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, $\displaystyle\int_{-4}^{4}\left(t-2\right)^2\delta'\left(-\frac13t+\frac12\right)dt=\int_{\frac43}^{-\frac43}(-3x-2)^2\delta'(x+\frac12)(-3dx)$


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 $\displaystyle a = -\frac12$.

$\displaystyle f'(x)=2\left(3x+2\right)\cdot3=18x+12 \Rightarrow -f'\left(-\frac12\right)=-3$


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.

  • $\begingroup$ Your calculation seems to be correct. $\endgroup$ – MathematicalPhysicist Aug 25 '15 at 10:40
  • $\begingroup$ Thanks, you mean, may the solution be wrong? $\endgroup$ – Danny_Kim Aug 25 '15 at 10:41
  • $\begingroup$ What is the name of the text book ? The Openhiem's Signal and System ..? $\endgroup$ – Cardinal Aug 25 '15 at 10:54
  • $\begingroup$ No, this is a Korean book. But the writer referred to many books. i.e. books by Oppenheim, Hsu, H.P., Phillips, Gordon, Soliman, etc. $\endgroup$ – Danny_Kim Aug 25 '15 at 17:42
  • $\begingroup$ @Emar You missed a crucial minus sign. $\endgroup$ – Ian Aug 25 '15 at 18:34

To solve this, it's easier to go by parts (treating the delta function formally as a function, which is in fact justifiable).

$$I = \int_{-4}^4 \underbrace{(t-2)^2}_{u} \underbrace{\delta'(-\frac{1}{3} t + \frac{1}{2} )}_{d v} dt$$

$$\dfrac{du}{dt} = 2(t-2)$$

$$v = -3 \delta(-\frac{1}{3} t + \frac{1}{2})$$

Therefore $$I = \left[\underbrace{-3 \delta(-\frac{1}{3} t + \frac{1}{2})}_{v} \underbrace{(t-2)^2}_u \right]_{-4}^4 - \int_{-4}^4 \underbrace{2(t-2)}_{du} \cdot \underbrace{-3 \delta(-\frac{1}{3} t + \frac{1}{2})}_{v} dt = 6 \int_{-4}^4 (t-2) \delta(\color{blue}{-\frac{1}{3} t + \frac{1}{2}}) dt$$

Now, a substitution $u = \color{blue}{-\frac{1}{3} t + \frac{1}{2}}$, so $du = \color{red}{-\frac{1}{3}} dt$:

$$I = 6 \times \color{red}{-3} \int_{11/6}^{-5/6} (\frac{3}{2} - 3 u - 2) \delta(u) du = -18 \int_{-5/6}^{11/6} (\frac{1}{2}-3u) \delta(u) du$$

Finally, that is clearly $-18 \times \frac{1}{2} = -9$.

That is essentially the construction of the formula you found in Mathworld.

  • $\begingroup$ Thank you for your meticulous solution. $\endgroup$ – Danny_Kim Aug 25 '15 at 11:04

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

$$\int_{-\infty}^{\infty} f(x) \delta^{(n)}(x-a) \, dx = (-1)^n f^{(n)}(a)$$.


The way to obtain $$ \int_{-\infty}^\infty f(x)\,\delta'(x-a)\,dx = -f'(a) $$ is to integrate by parts: $$ \int_{-\infty}^\infty f(x)\,\delta'(x-a)\,dx = \int_{-\infty}^\infty \left[\, f(x)\,\delta(x-a) \,\right]'\,dx - \int_{-\infty}^\infty f'(x)\,\delta(x-a)\,dx = \left[\, f(x)\,\delta(x-a) \,\right]\,\Big|_{-\infty}^{+\infty} - f'(a) = -f'(a) $$

  • $\begingroup$ +1. I'll comment that strictly speaking, this "integration by parts" calculation provides the definition of the distributional derivative. Its resemblance to ordinary integration by parts motivates the definition, but nevertheless it is just the definition. $\endgroup$ – Ian Aug 25 '15 at 10:49
  • $\begingroup$ Oh my god, thank you very much. I totally understand how the equation was induced. $\endgroup$ – Danny_Kim Aug 25 '15 at 10:53
  • $\begingroup$ @Ian True but I didn't want to get into the theory of distributions, so as not to potentially confuse the OP with more information than he needs at the moment. Let him get the "mechanics" of working with the Dirac distribution, then teach him why it's correct. :) $\endgroup$ – wltrup Aug 25 '15 at 10:53
  • $\begingroup$ In engineering, we have only derivatives!! I mean, I never heard anything about the distributional derivative. $\endgroup$ – Cardinal Aug 25 '15 at 10:53
  • $\begingroup$ @Danny_Kim Glad it helped you. $\endgroup$ – wltrup Aug 25 '15 at 10:56

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