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From the following answer of the post Fourier transform of derivate

Because you are assuming that $f'\in L^1(\mathbb{R})$, then the following limits exist $$ \lim_{x\rightarrow\pm\infty}f(x)=\lim_{x\rightarrow\pm\infty}\int_{0}^{x}f'(t)dt+f(0). $$ The above limits must be $0$ because $f \in L^1(\mathbb{R})$; otherwise, for example, if $\lim_{x\rightarrow\infty}f(x)=L \ne 0$, then $|f(x)| \ge L/2$ for large enough $x$, which would contradict the absolute integrability of $f$ on $[0,\infty)$. Because these limits must be $0$, then the evaluation terms below vanish: \begin{align} \int_{-\infty}^{\infty}f'(t)e^{-ist}dt & = \left.f(t)e^{-ist}\right|_{t=-\infty}^{\infty}-\int_{-\infty}^{\infty}f(t)\frac{d}{dt}e^{-ist}dt \\ & =is\int_{-\infty}^{\infty}f(t)e^{-ist}dt. \end{align}

I don't understand the part :

The above limits must be $0$ because $f \in L^1(\mathbb{R})$; otherwise, for example, if $\lim_{x\rightarrow\infty}f(x)=L \ne 0$, then $|f(x)| \ge L/2$ for large enough $x$, which would contradict the absolute integrability of $f$ on $[0,\infty)$

i.e, I don't understand what means "which contradict the absolute integrability of $f$" if one takes a limit different from $0$ ?

Let's take for example the function $f(x)=1/x +1$ : this function tends to $1$ when $x\to\infty$ : should one conclude that $f$ is not absolute integrable ?

I would like to prove with propositional logic that if I assume a limit $L$ different from $0$, then we can deduce a contradiction on absolute integrability.

For this, I would start by the definition of the limit $L\neq 0$ of $f$ when $x\to\infty$:

$\forall \epsilon > 0$,$\exists\,a$ such that $x \geqslant a \Longrightarrow f(x) \geqslant L - \epsilon$

How can I follow the reasoning to prove the contradiction with $\int_{0}^{+\infty}|f(x)|\text{d}x < +\infty$ ?

Regards

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    Which would contradict the absolute integrability of $f$. Indeed, if $|f(x)| \ge L/2 > 0$ were to hold for some $L > 0$ and for all $x \ge M$ for some $M$, then $\int_{M}^{\infty}|f(x)|dx = \infty$ would hold. And that would contradict the absolute integrability of $f$. – Disintegrating By Parts Sep 24 '18 at 00:57
  • -@DisintegratingByParts thanks for your answer. My issue is that I don't see how to pass from proposition (1) : $x \geqslant a \Longrightarrow f(x) \geqslant L - \epsilon$ to conclusion (2) : $\int_{a}^{\infty}|f(x)|dx = \infty$, have you get a way to start from (1) and obtain (2) (I mean using the logic propositional (1)) ? –  Sep 24 '18 at 10:13
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    $\int_a^{\infty}|f(x)|dx \ge \int_{a}^{\infty}L/2 dx = \infty$. – Disintegrating By Parts Sep 24 '18 at 15:23
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    -@DisintegratingByParts obviously ! regards –  Sep 24 '18 at 16:45

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