How is $\int_{-T_0/2}^{+T_0/2} \delta(t) \cos(n\omega_0 t)dt=1$ and $\int_{-T_0/2}^{+T_0/2} \delta(t) \sin(n\omega_0 t)=0$?

Question

In our Circuit Theory class we came across this function: $$f(t) = \begin{cases} \delta(t-nT_0), & \text{if $t=nT_0$ where $n\in\Bbb{Z}$} \\[2ex] 0, & \text{otherwise} \end{cases}$$

$\delta$ is presumably the dirac-delta function. And we can see that the time period of the function is $T_0$.

One can find the Fourier series for such a periodic function.

$$f(t)=a_0 + a_n \sum_{n=1}^{\infty} \cos(n\omega_0 t) + b_n \sum_{n=1}^{\infty} \sin(n\omega_0 t)$$ where $\omega_0=\frac{2\pi}{T_0}$.

Now our professor wrote,

$$a_0=\frac{1}{T_0}\int_{-T_0/2}^{+T_0/2} f(t) dt = \frac{1}{T_0}\int_{-T_0/2}^{+T_0/2} \delta(t) dt=\frac{1}{T_0}$$

I think this step is sort of justified because $\frac{1}{T_0}\int_{-\infty}^{+\infty} \delta (t) dt \approx \frac{1}{T_0}\int_{-T_0/2}^{+T_0/2} \delta (t) dt$ since $\delta(t)$ blows up only at $t=0$, and in other places it vanishes.

But then my main confusion occurred when he wrote:

$$a_n=\frac{2}{T_0}\int_{-T_0/2}^{+T_0/2} f(t) \cos(n\omega_0 t) dt = \frac{2}{T_0}$$

$$b_n=\frac{2}{T_0}\int_{-T_0/2}^{+T_0/2} f(t) \sin(n\omega_0 t) dt= 0$$

I'm not sure how $\int_{-T_0/2}^{+T_0/2} \delta(t) \cos(n\omega_0 t)dt=1$ and $\int_{-T_0/2}^{+T_0/2} \delta(t) \sin(n\omega_0 t)=0$.

Are these standard properties of the delta function? If not, how to find the value of those integrals?

I couldn't find anything related on the Wiki page.

Answer 1

$$\int_{-T_0/2}^{+T_0/2} \delta(t) \cos(n\omega_0 t)dt=1$$ $$\int_{-T_0/2}^{+T_0/2} \delta(t) \sin(n\omega_0 t)=0$$

Are these standard properties of the delta function? If not, how to find the value of those integrals?

The answer is yes.

One may recall that

$$ \int_{I}\, \delta(t)\,f(t)\,dt=f(0). $$

with $0\in I$.