# Indefinite integrals

1. Apr 13, 2005

### cepheid

Staff Emeritus
Well, this is an embarrassingly elementary question, but in my lecture slides (for an electrical engineering course, not a math course) the prof suddenly springs this claim on us:

The true definition of an indefinite integral is:

$$\int{f(t)dt} \equiv \int_{-\infty}^{t} {f(\tau) d\tau}$$

Well, I've never seen this before in my calculus text, and my attempt to make sense of it doesn't go so well. From what we know already, if F(t) is an antiderviative of f(t), then the left hand side becomes:

$$\int{f(t)dt} = F(t) + C$$

In comparison, from what I know of improper integrals, the right hand side should be:

$$\int_{-\infty}^{t} {f(\tau) d\tau} = \lim_{a \rightarrow -\infty} \int_{a}^{t} {f(\tau) d\tau}$$

$$= F(t) - [\lim_{a \rightarrow -\infty} F(a)]$$

So this "definition" is only true if the limit of the function F(t) as t approaches negative infinity exists and is either constant or zero. Why should this be true in general?

2. Apr 13, 2005

### dextercioby

Nope,the definition of the indefinite integrals (is that an oxymoron,or what??) of real valued functions over a certain domain of $\mathbb{R}$ is simply (original notation)

$$I(f(x))=:\left \{F(x)+C\left|\right\frac{dF(x)}{dx}=f(x),\frac{dC}{dx}=0\right\}$$

So the indefinite integral of a function is not a function,but a set of functions...

Daniel.

3. Apr 13, 2005

### Galileo

I don't know why your prof. said that, but I suspect he introduces it, because it would be important to realize in his notation for the remainder of the course.
Since it's an engineering class, I guess he can do that :uhh:

The biggest difference to note between a definite and an indefinite integral is that the former is a number, while the second is a function, or family of functions (a general notation for an antiderivative).

4. Apr 13, 2005

### cepheid

Staff Emeritus
Haha, yes, our engineering classes are quite ridiculous sometimes compared to our math and physics classes since they take so many liberties with the math! So not to worry, I'm not in such dire straits as it first may seem (for instance, I'm well aware of the difference between indefinite and definite integrals), its just that I came across this and it seemed totally off the wall. The only thing I can think of is that since he introduced it in the context of Laplace transforms (this was in the course of him trying to show us what taking the Laplace transform of the integral of a function results in), and all of the functions in the time domain that he seems to want us to worry about are zero for t less than zero! (If they aren't so by definition, then he just multiplies them by the unit step function which "steps up" at zero, so that they are).

I like Daniel's definition. I have not seen it expressed that way before, though it makes perfect sense.

Thanks everyone.

5. Apr 13, 2005

### dextercioby

U'll have to take care and 'report anything suspicious'...

Daniel.