Given an integral$$\int_1^T \frac{f(t)}{t} \, {\rm d}t$$where $f(t)$ is oscillating and I want to make an estimate I can do the following$$\left|\int_1^T \frac{f(t)}{t} \, {\rm d}t\right| \leq \int_1^T \left|\frac{f(t)}{t}\right| \, {\rm d}t \leq \int_1^T |f(t)| \, {\rm d}t$$but this is not enough. Is this even possible:$$\left|\int_1^T \frac{f(t)}{t} \, {\rm d}t\right| \leq \left|\int_1^T f(t) \, {\rm d}t \right|\, ? $$
$\endgroup$ 51 Answer
$\begingroup$The last step can only be done on intervals with fixed sign. So, you have to split an oscillating function into positive and negative domains, and you get
$$\int_{x_1}^{x_2}\frac{|f(t)|}{t}dt\leq \frac{1}{x_1}\int_{x_1}^{x_2}\left|f(t)\right|dt$$where $x_2>x_1>0$.
The entire integral is then an alternating sum of such terms -- this is the reason why it doesn't work in general. In alternating sums, if you change the magnitude of some terms, you can get any value (imagine if the inequality was exact for positive terms but for negative, they just become zero?)
Let's define
$$a_n=\int_{x_n}^{x_{n+1}}\frac{|f(t)|}{t}dt$$$$b_n=\int_{x_n}^{x_{n+1}}|f(t)|dt$$where $x_n$ are zeros of $f(t)$.
Your statement is, that
$$\sum (-1)^n b_n$$is bounded, and you want to prove that$$\sum (-1)^n a_n$$is bounded, too.
But that is exactly the Dirichlet test that says (bounded alternating)×(monotonically decreasing to zero)=(convergent).
In our case, the bounded alternating series is $(-1)^n b_n$ and the monotonically decreasing to zero is $1/x_n$, resulting in a statement for $a_n$ which is bounded by the product of these two.
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