# evaluating a Dirac delta function in an integral

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

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

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 $$e^{-\alpha\cdot 10^3} + \sin(5\pi\cdot 0)= e^{-1000\alpha}.$$

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

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: $$e^{-1000a}+0$$