- #1
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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:
[tex] \int{f(t)dt} \equiv \int_{-\infty}^{t} {f(\tau) d\tau} [/tex]
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:
[tex] \int{f(t)dt} = F(t) + C [/tex]
In comparison, from what I know of improper integrals, the right hand side should be:
[tex] \int_{-\infty}^{t} {f(\tau) d\tau} = \lim_{a \rightarrow -\infty} \int_{a}^{t} {f(\tau) d\tau} [/tex]
[tex] = F(t) - [\lim_{a \rightarrow -\infty} F(a)] [/tex]
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?
The true definition of an indefinite integral is:
[tex] \int{f(t)dt} \equiv \int_{-\infty}^{t} {f(\tau) d\tau} [/tex]
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:
[tex] \int{f(t)dt} = F(t) + C [/tex]
In comparison, from what I know of improper integrals, the right hand side should be:
[tex] \int_{-\infty}^{t} {f(\tau) d\tau} = \lim_{a \rightarrow -\infty} \int_{a}^{t} {f(\tau) d\tau} [/tex]
[tex] = F(t) - [\lim_{a \rightarrow -\infty} F(a)] [/tex]
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?