I am struggling to solve several problems in my 'Signals and Systems' textbook. However, I just met a confused problem.
Q) $\displaystyle\int_{-4}^{4}\left(t-2\right)^2\delta'\left(-\frac13t+\frac12\right)dt$
I tried to solve with the method, 'integration by substitution'.
$\displaystyle-\frac13t = x \Leftrightarrow t=-3x \Leftrightarrow dt = -3dx$
so, $\displaystyle\int_{-4}^{4}\left(t-2\right)^2\delta'\left(-\frac13t+\frac12\right)dt=\int_{\frac43}^{-\frac43}(-3x-2)^2\delta'(x+\frac12)(-3dx)$
$\displaystyle=3\int_{-\frac43}^{\frac43}(3x+2)^2\delta'(x+\frac12)dx$
Then, I found an equation on the internet,
(19) in http://mathworld.wolfram.com/DeltaFunction.html
,which is
$\displaystyle\int_{-\infty}^\infty f(x) \delta'(x-a)dx=-f'(a)$.
Actually, I failed to understand how the equation above is induced. :(
However, I can apply it.
So, I assumed that $f(x) = (3x+2)^2$ and $\displaystyle a = -\frac12$.
$\displaystyle f'(x)=2\left(3x+2\right)\cdot3=18x+12 \Rightarrow -f'\left(-\frac12\right)=-3$
$\displaystyle\therefore3\int_{-\frac43}^{\frac43}(3x+2)^2\delta'(x+\frac12)dx=3\cdot\left(-f'\left(-\frac12\right)\right)=3\cdot(-3)=-9$
But, the answer is 3 in the solution of this book.
Are there any errors in my solving process?
Thanks a lot for the kind answers.
Best Answer
The Dirac Delta function $\delta(x)$ is very cool in the sense that
$$ \delta(x) = \begin{cases} +\infty, \, & x =0 \\ 0, \, & x \ne 0 \end{cases} $$
Its unique characteristics do not end there though, because when integrating the Dirac Delta function we would get
$$ \int_{-\infty}^\infty \delta(x) dx = 1$$
Or, if we have another function $f(x)$ multiplied to the Dirac Delta function and integrating them we would get
$$ \int_\infty^\infty f(x) \delta(x) \, dx = f(x) \int_{-\infty}^\infty \delta(x) \, dx = f(0) $$
$$\\$$ Since
$$ \int_{-\infty}^\infty \delta(x) \, dx = \begin{cases} 0, \, & x \ne 0 \\ 1, \, & x = 0 \end{cases}$$ Therefore in the previous integral we would have $$\int_{-\infty}^\infty f(x)\delta(x) \, dx = \int_{-\infty}^\infty f(0)\delta(0) \, dx = f(0)$$.
What if we have $\delta(x-a)$? It's the same thing! The only thing here is we need to satisfy the condition $x-a = 0$ such that $\int_{-\infty}^\infty \delta(x-a) \, dx = 1$. If the bounds of our integral though is not infinity, we need to make sure that the if we let $x = a$, $a$ would be in the bounds of the integral or else the integral would evaluate into zero.
$$\int_{-b}^b f(x)\delta(x-a) \, dx = \begin{cases} 0, & \text{if $b<a$ or $-b >a$ such that we cannot let $x \ne a$} \\ f(0), & \text{if $-b<a<b$ such that we can let $x = a$} \end{cases}$$.
What if we have $\int_{-\infty}^\infty f(x) \delta'(x-a) \, dx$? Let's then compute it by integration by parts.
$$\int_{-\infty}^\infty f(x) \delta'(x-a) \, dx = f(x)\delta(x-a)\bigg|_{-\infty}^\infty - \int_{-\infty}^\infty \delta'(x-a)f(x) \, dx = -f'(a)$$
Or in general
$$\int_{-\infty}^{\infty} f(x) \delta^{(n)}(x-a) \, dx = (-1)^n f^{(n)}(a)$$.