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

Question

Bug introduced in 9.0 or earlier and fixed in 10.4

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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

Screenshot

Wolfram Technical Support contacted, a support case with the identification [CASE:3433414] was created.

Answer 1

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]}

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