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Evaluate the following limit:
$$lim limits_nto infty,,, n!! intlimits_0^pi/2!! left(1-sqrt [n]sin x right),mathrm dx $$

I have done the problem .
My method:
First I applied L'Hôpital's rule as it can be made of the form $frac0 0$. Then I used weighted mean value theorem and using sandwich theorem reduced the limit to an integral which could be evaluated using properties of define integration .

I would like to see other different ways to solve for the limit.

You can use the following fact

$$f(x) = log sin x$$ is integrable in $(0, pi/2)$

and

$$int_0^pi/2 -log sin x textdx = fracpi log 22$$

Now by the mean value theorem (applied to $(sin x)^y$, as a function of $y$), we have that

for some $c in (0, frac1n)$

$$ dfrac1 - sqrt[n]sin xfrac1n = -(sin x)^c log sin x le -log sin x$$

Since $log sin x$ is integrable, by the dominated convergence theorem, we can take the limit inside the integral to get

$$lim_n to inftyint_0^pi/2 n(1 - sqrt[n]sin x)textdx = int_0^pi/2 lim_n to inftyn(1 - sqrt[n]sin x) textdx= int_0^pi/2 -log sin x textdx = fracpi log 22$$

This is equivalent to finding $lim_epsilon rightarrow 0 f(epsilon) over epsilon$, where $f(epsilon) = int_0^pi over 2 (1 - sin(x)^epsilon),dx$. Since $lim_epsilon rightarrow 0 sin(x)^epsilon = 1$, one has
$lim_epsilon rightarrow 0 f(epsilon) = 0$, and so by L'Hopital's rule you get
$$lim_epsilon rightarrow 0 f(epsilon) over epsilon = lim_epsilon rightarrow 0 f'(epsilon)$$
Differentiating under the integral sign gives
$$f'(epsilon) = -int_0^pi over 2 ln(sin(x))(sin(x))^epsilon,dx$$
The limit of this as $epsilon rightarrow 0$ is
$$-int_0^pi over 2 ln(sin(x)),dx$$
This integral is well-known (and I'm sure it's been done on this site), and the above is just
$$pi over 2ln(2)$$

This is only a different way to get it to the final integral , but here goes,
$sqrt[n]sin x=exp(dfraclog sin xn)=1+dfraclog sin xn+ o(frac1n)$

So, $1-sqrt[n]sin x=-dfraclog sin xn +o(frac 1 n)$

Using, this we get
$nint_0^frac pi 21-sqrt[n]sin xdx=int_0^frac pi 2 -log sin x dx +O(frac 1 n)=dfracpi log 22 +O(frac 1 n)todfracpi log 22$

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$dslim_n to inftybraces%
nint_0^pi/2,
bracks1 - root[n],sinparsx,rm dx: large ?$

beginalign
int_0^pi/2root[n]sinparsx,rm dx&=int_0^1t^1/n
,dd t over root1 - t^2
=int_0^1t^1/pars2npars1 - t^-1/2,half,t^-1/2,dd t
\[3mm]&=halfint_0^1t^1/pars2n - 1/2pars1 - t^-1/2,dd t
=half,rm Bpars1 over 2n + half,half
\[3mm]&=half,Gammapars1/bracks2n + 1/2Gammapars1/2 over Gammapars1/bracks2n + 1
endalign

When $dsn gg 1$:
beginalign
int_0^pi/2root[n]sinparsx,rm dx
&approx
rootpi over 2,Gammapars1/2 + Gammapars1/2Psipars1/2/pars2n
over Gammapars1 + Gammapars1Psipars1/pars2n
=pi over 2,1 + Psipars1/2/pars2n over 1 + Psipars1/pars2n
\[3mm]&approx
pi over 2,bracks1 + Psipars1/2 over 2n
bracks1 - Psipars1 over 2n
approx pi over 2,bracks1 + Psipars1/2 - Psipars1 over 2n
endalign

$$
color#00f%
lim_n to inftybracesnint_0^pi/2,bracks1 - root[n],sinparsx
,rm dx
=pi over 4,bracksPsipars1 - Psiparshalf
=color#00fhalf,pilnpars2
$$
since $dsPsipars1 = -gamma$ and
$dsPsiparshalf = -gamma - 2lnpars2$. See
this table.

You can have a close form solution; infact if $Re(1/n)>-1$ you have that the integral collapse in:
$$int_0^pi/2left[1-(sin(x))^1/nright]dx=frac12 left(pi -frac2 sqrtpi n Gamma
left(fracn+12 nright)Gamma
left(frac12 nright)right)$$
So we define:
$$y(n)=fracn2 left(pi -frac2 sqrtpi n Gamma
left(fracn+12 nright)Gamma
left(frac12 nright)right)$$
And performing the limit:
$$lim_n rightarrow + inftyy(n)=-frac14 pi left[gamma +psi
^(0)left(frac12right)right]$$

Result

Let

$$f(n) = n int_0^fracpi2 (1- sin(x)^frac1n), dx$$

then

$$lim_nto infty , f( n)=fracpi2 log(2) simeq 1.088793045151801$$

Derivation 1

Attempting to perform the limit directly under the integral we write

$$n(1-sin(x)^frac1n) = nleft(1-expleft(frac1n logleft(sin(x)right)right)right)\
simeq nleft(1-1 - (frac1n) log(sin(x))+O(frac1n^2)right) \
= - log(sin(x)) + O(frac1n)$$

and the limiting integral becomes

$$- int_0^fracpi2 log(sin(x)) = fracpi2 log(2)$$

Derivation 2

The transformation $sin(x) to t$, $dx to fracdtsqrt1-t^2$leads to

$$f(n)= int_0^1 n left(1-t^1/nright) frac1sqrt1-t^2 , dt$$

Now performing the limit under the integral sign gives

$$lim_nto infty , n left(1-t^1/nright)=-log (t)$$

so that the integral becomes

$$- int_0^1 fraclog (t)sqrt1-t^2 , dt=fracpi2 log(2)$$

which is the announced result.