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$\int_{-\infty }^{\tau} \! x(\tau)\delta(a\tau-b\tau') \, d\tau'=\int_{-\infty }^{(a+b)\tau} \! x\left(\frac{\tau''-b}{b}\right)\delta(\tau'')\frac{1}{b} \, d\tau \mapsto 0$ for $\tau<0, \frac{1}{b}x(\frac{-a}{b}\tau )$ for $\tau>0 $

Can someone write me steps how they got the solution for case $\tau>0$ and why is upper limit $(a+b)\tau$?

I know that they used substitution $\tau''=a\tau-b\tau'$, but I am getting lost trying to get this solution and can't get this upper limit.

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  • $\begingroup$ I don't understand the notations in the first line. $\endgroup$ – Jack Mar 6 '17 at 15:27
  • $\begingroup$ @AJMansfield There is indeed a factor of $b$ in there. $\delta(a\tau-b\tau')=\delta \left ( (-b) \left ( \tau' - \frac{a\tau}{b} \right ) \right )=\frac{1}{|b|} \delta \left ( \tau' - \frac{a\tau}{b} \right )$. $\endgroup$ – Ian Mar 6 '17 at 16:03
  • $\begingroup$ @Ian oh wait you are correct, I messed up on my calculation. $\endgroup$ – AJMansfield Mar 6 '17 at 16:12
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I think there are some mistakes here. The point of the first step is to make the argument of the delta function just be the variable of integration. So (using cleaner notation) we have

$$\int_{-\infty}^x f(y) \delta(ax-by) dy$$

and we change variables to $z=ax-by$. This turns the lower limit into $+\infty$ and the upper limit into $(a-b)x$. (I assume $b>0$.) Then $y=-\frac{z-ax}{b}$ and $\frac{dy}{dz}=-\frac{1}{b}$ So we have

$$-\frac{1}{b} \int_{+\infty}^{(a-b)x} f \left ( \frac{z-ax}{b} \right ) \delta(z) dz$$

Reversing the limits (so that the orientation is correct) changes it to:

$$\frac{1}{b} \int_{(a-b)x}^{+\infty} f \left ( \frac{z-ax}{b} \right ) \delta(z) dz.$$

Finally the integral is either $f \left ( -\frac{ax}{b} \right )$ if $(a-b)x<0$ or $0$ if $(a-b)x>0$. It is badly defined if $(a-b)x=0$.

Generally speaking, if $f$ is a continuously differentiable function which only has simple roots $r_i$, then $\delta(f(x))=\sum_i \frac{1}{|f'(r_i)|} \delta(x-r_i)$, and so $\int_a^b g(x) \delta(f(x)) dx = \sum_{i : r_i \in (a,b)} \frac{1}{|f'(r_i)|} g(r_i)$. (Again we must require that none of the $r_i$ be exactly equal to $a$ or $b$, otherwise things become badly defined again.)

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  • $\begingroup$ I dont understand part where solution is excatly $f(\frac{-ax}{b})$? How to get excatly that? $\endgroup$ – Ana Matijanovic Mar 6 '17 at 16:58
  • $\begingroup$ @AnaMatijanovic It is just the factor $f(...)$ with $z=0$. $\endgroup$ – Ian Mar 6 '17 at 16:58
  • $\begingroup$ Okay. How did you get $(a-b)x$? If I find $x$ from $z=ax-by$ I don't get that? $\endgroup$ – Ana Matijanovic Mar 6 '17 at 17:03
  • $\begingroup$ @AnaMatijanovic $(a-b)x$ is what you get when you substitute $y=x$ into $z=ax-by$. Notice that my variable of integration is $y$ at first. $\endgroup$ – Ian Mar 6 '17 at 17:04
  • $\begingroup$ @AnaMatijanovic Here $y=x$ came from the upper limit of the original integration. $\endgroup$ – Ian Mar 6 '17 at 17:10

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