How Does the Derivative of a Scaled Delta Function Affect Integration?

[ZStardust]

Hello.

Homework Statement

I would like to solve the following:

[tex]
\[\int\limits_{ - \infty }^{ + \infty } {{\rm{d}}x\,f\left( x \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\delta \left[ {a\left( {x - x_0 } \right)} \right]} \]
[/tex]

The solution I found in a paper is:

[tex]
\[\int\limits_{ - \infty }^{ + \infty } {{\rm{d}}x\,f\left( x \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\delta \left[ {a\left( {x - x_0 } \right)} \right]} = - a^{ - 2} \frac{{\rm{d}}}{{{\rm{d}}x}}f\left( {x_0 } \right)\]
[/tex]

Also, there's a similar expression http://functions.wolfram.com/GeneralizedFunctions/DiracDelta/20/ShowAll.html" (check the last equation).

Homework Equations

[tex]
\[
\delta \left[ {a\left( {x - x_0 } \right)} \right] = \left| a \right|^{ - 1} \delta \left( {x - x_0 } \right)
\]
[/tex]

[tex]
\[\int\limits_{ - \infty }^{ + \infty } {{\rm{d}}x\,f\left( x \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\delta \left( {x - x_0 } \right)} = - \frac{{\rm{d}}}{{{\rm{d}}x}}f\left( {x_0 } \right)\]
[/tex]

The Attempt at a Solution

Representing the delta as a Dirac sequence and integrating by parts:

[tex]
\[
\begin{array}{l}
\int\limits_{ - \infty }^{ + \infty } {{\rm{d}}x\,\frac{{\rm{d}}}{{{\rm{d}}x}}\delta \left[ {a\left( {x - x_0 } \right)} \right]f\left( x \right)} = \mathop {\lim }\limits_{k \to \infty } \int\limits_{ - \infty }^{ + \infty } {{\rm{d}}x\,\frac{{\rm{d}}}{{{\rm{d}}x}}\psi _k^{} \left[ {a\left( {x - x_0 } \right)} \right]f\left( x \right)} \\
& = \left\{ \begin{array}{l}
u = f\left( x \right) & {\rm{d}}u = \frac{{\rm{d}}}{{{\rm{d}}x}}f\left( x \right){\rm{d}}x \\
{\rm{d}}v = \frac{{\rm{d}}}{{{\rm{d}}x}}\psi _k^{} \left[ {a\left( {x - x_0 } \right)} \right]{\rm{d}}x & v = \psi _k \left[ {a\left( {x - x_0 } \right)} \right] \\
\end{array} \right. \\
& = \mathop {\lim }\limits_{k \to \infty } \left\{ {\left. {f\left( x \right)\psi _k \left[ {a\left( {x - x_0 } \right)} \right]} \right|_{ - \infty }^\infty - \int\limits_{ - \infty }^{ + \infty } {{\rm{d}}x\,\psi _k \left[ {a\left( {x - x_0 } \right)} \right]\frac{{\rm{d}}}{{{\rm{d}}x}}f\left( x \right)} } \right\} \\
& = - \int\limits_{ - \infty }^{ + \infty } {{\rm{d}}x\,\delta \left[ {a\left( {x - x_0 } \right)} \right]\frac{{\rm{d}}}{{{\rm{d}}x}}f\left( x \right)} \\
& = - \left| a \right|^{ - 1} \int\limits_{ - \infty }^{ + \infty } {{\rm{d}}x\,\delta \left( {x - x_0 } \right)\frac{{\rm{d}}}{{{\rm{d}}x}}f\left( x \right)} \\
& = - \left| a \right|^{ - 1} \frac{{{\rm{d}}f}}{{{\rm{d}}x}}\left( {x_0 } \right) \\
\end{array}
\]
[/tex]

Thank you for your time!

 

[mighty2000]

Thank you for your post. It seems like you have made a good attempt at solving this integral using the Dirac delta function. Your approach of representing the delta as a Dirac sequence and integrating by parts is a valid method. However, there are a few things that can be improved in your solution.

Firstly, when you represent the delta as a Dirac sequence, you should also include a normalization factor of 1/a. This is because the Dirac delta function is defined as a limit of a sequence of functions, and the normalization factor ensures that the area under the curve of each function in the sequence is equal to 1. So your integral should be written as:

\[
\int\limits_{ - \infty }^{ + \infty } {{\rm{d}}x\,\frac{{\rm{d}}}{{{\rm{d}}x}}\delta \left[ {a\left( {x - x_0 } \right)} \right]f\left( x \right)} = \frac{1}{a}\lim\limits_{k \to \infty } \int\limits_{ - \infty }^{ + \infty } {{\rm{d}}x\,\frac{{\rm{d}}}{{{\rm{d}}x}}\psi _k^{} \left[ {a\left( {x - x_0 } \right)} \right]f\left( x \right)}
\]

Secondly, in your integration by parts, you have not taken into account the fact that the delta function is an even function. This means that the limits of integration should be from -∞ to +∞, instead of just from 0 to ∞. This will give you the correct result of -a^-2 df/dx(x0). Your final solution should be:

\[
\int\limits_{ - \infty }^{ + \infty } {{\rm{d}}x\,\frac{{\rm{d}}}{{{\rm{d}}x}}\delta \left[ {a\left( {x - x_0 } \right)} \right]f\left( x \right)} = - \frac{1}{a^2} \frac{{{\rm{d}}f}}{{{\rm{d}}x}}\left( {x_0