proof of Delta dirac sifting

Question

I need to proof this expression : $\int_{-\infty}^\infty \delta(x-a)f(x) \, dx=f(a) $

Starting with this one: $\int_{-\infty}^\infty \delta(x)f(x) \, dx=f(0) $

Thanks in advance

Answer 1

The key to this problem is $u$-substitution in order to change $\delta(x-a)$ to $\delta(u)$. Thus, we can define $u=x-a \rightarrow x=u+a$. Since $\frac{dx}{du}=1$, the integral now becomes:

$$\int_{-\infty}^\infty \delta(x-a)f(x)dx=\int_{-\infty}^\infty \delta(u)f(u+a)du$$

Now, by the identity of $\int_{-\infty}^{\infty} \delta(x)f(x)dx=f(0)$, we know that taking the integral of $\delta(u)$ with some other function is just the same as plugging $u=0$ into that function. In this case, that other function is $f(u+a)$, so by plugging in $u=0$, we get $f(0+a)$. Therefore:

$$\int_{-\infty}^\infty \delta(u)f(u+a)du=f(0+a)=f(a)$$