Calculate the limit of a sequence $lim_{nto infty} frac{I_{n}}{n^{n}}$

Question

Calculate $$lim_{nto infty} frac{1}{n^{n}}int_{0}^{n}x^narctan(sqrt[n]{x})dx$$

My approach: I was trying to solve first the integral, so for example for $n=1$, we have that
$$int xarctan(x)dx=frac{1}{2}x^{2}arctan(x)-frac{x}{2}+frac{1}{2}arctan(x)+c$$
and for $n=2$, we can see that $$int xarctan(sqrt{x})dx=frac{1}{6}(3(x^{2}-1)arctan(sqrt{x})-(x-3)sqrt{x}+c$$
and for $n=3$, we have
$$int xarctan(sqrt[3]{x})dx=frac{1}{30}(-3x^{5/3}+15(x^{2}+1)arctan(sqrt[3]{x})+5x-15sqrt[3]{x})+c$$

I was trying to see if by induction I could give a closed formula for the integral but the calculation of the integral does not show any pattern. How can I solve this limit?

Ninad Munshi · Answer

You can also use $t = left(frac{x}{n}right)^{n+1}$
$$implies lim_{ntoinfty} frac{n+1}{n}int_0^1 tan^{-1}left(n^{frac{1}{n}}t^{frac{1}{n^2+n}}right)dt to int_0^1 tan^{-1}(1)dt = frac{pi}{4}$$
by dominated convergence since the sequence is dominated by $g(t)=pi$.

Sangchul Lee · Answer

Substitute $x = n - u$ to rewrite
$$ frac{1}{n^n} int_{0}^{n} x^n arctan(x^{1/n}) , mathrm{d}x
= int_{0}^{infty} left(1 - frac{u}{n}right)^n arctan((n-u)^{1/n}) mathbf{1}_{[0,n]}(u) , mathrm{d}u. $$
Now by the dominated convergence theorem, this converges to
$$ int_{0}^{infty} lim_{ntoinfty} left(1 - frac{u}{n}right)^n arctan((n-u)^{1/n}) mathbf{1}_{[0,n]}(u) , mathrm{d}u
= int_{0}^{infty} e^{-u} arctan(1) , mathrm{d}u
= frac{pi}{4}. $$

math · Answer

Hint :
$u = frac{x}{n}$ gives you a simpler formula :
$$lim_{nto infty} int_{0}^{1}n u^narctan(sqrt[n]{nu})du$$

Calculate the limit of a sequence $lim_{nto infty} frac{I_{n}}{n^{n}}$

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