Let $f \in L^1$ with $f$ differentiable at zero and $f(0)=0$. Show $\int_{-\infty}^{\infty} \frac{f(x)}{x} dx$ exists.

Question

Is this proof good? Given the problem as stated. I first define,

$$ g(x,b) = \frac{f(x)}{x}e^{ibx} $$

Which has the following property,

$$ g_b(x,b) = if(x)e^{ibx} $$

And that,

$$ |g_b(x,b)| = |if(x)e^{ibx}| = |i||f(x)||e^{ibx}| \leq |f(x)| $$

where since $f \in L^1$ we have a bounding function of $g$. Thus, we can differentiate under the integral,

$$ \frac{d}{db} \int_{-\infty}^{\infty} g(x,b)dx $$

My question before I do that, is whether or notI can split the integral like this because $f(0)=0$ and $f$ is differentiable at zero,

$$ \int_{-\infty}^{\infty} g(x,b)dx = \int_{-\infty}^{0} g(x,b)dx + \int_{0}^{\infty} g(x,b)dx $$

So that,

$$ \frac{d}{db} \int_{-\infty}^{\infty} g(x,b)dx = \frac{d}{db}\int_{-\infty}^{0} g(x,b)dx + \frac{d}{db}\int_{0}^{\infty} g(x,b)dx $$

And thus,

$$ \frac{d}{db}\int_{-\infty}^{0} g(x,b)dx + \frac{d}{db}\int_{0}^{\infty} g(x,b)dx = \int_{-\infty}^{0} g_b(x,b) dx + \int_{0}^{\infty} g_b(x,b)dx $$

And since $g_b$ is bounded by $f$ which is integrable, the two integrals exists. Thus, setting $b=0$, we attain the desired result.

Answer 1

Let me restate your question:

If $f$ is Lebesgue integrable, $f(0)=0$, and $f'(0)$ is finite, then $\frac{f(x)}x$ is Lebesgue integrable.

I didn't read your solution because it seemed needlessly complicated, so here's mine:

Obviously $\frac{f(x)}x$ is measurable, so we only have to find some bounds.

Firstly, $\lim_{x\to0}\frac{f(x)}x=\lim_{x\to0}\frac{f(x)-f(0)}x=f'(0)\in\mathbb R$, so it must be bounded on some neighborhood of $0$, namely there are $\delta>0$ and $K>0$ such that for all $x\in(-\delta,\delta)\setminus\{0\}$ we have $\Bigl|\frac{f(x)}x\Bigr|<K$. Therefore it is Lebesgue integrable on $(-\delta,\delta)$.

Secondly, $\Bigl|\frac{f(x)}x\Bigr|\le\frac{|f(x)|}\delta$ on $(-\infty,-\delta]\cup[\delta,\infty)$, so it's also Lebesgue integrable there since $\frac{f(x)}\delta$ is.