$int_0^T f(t) e^-delta t dt ll delta^-1$ and $f(t)$

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I remember reading in a book (a very famous number theorist) that
[
int_0^T f(t) e^-delta t dt ll delta^-1, quad f geq 0
]
implies that
[
int_0^Tf ll T.
]



If we combine this with another inequality (whose proof I want to read
but can't find resource online)
[
f(u) ll int_-log u^log u f(u + t) e^-tdt
]
then it seems that we actually have that
[
int_0^T f(t) e^-delta t dt ll delta^-1, quad f geq 0
]
implies $f$ grows at most polynomially, don't we?



My other question is, is the inequality
[
f(u) ll int_-log u^log u f(u + t) e^-tdt
]
valid when $f(t) ll e^$?




calculus inequality analytic-number-theory




share|cite|improve this question























    up vote
    0
    down vote

    favorite












    I remember reading in a book (a very famous number theorist) that
    [
    int_0^T f(t) e^-delta t dt ll delta^-1, quad f geq 0
    ]
    implies that
    [
    int_0^Tf ll T.
    ]



    If we combine this with another inequality (whose proof I want to read
    but can't find resource online)
    [
    f(u) ll int_-log u^log u f(u + t) e^-tdt
    ]
    then it seems that we actually have that
    [
    int_0^T f(t) e^-delta t dt ll delta^-1, quad f geq 0
    ]
    implies $f$ grows at most polynomially, don't we?



    My other question is, is the inequality
    [
    f(u) ll int_-log u^log u f(u + t) e^-tdt
    ]
    valid when $f(t) ll e^$?




    calculus inequality analytic-number-theory




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I remember reading in a book (a very famous number theorist) that
      [
      int_0^T f(t) e^-delta t dt ll delta^-1, quad f geq 0
      ]
      implies that
      [
      int_0^Tf ll T.
      ]



      If we combine this with another inequality (whose proof I want to read
      but can't find resource online)
      [
      f(u) ll int_-log u^log u f(u + t) e^-tdt
      ]
      then it seems that we actually have that
      [
      int_0^T f(t) e^-delta t dt ll delta^-1, quad f geq 0
      ]
      implies $f$ grows at most polynomially, don't we?



      My other question is, is the inequality
      [
      f(u) ll int_-log u^log u f(u + t) e^-tdt
      ]
      valid when $f(t) ll e^$?




      calculus inequality analytic-number-theory




      share|cite|improve this question











      I remember reading in a book (a very famous number theorist) that
      [
      int_0^T f(t) e^-delta t dt ll delta^-1, quad f geq 0
      ]
      implies that
      [
      int_0^Tf ll T.
      ]



      If we combine this with another inequality (whose proof I want to read
      but can't find resource online)
      [
      f(u) ll int_-log u^log u f(u + t) e^-tdt
      ]
      then it seems that we actually have that
      [
      int_0^T f(t) e^-delta t dt ll delta^-1, quad f geq 0
      ]
      implies $f$ grows at most polynomially, don't we?



      My other question is, is the inequality
      [
      f(u) ll int_-log u^log u f(u + t) e^-tdt
      ]
      valid when $f(t) ll e^$?





      calculus inequality analytic-number-theory





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 22 at 9:01









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