If I define the cosine and sine Fourier transform as (skipping constant prefactors $(2\pi)^{0.5}$):

$$\mathcal{F}_C\{f(x)\}=\int_0^{\infty}\,f(x)\,\cos(\omega x)\,dx$$


$$\mathcal{F}_S\{f(x)\}=\int_0^{\infty}\,f(x)\,\sin(\omega x)\,dx$$

with $\omega>0$ and $f(x)$ is a real continous function such as $\int_0^{\infty}\,f(x)\,dx=1$ and $\lim_{x \to +\infty}\,f(x) = 0$. Now I would like to calculate the following limits:

1.$$\lim_{\omega \to 0}\,\mathcal{F}_{C}\{f(x)\}=?$$

2.$$\lim_{\omega \to 0}\,\mathcal{F}_{S}\{f(x)\}=?$$

3.$$\lim_{\omega \to +\infty}\,\mathcal{F}_{C}\{f(x)\}=?$$

4.$$\lim_{\omega \to +\infty}\,\mathcal{F}_{S}\{f(x)\}=?$$

Limits 1 and 2 are easily calculated to be equal to 1 and 0, respectively. This result is obtained by interchanging the integral and limit operators. However, since the limits $\omega \to +\infty$ for $\cos(\omega x)$ and $\sin(\omega x)$ do not exist, I cannot use the same trick for limits 3 and 4.

By using $f(x)=\exp(-x)$ as test function, I see that both limits 3 and 4 goes to 0.

How can I prove this for a general $f(x)$?


The Riemann-Lebesgue Lemma states that if $f$ is $\mathscr{L}^1$ integrable, then

$$\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)e^{i\omega t}dt=0$$.

Obviously, for real-valued $f$, we have

$$\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)e^{i\omega t}dt=\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)\cos(\omega t)dt+i\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)\sin(\omega t)dt=0.$$

Thus, this implies that both

$$\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)\cos(\omega t)dt=0$$


$$\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)\sin(\omega t)dt=0.$$


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