3
$\begingroup$

I have a problem evaluating this integral $$ \int\limits_{-\infty}^\infty \left[ e^{-at^3} \delta(t-10) + \sin (5 \pi t) \delta(t) \right] \mathrm{d}t $$

Can you please help me evaluate the integral?

$\endgroup$
  • $\begingroup$ You can use $\LaTeX$ to format equations, no need for oversized pictures. $\endgroup$ – Ruslan May 6 '14 at 15:29
5
$\begingroup$

The defining property of the Dirac delta is that $\langle \delta, f \rangle = \int \delta(t)f(t) dt=f(0)$ (for sufficiently nice $f$). It follows that $\int \delta(t-x)f(t)dt=\int \delta(t)f(t+x)dt = f(x)$; i.e., the shifted delta $\delta(t-x)$ "pulls out" the value $f(x)$. So your result is just $$e^{-\alpha\cdot 10^3} + \sin(5\pi\cdot 0)= e^{-1000\alpha}.$$

$\endgroup$
  • $\begingroup$ What does "sufficiently nice" mean in this context? $\endgroup$ – Ruslan May 6 '14 at 15:31
  • $\begingroup$ Continuous at the origin is fine in this context. $\endgroup$ – mjqxxxx May 6 '14 at 16:28
3
$\begingroup$

You can separate this integral like this (assuming the integrand is $dt$):

$$ \int\limits_{-\infty}^{+\infty} e^{-at^3} \delta(t-10) dt + \int\limits_{-\infty}^{+\infty} \sin{ (5\pi t) \delta(t) } dt $$

The expression $\delta(t-P)$ means you just can just substitute for the entire integral the function under the integral, putting the $P$ from the delta in place of the argument $t$. So in you example:

$$ e^{-at^3} \textrm{(where we replace } t \textrm{ with } 10 \textrm{) and } \sin(5\pi t) \textrm{(where we replace } t \textrm{ with } 0 \textrm{)} $$

So the final expression is: $$ e^{-1000a}+0 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.