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- Thread starter math&science
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Remember the defining property of the Dirac delta:

[tex] \int_{-\infty}^{\infty} \delta(x)dx = 1[/tex]

Thus

[tex] \int_{-\infty}^{\infty} \delta(ax)dx = \frac{1}{a} \int_{-\infty}^{\infty} \delta(y)dy = \frac{1}{a}[/tex]

So we can think of multipling the argument by*a* as being the same thing as dividing the function by *a*:

[tex]\delta(ax) = \frac{1}{a}\delta(x)[/tex]

[tex] \int_{-\infty}^{\infty} \delta(x)dx = 1[/tex]

Thus

[tex] \int_{-\infty}^{\infty} \delta(ax)dx = \frac{1}{a} \int_{-\infty}^{\infty} \delta(y)dy = \frac{1}{a}[/tex]

So we can think of multipling the argument by

[tex]\delta(ax) = \frac{1}{a}\delta(x)[/tex]

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