Why do singularities disappear when performing convolutions with the Dirac Delta function?

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.

• Please edit your question to include the actual code, not an image of it, so that readers can run the code without retyping it. Sep 26 '15 at 23:40
• While the rule that integrating $f(t)$ against the $\delta$ function yeilds $f(0)$ applies only to suitably nice functions, I think it's a bug. It happens in general, too, to Integrate[DiracDelta[t - a]/(t - a), {t, -Infinity, Infinity}], even for symbolic a. The Integrate 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. Sep 27 '15 at 2:42
• @MichaelE2 I agree it's a bug and I added the corresponding tag. The expected result can be see here: Integrate[DiracDelta[t]/t,{t,-Infinity,Infinity}].
– Jens
Sep 27 '15 at 5:37
• Wolfram Technical Support contacted, a support case with the identification [CASE:3433414] was created. Sep 27 '15 at 8:57
• Just to be clear: the second integral is correct, since it agrees with the formal definition in the case $f(x) = 1/x$ and $a = -1$. The first one is definitely screwy, though. Sep 28 '15 at 14:06

With Limit[] seems to work.

Limit[Integrate[1/(t + 1 + \[Epsilon])*
DiracDelta[t + 1], {t, -\[Infinity], \[Infinity]}], \[Epsilon] -> 0]

$\infty$

Limit[Integrate[1/(t + 1 - \[Epsilon])*
DiracDelta[t + 1], {t, -\[Infinity], \[Infinity]}], \[Epsilon] -> 0]

$-\infty$

we have at the same point: $(\infty\ \text{and} -\infty) \to \text{undefined}$

Or:

{Limit[Integrate[
1/(t + 1 + \[Epsilon])*
DiracDelta[t + 1], {t, -\[Infinity], \[Infinity]}], \[Epsilon] ->
0, Direction -> 1],
Limit[Integrate[
1/(t + 1 + \[Epsilon])*
DiracDelta[t + 1], {t, -\[Infinity], \[Infinity]}], \[Epsilon] ->
0, Direction -> -1]}

$\{-\infty ,\infty \}$