# Definite integrals with -infinity low bound

1. Feb 21, 2012

### 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?

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

?

2. Feb 21, 2012

### 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:

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

3. Feb 21, 2012

### 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.

4. Feb 21, 2012

### 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.