Find this limit of an integral Find $$\lim_{n\to\infty}n\left(\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}dx-\frac{\pi}{2}\right)$$
it is obvious
$$\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}})dx=B\left(\dfrac{n+1}{2n},\dfrac{2n+1}{4n}\right)$$
 A: You are missing a crucial factor $2$. You want to compute
$$ \lim_{n\to +\infty}n^2\left[\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\right)\,\Gamma\left(\frac{1}{2}+\frac{1}{4n}\right)}{2\,\Gamma\left(1+\frac{3}{4n}\right)}-\frac{\Gamma\left(\frac{1}{2}\right)^2}{2}\right] $$
hence it is enough to exploit the Taylor series of $\Gamma(z)$ in neighbourhoods of $z=\frac{1}{2}$ and $z=1$:
$$ \Gamma(1+x)=1-\gamma x + \frac{\gamma^2+\zeta(2)}{2} x^2+O(x^3) $$
$$ \Gamma\left(\frac{1}{2}+x\right)=\sqrt{\pi}-\sqrt{\pi}(\gamma+\log 4)x + \sqrt{\pi}\frac{\pi^2+2(\gamma+\log 4)^2}{4} x^2+O(x^3) $$
to get that the previous limit is just $-\infty$. However, if $n^2$ is replaced by $n$ the limit becomes $-\frac{3\pi\log 2}{4}$. The previous Taylor series are derived from  $\Gamma'(x)=\psi(x)\Gamma(x)$, $\Gamma''(x)=\psi'(x)\Gamma(x)+\psi(x)^2\Gamma(x)$ and the well-known values of $\psi(1)$ and $\psi\left(\frac{1}{2}\right)$.
The first terms of the asymptotic expansion of the integral are given by:
$$ \int_{0}^{\pi/2}(\sin x)^{\frac{1}{n}}(\cos x)^{\frac{1}{2n}}\,dx = \frac{\pi}{2}-\frac{3\pi\log 2}{4n}+\frac{\pi^3+36 \pi\log(2)^2}{64 n^2}+O\left(\frac{1}{n^3}\right).$$
A: $\lim_{n\to\infty}n^2\left(\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}dx-\frac{\pi}{2}\right)
$
I'll proceed very naively.
Since,
for large $n$,
$z^{1/n}
=e^{\ln z/n}
=1+\frac{\ln z}{n}+\frac{\ln^2 z}{2n^2}+O(\frac1{n^3})
=1+\frac{\ln z}{n}+O(\frac1{n^2})
$,
$\begin{array}\\
(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}
&=(1+\frac{\ln \sin x}{n}+O(\frac1{n^2}))(1+\frac{\ln \cos x}{2n}+O(\frac1{n^2}))\\
&=1+\frac{2\ln (\sin x)+\ln(\cos x))}{2n}+O(\frac1{n^2})\\
&=1+\frac{\ln (\sin^2 x\cos x))}{2n}+O(\frac1{n^2})\\
\end{array}
$
so
$\begin{array}\\
\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}dx
&=\int_{0}^{\frac{\pi}{2}}(1+\frac{\ln (\sin^2 x\cos x)}{2n}+O(\frac1{n^2}))dx\\
&=\frac{\pi}{2}+\frac1{2n} \int_{0}^{\frac{\pi}{2}}\ln (\sin^2 x\cos x)dx +O(\frac1{n^2})\\
&=\frac{\pi}{2}+\frac1{2n} (-\frac32 \pi\ln 2) +O(\frac1{n^2})
\qquad\text{(according to Wolfy)}\\
&=\frac{\pi}{2} -\frac{3\pi \ln 2}{4n}+O(\frac1{n^2})\\
\text{so}\\
n(\int_{0}^{\frac{\pi}{2}}(\sin{x})^{\frac{1}{n}}(\cos{x})^{\frac{1}{2n}}dx
-\frac{\pi}{2})
&= -\frac{3\pi \ln 2}{4}+O(\frac1{n})\\
\end{array}
$
Therefore
the limit in the question
is $\infty$
(unless I have made a mistake,
in which case
more terms may need to be taken
in the expansion).
