Definite integrals with -infinity low bound

[Salt]

I see equations of the form,

y=\int_{-\infty }^{t}{F\left( x \right)}dx

a lot in my texts.

What exactly does it mean? From the looks of it, it just means there is effectively no lower bounds.

I looked up improper integrals, but I can't say I really understand what is going on.

So when evaluating,

If \frac{d\left( f\left( x \right) \right)}{dx}=F\left( X \right)

Do I just take the lower bound term - that I have to subtract - to be the f(x) as x approaches -infinity? Do I set the lower bound term to 0?

y=\int_{-\infty }^{t}{F\left( x \right)}dx\; =\; f\left( t \right)-0=f\left( t \right)

?

 

[jbunniii]

It's defined this way:

\int_{-\infty}^t F(x) dx = \lim_{a \rightarrow -\infty} \int_{a}^{t} F(x) dx

and you would evaluate it as follows:

y = \int_{-\infty}^t F(x) dx = \lim_{a \rightarrow -\infty} \int_{a}^{t} F(x) dx = f(t) - \lim_{a \rightarrow -\infty} f(a)

 

[Salt]

OK I just saw this in my text (paraphrasing),

y\left( t \right)=\int_{-\infty }^{t}{x\left( \tau \right)}d\tau

therefore

x\left( t \right)=\frac{dy}{dt}​

So y\left( t \right)=\int_{-\infty }^{t}{x\left( \tau \right)}d\tau is just like a non-definite integral?

y\left( t \right)=\int_{-\infty }^{t}{x\left( \tau \right)}d\tau =\; \int_{}^{}{x\left( t \right)dt} ?

BTW I'm studying an engineering text, so maybe it's "shortcuted" somehow.

 

[Char. Limit]

I think it's pretty easily provable that in order for a definite integral with a lower bound at -infinity to exist, the function being integrated needs to tend to zero as the variable being integrated tends to -infinity. (Note that it doesn't work the other way! Try -1/x for proof!) So the derivative of the integral would be just the integrand, assuming that the integrand satisfies the conditions.