Dirac Delta Integration for Exponential Functions

[Jncik]

Homework Statement

find

\int_{-\infty}^{+\infty} x(t) \delta (\beta t - t_{0}) dt

for x(t) = e^{a t} u(t)

there is no information conserning a, β, or t_{0}...

The Attempt at a Solution

assuming that t_{0} is a constant\int_{-\infty}^{+\infty} x(t) \delta (\beta t - t_{0}) dt = <br /> \int_{-\infty}^{+\infty} e^{at}u(t) \delta (\beta t - t_{0}) dt <br /> = \int_{0}^{+\infty} e^{at} \delta (\beta t - t_{0}) dt <br /> = \int_{0}^{+\infty} e^{\frac{a}{b}t_{0}} dt<br /> = +\infty<br />

 

[lanedance]

when you integrate, the delta function is only non-zero at t_0 and
\int f(t) \delta (t-t_0) = f(t_0)

 

[lanedance]

this is useful for the properties of shifting and translating the delta function
http://en.wikipedia.org/wiki/Dirac_delta_function#Properties

 

[Jncik]

thanks

hence the result is just

e^{\frac{a}{b}t_{0}}

it would be 0 if we didn't integrate from 0 to infinity but from k>0 to infinity

correct?

edit: actually it would not be 0 if the integration was from k>t0/β

 

[lanedance]

you've got the right idea on the k>t0/b stuff, but i think you need to be careful with the scaling properties of the delta function

\int f(t) \delta (\beta t-t_0) dt

now try making the substitution
u = \beta t-t_0

 

[Jncik]

u = \beta t - t_{0} =&gt; du = \beta dt <br />

also

t = \frac{u + t_{0}}{\beta}

hence

\int_{-t_{0}}^{+\infty} f(\frac{u+t_{0}}{\beta}) \delta(u) \frac{du}{\beta}

which will give us

\frac{1}{\beta} f(\frac{t_{0}}{\beta}) = \frac{e^{\frac{a t_{0}}{\beta}}}{\beta} t_{0}&gt;0

:S, seems complicated... is this correct? In my previous attempt I forgot to change the variables.. I thought that we just had to search when the function inside \delta is 0, and we just put the t for which δ is 0, in the function before δ..

 

[lanedance]

that looks good to me - you could do it either way if you remember the scaling & translation properties of the delta function.

Whilst slightly longer I prefer the substitution as you don't need to remember any rules, and we got that extra factor of 1/beta correct without needing to remember it.

 

[Jncik]

yes you're right, thanks for your help :)

 

[lanedance]

no worries;)