Find out the value of $ \ \int_{-\infty}^{\infty} 2 * \delta(x+1) dx \ $ , Diract-Delta function $ \delta \ $

Question

If $ \ \delta \ $ be the Dirac -Delta function , then find out the value of $ \ \int_{-\infty}^{\infty} 2 * \delta(x+1) dx \ $ ?

Answer:

I know that

$ \int_{-\infty}^{\infty} \delta (x-b) f(x) dx=f(b) \ $

Thus if $ f(x)=1 $ , then we should have

$ \int_{-\infty}^{\infty} \delta (x-b) \cdot 1 dx =1 \ $

Therefore,

$ \int_{-\infty}^{\infty} 2* \delta(x+1) dx=2 \int_{-\infty}^{\infty} \delta (x+1) dx=2\cdot 1=2 \ $

I need confirmation of my work.

Is there any help ?

Answer 1

Well, in general:

$$\lim_{\text{n}\to\infty}\space\mathscr{I}_{\space\text{n}}\left(\text{a},\text{b}\right):=\lim_{\text{n}\to\infty}\space\int_{-\text{n}}^\text{n}\text{f}\left(x\right)\cdot\delta\left(\text{a}\cdot x+\text{b}\right)\space\text{d}x=\frac{1}{\left|\text{a}\right|}\cdot\text{f}\left(-\frac{\text{b}}{\text{a}}\right)\tag1$$

For $\text{a}\in\mathbb{R}\space\wedge\space\text{b}\in\mathbb{R}$

---

So, in your problem we get (when $\text{a}=\text{b}=1$):

$$\lim_{\text{n}\to\infty}\space\mathscr{I}_{\space\text{n}}\left(1,1\right):=\lim_{\text{n}\to\infty}\space\int_{-\text{n}}^\text{n}\text{f}\left(x\right)\cdot\delta\left(1\cdot x+1\right)\space\text{d}x=\frac{1}{\left|1\right|}\cdot\text{f}\left(-\frac{1}{1}\right)=\text{f}\left(-1\right)\tag2$$

Now, when $\text{f}\left(x\right)=2$ we get:

$$\text{f}\left(-1\right)=2\tag3$$

So:

$$\lim_{\text{n}\to\infty}\space\int_{-\text{n}}^\text{n}2\cdot\delta\left(1\cdot x+1\right)\space\text{d}x=2\tag4$$