^{\pi /2} {[1 - (\sin x)^{1/r} ]dx} }}{{1/r}} = \mathop {\lim }\limits_{r \to \infty } \frac{{\int_0^{\pi /2} {[1 - e^{\ln (\sin x)/r} ]dx} }}{{1/r}},
$$
followed by L'Hospital's rule.
A: $$n\int_0^{\frac{\pi}{2}}(1-\sqrt[n]{\sin(x)})\mathrm dx=\pi\left(\frac{n}{2}-n^2 2^{1-\frac1{n}}\frac{\Gamma\left(\frac1{n}\right)}{\Gamma\left(\frac1{2n}\right)^2}\right)$$
$$\approx\pi\left(\ln\sqrt{2}-\left(\left(\ln\sqrt{2}\right)^2+\frac{\pi^2}{48}\right)\frac1{n}+O\left(\frac1{n^2}\right)\right)$$
Now, take the limit...
