[XI Liwen, typical problems and methods in mathematical analysis (2nd), Beijing: Higher Education Press, 2006] (page 436, t 4.5.14) If function $ p (t) $ can be accumulated on $ [0, + \ infty) $, and when $ t \ to + \ infty $, $ p (t) = O (t ^ N) $ ($ N $ is a positive integer ). and $ \ lm <0 $, proof: When $ t \ to + \ infty $, $ \ Bex \ int_t ^ {+ \ infty} p (\ Tau) e ^ {\ lm \ Tau} \ RD \ Tau = O (t ^ {n + 1}) e ^ {\ lm t }. \ EEx $
Proof: the original question is $ p (t) $ consecutive. by $ p (t) = O (t ^ N) $ (T \ to + \ infty) $ Zhi $ \ Bex \ forall \ ve> 0, \ exists \ t \ geq 1, \ st t \ geq t \ rA | p (t) | \ Leq \ ve T ^ n. \ EEx $ when $ t \ geq T $, $ \ beex \ Bea \ sev {\ int_t ^ {+ \ infty} p (\ Tau) e ^ {\ lm \ Tau} \ RD \ Tau} & \ Leq \ ve \ cdot \ int_t ^ {+ \ infty} \ Tau ^ n e ^ {\ lm \ Tau} \ RD \ Tau \ & <\ ve \ cdot C t ^ n e ^ {\ lm t }. \ EEA \ eeex $ here, the last step can be obtained through the segment points, and $ C $ depends on $ \ lm $, $ N $.
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