show that the sequence-indexed with $a_n$ , $${1\over{1+t^2}} - {e^{-ta_n}\over{(1+t^2)}}(\cos a_n + t\sin a_n)$$ is bounded from above by an integrable function for a sufficiently large $a_n$ ($\lim_{n\to\infty}a_n= + \infty$)
i tried this :
$(\cos a_n+t\sin a_n)\ge(-1-t)$
$1-e^{-ta_n}(\cos a_n+t\sin a_n)\le1+ {(1+t)e^{-ta_n}}$
${1\over{1+t^2}}(1-e^{-ta_n}(\cos a_n+t\sin a_n))\le{1\over{1+t^2}}(1+ {(1+t)e^{-ta_n}})$
and $g_n(t)={1\over{1+t^2}}(1+ {(1+t)e^{-ta_n}})$ is integrable ?
Consider $f_n(t)=\dfrac1{1+t^2} -\dfrac {\mathrm e^{-ta_n}}{1+t^2}(\cos a_n + t\sin a_n)$. The goal is not to find some integrable functions $(g_n)$ such that $f_n\leqslant g_n$, presumably on $(0,+\infty)$, for every $n$ large enough, but to find an integrable function $g$ such that $f_n\leqslant g$ on $(0,+\infty)$ for every $n$ large enough.
To do so, let $N$ such that $a_n\geqslant1$ for every $n\geqslant N$ and note that $\cos a_n + t\sin a_n\geqslant-(1+t)\geqslant-2(1+t^2)$ and that $\mathrm e^{-a_n t}\leqslant\mathrm e^{-t}$ for every $n\geqslant N$, hence $$ f_n(t)\leqslant\frac1{1+t^2}+2\mathrm e^{-t}=g(t), $$ and $g$ is obviously integrable on $(0,+\infty)$.
If, as I suspect, the goal of the exercise is actually to find an integrable function $g$ such that $|f_n|\leqslant g$ on $(0,+\infty)$ for every $n$ large enough, use $|\cos a_n + t\sin a_n|\leqslant1+t\leqslant2(1+t^2)$ $$ |f_n(t)|\leqslant\frac1{1+t^2}+2\mathrm e^{-t}=g(t). $$ Finally, note that somewhat more refined arguments allow to show that, irrespectively of the value of $a_n$ and for every $n$, for every $t\geqslant0$, $$ |f_n(t)|\leqslant\frac3{1+t^2}. $$