Antiderivative of little oThis question has already been asked but non of the answers I found were correct/thorough enough.For x near zero and n &gt; 1, is the antiderivative of o ( x n ) equal to o ( x n + 1 ) and how do we prove it?Little o is defined as:f is o(g) near zero if and only if there is a function u that tends to zero when x nears zero and satisfies f = u g ,, or for any ϵ &gt; 0, there exists some a &gt; 0 such that ∣ f ( x ) ∣≤ ϵ | g ( x ) | whenever 0 &lt; | x | &lt; a.I used both definitions but couldn't get anywhere, integration by parts also didn't help, I know that it is not true for the derivative that: the derivative of o ( x n ) equal to o ( x n − 1 ) but I'm not sure about integration.

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Antiderivative of little oThis question has already been asked but non of the answers I found were correct/thorough enough.For x near zero and   n  &amp;gt;  1, is the antiderivative of   o  (      x    n    ) equal to   o  (      x          n      +      1        ) and how do we prove it?Little o is defined as:f  is o(g) near zero if and only if there is a function u that tends to zero when x nears zero and satisfies   f  =  u  g  ,, or for any   ϵ  &amp;gt;  0, there exists some   a  &amp;gt;  0 such that   ∣  f  (  x  )  ∣≤  ϵ      |    g  (  x  )      |   whenever   0  &amp;lt;      |    x      |    &amp;lt;  a.I used both definitions but couldn&#039;t get anywhere, integration by parts also didn&#039;t help, I know that it is not true for the derivative that: the derivative of   o  (      x    n    ) equal to   o  (      x          n      −      1        ) but I&#039;m not sure about integration.

jcholewa39v8f · Accepted Answer

Step 1Yes, there&#039;s no trouble with this, the key point is that integration preserves inequalities (which is very false for differentiation); that is, if   f  (  x  )  ≤  g  (  x  ) then       ∫    a    b    f  (  x  )    d  x  ≤      ∫    a    b    g  (  x  )    d  x.More formally, if   f  (  x  )  =  o  (      |    x            |        n    ), then by definition we have that for any   ϵ  &amp;gt;  0 there exists   δ  &amp;gt;  0 such that if   0  &amp;lt;      |    x      |    &amp;lt;  δ then       |    f  (  x  )      |    &amp;lt;  ϵ      |        x    n        |  . Define  F  (  x  )  =      ∫    0    x    f  (  t  )    d  t  .Step 2Then for any   ϵ  &amp;gt;  0 there exists   δ  &amp;gt;  0 such that if   0  &amp;lt;      |    x      |    &amp;lt;  δ then       |    f  (  x  )      |    &amp;lt;  ϵ  (  n  +  1  )      |        x    n        |  , hence such that      |    F  (  x  )      |    =      |          ∫      0      x        f    (    t    )        d    t    |    ≤      ∫    0    x        |    f  (  t  )      |      d  t  ≤      ∫    0    x    ε  (  n  +  1  )      |    t            |        n      d  t  =  ε      |    x            |              n      +      1        .So   F  (  x  )  =  o  (      |    x            |              n      +      1        ). We didn&#039;t really need to insert the   n  +  1 prefactor but it&#039;s nice to make the final bound come out to exactly what it needs to be.

Antiderivative of little o This question has already been asked but non of the answers I found were

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