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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 ?

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That's true. But is it terribly clear why $g_n(t)$ is integrable? (BTW, there's a missing minus sign in the final line) – user108903 Jan 19 '13 at 0:49
for me , i didnt understand the concept of "an integrable function for a sufficiently large $a_n$" , i just wanted to prove that a bounded integrable function exist , even if my solution is correct it is not complete ! the teacher solution goes to calculate the limit of $g_n$ then replacing $te^{-ta_n}$ with an $\varepsilon \, \forall n \ge N$ , i didn't understand why $te^{u}$ form and not $e^{u}$ and why he did all of that ! – Lofaif Jan 19 '13 at 1:10
It sounds like the $\epsilon$ business is directed towards an application of the dominated convergence theorem in order to show that the limit of the integrals is zero. – user108903 Jan 19 '13 at 1:20
yeah correct ! , he worked in his solution with $f_n(t)={1\over{1+t^2}}e^{-ta_n}(cosa_n+tsina_n)$ and he said : lets prove that : $\int_0^\infty f_n(t) dt=0$ .. and he proved that but our sequence is different , i didnt understand why he cut the sequence into two part , and i just tried to prove it as above ! – Lofaif Jan 19 '13 at 1:34
up vote 0 down vote accepted

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 $t\geqslant0$, $$ |f_n(t)|\leqslant\frac3{1+t^2}. $$

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