Integrating with the Dirac delta distribution

  • #1
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Summary:

In the definite integral of a delta function, how narrow can the interval be?
Given
\begin{equation}
\begin{split}
\int_{y-\epsilon}^{y+\epsilon} \delta^{(2)}(x-y) f(x) dx &= f^{(2)}(y)
\end{split}
\end{equation}
where ##\epsilon > 0##

Is the following also true as ##\epsilon \rightarrow 0##
\begin{equation}
\begin{split}
\int_{y-\epsilon}^{y+\epsilon} \delta^{(2)}(x-y) f(x) dx &\rightarrow \delta^{(2)}(x-y) f(x)
\\
&= f^{(2)}(y)
\end{split}
\end{equation}

If not, why?
 

Answers and Replies

  • #2
PeroK
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The integral is constant for ##\epsilon > 0##, hence the limit as ##\epsilon \rightarrow 0^+## is ##f(y)##.

I'm not sure that the exponent ##^{(2)}## means.
 
  • #3
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##\delta^{(2)}(x)= \delta''(x)##

My question: ##\delta''(x-y) f(y) \stackrel{?}{=} f''(y)##
 
  • #4
PeroK
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##\delta^{(2)}(x)= \delta''(x)##
The same applies. If it has that value for all ##\epsilon > 0## then the right-hand limit exists and is equal to the constant value. That doesn't, however, mean that you can set ##\epsilon = 0##.
 
  • #5
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For any value of x other than y, the equation is zero. The equation can be written as follows:
\begin{equation}
\begin{split}
\delta(x-y) f(x) &= \begin{array}{cc}
f(y) & \text{ if } x=y \\
0 & \text{ if } x \neq y
\end{array}
\end{split}
\end{equation}

It is simply a simplification of this notation to write ##\delta(x-y) f(x)=f(y)##

Frankly, any function can be written in this fashion (and the same for any derivative), such that
\begin{equation}
\begin{split}
f(y) &= \delta(x-y) f(x)
\end{split}
\end{equation}

I don't see why the integral is necessary.
 
  • #6
PeroK
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##\delta^{(2)}(x)= \delta''(x)##

My question: ##\delta''(x-y) f(y) \stackrel{?}{=} f''(y)##
That would imply that (for all ##x, y## and functions ##f##) we have ##\delta''(x-y)= \frac{f''(y)}{f(y)}##.

Which can't be right at all.
 
  • #7
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It is true that ##\delta''(x) * f(x) = \int_{y-\epsilon}^{y+\epsilon} \delta''(x-y) f(x) dx = f''(y)##

Thus, why would it not also be true that:
\begin{equation}
\begin{split}
\delta''(x-y) f(x) = \begin{array}{cc}
f''(y) & \text{ if } x=y \\
0 & \text{ if } x\neq y
\end{array}
\end{split}
\end{equation}
which can be simplified to ##\delta''(x-y) f(x) = f''(y)##
 
  • #8
PeroK
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It is true that ##\delta''(x) * f(x) = \int_{y-\epsilon}^{y+\epsilon} \delta''(x-y) f(x) dx = f''(y)##

Thus, why would it not also be true that:
\begin{equation}
\begin{split}
\delta''(x-y) f(x) = \begin{array}{cc}
f''(y) & \text{ if } x=y \\
0 & \text{ if } x\neq y
\end{array}
\end{split}
\end{equation}
which can be simplified to ##\delta''(x-y) f(x) = f''(y)##
It makes no sense. You can't equate a function of two variables with a function of one variable. I've already proved that it's false. What more can one say? It obviously cannot be true.
 
  • #9
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You are right, except that ##x## and ##y## can be considered dummy variables in the convolution ##\delta'' * f = f''##
 
  • #10
vela
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Frankly, any function can be written in this fashion (and the same for any derivative), such that
$$ f(y) = \delta(x-y) f(x) $$
No, it can't. You can only say
$$\int_{-\infty}^\infty \delta(x-y) f(x)\,dx = f(y).$$ Note that inclusion of the integral eliminates the problem @PeroK noted. Both sides are functions of only ##y##.

Moreover, sloppily speaking, one has
$$\delta(x) = \begin{cases}
+\infty & x = 0 \\
0 & x \ne 0
\end{cases}$$ so
$$f(x)\delta(x-y) = \begin{cases}
\operatorname{sgn}(f(y))\cdot \infty & x = y \\
0 & x \ne 0
\end{cases}$$ Only by integrating do you get a finite result.
 
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