Bug introduced in 9.0 or earlier and fixed in 10.4
I am attempting to perform a convolution involving the Dirac Delta function:
$\int_{-\infty}^{\infty} \frac{1}{t+1} \cdot \delta(t+1)\ dt$
I would expect that the result of this integral is undefined; however, Mathematica says that it is $-1$.
Strangely, Mathematica also says that the following similar integral also evaluates to $-1$:
$\int_{-\infty}^{\infty} \frac{1}{t} \cdot \delta(t+1)\ dt$
This makes no sense to me based on Wolfram's definition of the Dirac function, which states that:
$\int_{-\infty}^{\infty} f(x) \cdot \delta(x-a)\ dx = f(a)$
Can anyone explain Mathematica's rational to me?
Mathematica Code
Integrate[1/(tao + 1) * DiracDelta[tao + 1], {tao, -Infinity, Infinity}]
-1
Integrate[1/(tao) * DiracDelta[tao + 1], {tao, -Infinity, Infinity}]
-1
Wolfram Technical Support contacted, a support case with the identification [CASE:3433414] was created.
Integrate[DiracDelta[t - a]/(t - a), {t, -Infinity, Infinity}]
, even for symbolica
. TheIntegrate
code translates the function $f(t) = 1/(t+a)$ to $f^*(t)=1/t$, but it does not translate the location of the singularity $t=a$ to $t = 0$; so you end up with $f^*(a) = 1/a$ for the result instead of either $0$ or an error message. $\endgroup$ – Michael E2 Sep 27 '15 at 2:42Integrate[DiracDelta[t]/t,{t,-Infinity,Infinity}]
. $\endgroup$ – Jens Sep 27 '15 at 5:37