convolution integral limits There are 2 types of convolution:

*

*The limit of the integral is from minus infinity to plus infinity


*The limit is from zero to t.
When we use the first and when we use the second?
$$\int f(\tau)g(t-\tau)\; d\tau$$
the second kind of convolution in my question there is in page 346 of the book titled "elementary differential equations and boundary value problems" by William Boyce and Richard DiPrima. You can see the picture below.

 A: Suppose that $f$ and $g$ are defined such that they only can take non-zero values on the positive reals, but are zero everywhere on the negative reals. Then
$$\int_{-\infty}^{\infty} f(\tau)g(t-\tau)\; d\tau = \int_0^t f(\tau)g(t-\tau)\; d\tau$$
I think the second case you came across is just a manifestation of your functions being defined that way. That's often implicit in the theory of Laplace transforms for instance or in probability theory (think of eg. the exponential distribution).
A: Convolution is defined as an integral over the whole space. The second "type" of convolution sometimes appear in the theory of differential equations. For instance, it appears in the so-called Duhamel's formula.
Although I think that the term "convolution" is a strong abuse in the last case, I've personally heard some colleagues using it.
A: The convolution of real functions is usually defined as 
$$(f * g)(t) := \int_{-\infty}^{\infty} f(\tau)g(t-\tau) d\tau.$$
I haven never seen a definition using a finite upper bound, except for periodic functions with period $T$ you will often see
$$(f * g)(t) := \frac 1 T \int_a^{a+T} f(\tau)g(t-\tau) d\tau.$$
