My textbook says: "Let $f$ be piecewise continuous on every interval $[-L,L]$ and suppose that $\int_{-\infty}^{\infty}|f(x)| dx$ converges. Then the Fourier integral of $f$ converges to $\frac{1}{2}[f(x+)+f(x-)]$ at each $x$ at which $f$ has a right and left derivative there."
It defines the Fourier integral as $$\\f(t) = \int_0^ \infty A( \omega ) \cos( \omega t) + B( \omega ) \sin( \omega t) d \omega$$ where $$\displaystyle A(\omega ) = \frac{1}{\pi} \int_{- \infty }^\infty f(t) \cos( \omega t) dt \\\displaystyle B(\omega ) = \frac{1}{\pi} \int_{- \infty }^ \infty f(t) \sin( \omega t) dt.$$
It derives this expression from
$$
\begin{split}
f(x) = & \frac{1}{2\pi}\left(\int_{-L}^{L}f(t) dt\right) Δ\omega \\
&+\frac{1}{\pi}\sum_{n=0} ^{\infty}\left[\left(\int_{-L}^{L}f(t)\cos(\omega_nt) dt\right)\cos(\omega_nt)+\left(\int_{-L}^{L}f(t)\sin(\omega_nt) dt\right)\sin(\omega_nt)\right]Δ\omega
\end{split}
$$ where $$\omega_n=\frac{n\pi}{L}$$ $$Δ\omega=\omega_n-\omega_{n-1}=\frac{\pi}{L}$$ and let $Δ\omega$ approaches $0$ and $L$ approaches $\infty$.
Here, since (if the following integral is convergent)
$$
f(x)=\frac{1}{2\pi}\left(\int_{-L}^{L}f(t) dt\right) Δ\omega=0,$$ $f(x)$ becomes the previous expression without the first term and the book's theorem follows.
My question is, in the derivation above, only $\int_{-\infty}^{\infty}f(t) dt$ is required to be convergent, so why is the convergent of $\int_{-\infty}^{\infty}|f(x)| dx$, a stricter condition, necessary? Am I missing any details?
Best Answer
Too long for comment
Let us see an example in which $\int_{-\infty}^\infty f(t) \mathrm{d}t$ exists however $B(\omega) = \frac{1}{\pi}\int_{-\infty}^\infty f(t) \sin (\omega t) \mathrm{d} t$ does not exist everywhere.
Let $$f(t) = \left\{\begin{array}{cc} \frac{\sin t}{t} & t\ge \pi \\[4pt] 0 & t < \pi. \end{array} \right.$$
$f(t)$ is continuous on $(-\infty, \infty)$. We have $$\int_{-\infty}^\infty f(t) \mathrm{d}t = \int_{\pi}^\infty \frac{\sin t}{t} \mathrm{d}t = -\mathrm{Si}(\pi)+ \frac{\pi}{2}.$$
However, $$B(1) = \frac{1}{\pi}\int_{\pi}^\infty \frac{\sin t}{t} \sin t \mathrm{d} t = \infty.$$