$ \int_{0}^{ +\infty} \exp\left( -\dfrac{t}{n^{\frac{1}{k} }}\right) \arctan t \; dt \sim \dfrac{ \pi n^{\frac{1}{k} }}{2} $ 
*

*$k \geq 5$

*$g_n(t)= \dfrac{ \exp\left( -\dfrac{t}{n^{\frac{1}{k} }}\right)}{1+t^2  }   $

*$g(t)= \dfrac{1}{1+t^2}$

*$K_n(t)= \int_{0}^{ +\infty} \exp\left( -\dfrac{t}{n^{\frac{1}{k} }}\right) \arctan t  \; dt$
We want to prove, without dominated convergence theorem, that

*

*$\exists A>0, \forall t >A, 0 \leq g(t) - g_n(t) \leq \dfrac{ 1 }{ n^{ \frac{1}{2k} }} $

*$K_n \sim \dfrac{ \pi n^{\frac{1}{k} }}{2}$

My attempt :
$ t^2 \exp\left( -\dfrac{t}{n^{\frac{1}{k} }}\right) \arctan t  \underset{ t \to \infty}{ \sim}  \dfrac{\pi}{2} t^2 \exp\left( -\dfrac{t}{n^{\frac{1}{k} }}\right)\underset{ t \to \infty}{ \to}0 $
Therefore $K_n < \infty$
By integration by parts, $ K_n= n^{ \frac{1}{k} } \int_{0}^{ + \infty} g_n(t) dt$
$ \forall u >0, \exists 0<c<u, e^{-u} = 1 - u+ \dfrac{u^2}{2!} - \dfrac{u^3}{3!} e^{c} $
A Taylor formula.
 A: Making the change of integration variables $t =  n^{1/k}s$ yields
$$
K_n (t) = n^{1/k} \int_0^{ + \infty } {e^{ - s} \arctan \left( {n^{1/k} s} \right)ds} .
$$
The integrand is at most $\frac{\pi}{2} e^{-s}$, which is integrable. Thus by the dominated convergence theorem
$$
\mathop {\lim }\limits_{n \to  + \infty } \int_0^{ + \infty } {e^{ - s} \arctan \left( {n^{1/k} s} \right)ds}  = \int_0^{ + \infty } {e^{ - s} \frac{\pi }{2}ds}  = \frac{\pi }{2},$$
leading to the asymptotics in the title of the question.
Addendum. Note that
$$
\int_0^{1/n^{1/k} } {e^{ - s} \arctan \left( {n^{1/k} s} \right)ds}  \le \int_0^{1/n^{1/k} } {\frac{\pi }{2}ds}  \le \frac{\pi }{{2n^{1/k} }} \to 0.
$$
For the remaining part, we write
$$
\int_{1/n^{1/k} }^{ + \infty } {e^{ - s} \arctan \left( {n^{1/k} s} \right)ds}  = \int_{1/n^{1/k} }^{ + \infty } {e^{ - s} \frac{\pi }{2}ds}  - \int_{1/n^{1/k} }^{ + \infty } {e^{ - s} \arctan \left( {\frac{1}{{n^{1/k} s}}} \right)ds} .
$$
Here
$$
\int_{1/n^{1/k} }^{ + \infty } {e^{ - s} \frac{\pi }{2}ds}  \to \int_0^{ + \infty } {e^{ - s} \frac{\pi }{2}ds}  = \frac{\pi }{2}
$$
and
$$
\left| {\int_{1/n^{1/k} }^{ + \infty } {e^{ - s} \arctan \left( {\frac{1}{{n^{1/k} s}}} \right)ds} } \right| \le \frac{1}{{n^{1/k} }}\int_{1/n^{1/k} }^{ + \infty } {\frac{{e^{ - s} }}{s}ds}  \to 0,
$$
by L'Hôpital's rule.
A: For your curiosity.
You can make very nice things with
$$K=\int_0^\infty e^{-a t}\tan^{-1}(t)\,dt$$ Using the fact that
$$\frac{e^{-a t}}{ \left(t^2+1\right)}=\frac i 2\left(\frac{e^{-a t}}{t+i}-\frac{e^{-a t}}{t-i}  \right)$$ and obvious changes of variable lead to exponential integral functions. So, after using the bounds
$$K=\frac{i (e^{-i a} \text{Ei}(i a)- e^{i a} \text{Ei}(-i a))+\pi ( e^{i a}+  e^{-i a})}{2 a}$$ Using expansions around $a=0$ leads to
$$K=\frac{\pi }{2 a}+(\log (a)+\gamma -1)-\frac{\pi }{4}a-\frac{6\log (a)+6 \gamma -11}{36}  a^2+O\left(a^3\right)$$ Make $a=n^{-\frac 1k}$ to obtain your approximation.
