Integration of unit impulse function

Question

Can someone please show me the integration steps?

$$\frac{1}{2}\int_{-\frac{1}{2}}^{\frac{3}{2}}\big[\delta(t)-2\delta(t-1)\big]e^{-jk\omega t}\ dt=\frac{1}{2}-e^{-jk\omega}$$

Answer 1

Hint: Use the following property of the delta function

$$\int_{-\infty}^{\infty}f(t)\delta(t-t_0)dt=f(t_0).$$

Answer 2

Well, when $\text{n}\cdot\left(1+\text{a}\right)>0$:

$$\int_{-\text{n}}^{\text{a}\cdot\text{n}}\text{n}\cdot\text{y}\left(t\right)\cdot\left\{\delta\left(t\right)-\frac{\delta\left(t-2\text{n}\right)}{\text{n}}\right\}\space\text{d}t=\text{n}\cdot\text{y}\left(0\right)-\text{y}\left(2\text{n}\right)\cdot\theta\left(\text{n},\text{n}\cdot\left(\text{a}-2\right)\right)\tag1$$

Where $\theta\left(x\right)$ is the Heaviside Theta function.

And, when $\text{a}>2$ (that is true in your case):

$$\int_{-\text{n}}^{\text{a}\cdot\text{n}}\text{n}\cdot\text{y}\left(t\right)\cdot\left\{\delta\left(t\right)-\frac{\delta\left(t-2\text{n}\right)}{\text{n}}\right\}\space\text{d}t=\text{n}\cdot\text{y}\left(0\right)-\text{y}\left(2\text{n}\right)\tag2$$

So, in your case we have $\text{n}=\frac{1}{2}$:

$$\int_{-\text{n}}^{\text{a}\cdot\text{n}}\text{n}\cdot\text{y}\left(t\right)\cdot\left\{\delta\left(t\right)-\frac{\delta\left(t-2\text{n}\right)}{\text{n}}\right\}\space\text{d}t=\frac{1}{2}-\exp\left(-\omega \text{k}i\right)\tag3$$