1

Evaluate $$\int_0^\infty f(x)\delta(x-1)dx$$where$$f(x)=\begin{cases}x^2,&0\le x<1\\\sin 2,&x=1\\x,&x>1\end{cases}$$

Attempt

Since the function is discontinuous at $1$, I couldn't directly say the answer would be $f(1)$. I considered the $\delta$-sequence$$\delta_k(x-1)=\begin{cases}\frac k2,& |x-1|<1/k\\0,&\text{otherwise}\end{cases}\\\int_0^\infty f(x)\delta_k(x-1)dx=\frac k2\Big[\int_{1-1/k}^1x^2dx+\int^{1+1/k}_1xdx\Big]\\=1+\frac12\Big(\frac1{3k^2}-\frac1{2k}\Big)$$

Then taking the limit as $k\to\infty$, I got the answer as $1$, which seems correct intuitively as it is the limit of the function at $x=1$. But the answer key gives the answer $\sin 2=f(1)$.

Is the key wrong or am I missing something?

Edit

The book defines the Dirac $\delta$ function (also called impulse function) $\delta(t)$ as the limit of a sequence of functions $\{\delta_k(t)\}$, where $$\int_{-\infty}^\infty\delta_k(t)dt=1$$For example,$$\delta_k(t)=\begin{cases}\frac k2,&|t|<1/k\\0,&\text{otherwise}\end{cases}$$is one such sequence. Then it goes on to say that the Dirac $\delta$ function can be understood as the generalized function $$\delta(t)=\begin{cases}0,&t\ne0\\\infty,&t=0\end{cases}$$

For continuous functions in $[0,\infty)$ and $a\ge0$,$$\int_0^\infty f(t)\delta(t-a)dt=f(a)$$

Shubham Johri
  • 17,739
  • What is your definition of $\delta$? (for discontinuous functions, I mean) – Klaus Feb 19 '19 at 13:59
  • I'm not sure whether the initial integral is well defined in distribution theory. Suppose you modify it slightly so that the limit of $f$ for $x \rightarrow 1$ differs for $x>1$ and $x < 1$. What value do you expect to see? A discontinuous function is not a suitable test function as far as I know. – quarague Feb 19 '19 at 14:07
  • I don't see how you can apply $\delta$ to that function since it's not in $C^{\infty}$? –  Feb 19 '19 at 14:22
  • @Klaus Please check the edit – Shubham Johri Feb 19 '19 at 15:30
  • @quarague Please check the edit. I encountered this function while studying Laplace transform. In case the $\delta$ function is centred at a point of jump discontinuity $x=a$ of $f(x)$, I would expect the integral$$\frac k2\Big[\int_{a-1/k}^af(x)dx+\int_a^{a+1/k}f(x)dx\Big]$$to converge to the average of the left and right hand limits of $f(x)$ at $x=a$. Does this seem right? – Shubham Johri Feb 19 '19 at 15:36
  • @ShubhamJohri I don't think this is well-defined. Wouldn't you get different results (for discontinuous functions) if you take different sequences $\delta_k$? Take, for example, a sequence that is not symmetric around $0$. – Klaus Feb 19 '19 at 15:52
  • 1
    Which text book are you using? Defining $\delta$ as a limit of functions seems shady to me and your example shows how this runs into trouble. Looking at the wikipedia article, if you define it through measure theory, you get $f(1)$ as the answer immediately per definition. If you define it via integrals with test functions which is what I was thinking about, this looks more complicated but the answer should be the same as the definitions are equivalent. – quarague Feb 19 '19 at 15:54
  • @Klaus In fact, the choice of regularization does not impact the result. Please see my posted solution equipped with an example that uses an asymmetrically placed regularized Dirac Delta. – Mark Viola Feb 22 '19 at 18:47
  • @quarague I suggest you read about regularization of the Dirac Delta. Moreover, the answer is NOT $f(1)$ from measure theory or otherwise. The answer is $\lim_{x\to 1}f(x)$ which equals $1$ here as $f$ has a removable singularity at $1$. See my posted solution for additional explanation. – Mark Viola Feb 22 '19 at 18:49
  • @MarkViola Sure, that much was clear. My point was that if you take a properly discontinuous function, the limit may actually depend on the Dirac sequence and therefore the Dirac delta in the OP is not well-defined. Of course, the function at hand is equivalent to a continuous function and therefore the integral cannot depend on which value you choose at $x = 1$. – Klaus Feb 25 '19 at 11:10

1 Answers1

1

The function $f$ has a removable discontinuity at $1$. Therefore, we have

$$\langle \delta_{1},f\rangle=\lim_{x\to 1}f(x)$$

To see this, we denote $\delta_n(x)$ as a regularization of the Dirac Delta, which for any suitable test function, $\phi(x)$, satisfies

$$\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x)\phi(x)\,dx=\phi(0)$$

Now, suppose we have a function $f(x)$ that is of compact support and is smooth everywhere except at $1$ where it has a removable discontinuity. Let $g(x)$ be defined as

$$g(x)=\begin{cases}f(x)&,x\ne 1\\\\\lim_{x\to 1}f(x)&, x=1\end{cases}$$

Then, since $g$ is a suitable test functions we see that

$$\begin{align} \lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x-1)f(x)\,dx&=\lim_{n\to \infty}\int_{-\infty}^\infty \delta_n(x-1)g(x)\,dx\\\\ &=g(1)\\\\ &=\lim_{x\to 1}f(x) \end{align}$$

And we are done!

EXAMPLE:

As an example, suppose $\delta_n(x)$ is the asymmetrically centered pulse function, which for $a\in(0,1)$ given by

$$\delta_n(x)=\begin{cases}n&,x\in[-a/n,(1-a)/n]\\\\0&,\text{otherwise}\end{cases}$$

Then, we have

$$\begin{align} \lim_{n\to\infty}\int_{-\infty}^\infty \delta_n(x-1)f(x)\,dx&=\lim_{n\to\infty}\left(n\int_{1-a/n}^{1+(1-a)/n} f(x)\,dx\right)\\\\ &=\lim_{n\to\infty}\left(n\int_{1-a/n}^{1} f(x)\,dx\right)+\lim_{n\to\infty}\left(n\int_{1}^{1+(1-a)/n} f(x)\,dx\right)\\\\ &=a\lim_{x\to 1^-}f(x)+(1-a)\lim_{x\to 1^+}f(x)\\\\ &=\lim_{x\to 1}f(x) \end{align}$$

since the discontinuity at $1$ is removable and therefore the right-side and left-side limits are equal.

NOTE:

It is of paramount importance to understand that if $f$ has a removable discontinuity at $x_0$, then the functional $\langle \delta_{x_0},f\rangle =\lim_{x\to x_0}f(x)$ but if $f$ has a jump discontinuity at $x_0$, then the functional $\langle \delta_{x_0},f\rangle$ is not defined.

In fact, I showed in This Answer, that if $H$ is he Heaviside Function, then $\langle \delta_0,H\rangle $ is meaningless.

Mark Viola
  • 184,670