# Fourier series of Heaviside step function?

Let us say we have the Heaviside unit step function $$\Theta(t-t^\prime)$$. I want to calculate its Fourier series $$\Theta(t-t^\prime)=\frac{1}{T}\sum_{n,m}\Theta_{\omega_n,\omega_m}e^{-i\omega_n t}e^{-i\omega_m t^\prime}\, ,$$ where $$\omega_n=\frac{2\pi}{T}n$$. So (according to my understanding) I have to execute the following integral $$\Theta_{\omega_p,\omega_l}=\frac{1}{T}\int_0^Tdt\int_0^Tdt^\prime \Theta(t-t^\prime)e^{i\omega_p t}e^{i\omega_l t^\prime}\, .$$ The unit step function affects the integration boundaries, such that $$\Theta_{\omega_p,\omega_l}=\frac{1}{T}\int_0^Tdt\int_0^tdt^\prime e^{i\omega_p t}e^{i\omega_l t^\prime}\, .$$ I calculated this integral for four different cases of $$(p,l)\in \{(0,0),(p,0),(0,l),(p,-p)\}$$ because all other cases are $$0$$. I get the following result $$\Theta_{\omega_p,\omega_l}=\frac{1}{T}\int_0^Tdt\int_0^tdt^\prime e^{i\omega_p t}e^{i\omega_l t^\prime}= \begin{cases} &\frac{1}{2}\, , \quad &p=l=0\\ &-\frac{i}{\omega_p}\, , &l=0\neq p\\ &\frac{i}{\omega_l}\, , &p=0\neq l\\ &\frac{i}{\omega_p}\, , &p=-l\neq 0\, . \end{cases}$$ And then I plug it into the sum and get the following result $$\Theta(t-t^\prime)=\frac{1}{2}+\frac{i}{T}\sum_{n\neq 0}\frac{1}{\omega_n}\left(e^{-i\omega_n t^\prime}-e^{-i\omega_n t}+e^{-i\omega_n (t-t^\prime)}\right)\, .$$ However, I am not sure if that is correct because the step function depends on the difference $$t-t^\prime$$, so I expected that the Fourier coefficients would also only depend on the frequency differences. Is there a flaw in the calculation?

The second question is: Since depends on the difference $$t-t^\prime$$, can one just write the Fourier series in the form of $$\Theta(t-t^\prime)=\frac{1}{T}\sum_{n,m}\Theta_{\omega_n-\omega_m}e^{-i\omega_n (t-t^\prime)}e^{-i\omega_m (t-t^\prime)}\, ,$$ or something similar, where can read out that it only depends on the time difference??

THX

• It is not a double Fourier series but just a single one. – Jon Nov 28 '18 at 15:00
• Ok, so you mean it is – mr. curious Nov 28 '18 at 15:56
• $$\Theta(t-t^\prime)=\frac{1}{\sqrt{T}}\sum_n\Theta_{\omega_n}e^{-i\omega_n(t-t^{prime})}$$? I'm asking because I have two time variables, so I will have to integrate by $t$ and $t^\prime$? – mr. curious Nov 28 '18 at 15:57
• The idea is that you just transform $\Theta(t)$ then, you can translate to whatever you want. The point is that, as far as I can say, just the Fourier transform has a meaning. $\Theta$ belongs to a distribution space that is meaningful only under integration sign.Check math.stackexchange.com/questions/73922/…. – Jon Nov 28 '18 at 17:17