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If I'm taking the derivative with respect to $t$ of the integral: $$\int_{-\infty}^t (t-\tau)u(\tau)d\tau$$ Does the Fundamental theorem of Calculus result in the answer of $0$?

I am trying to write the system in implicit form which is why I'm asking. I believe the implicit form would then be: $\dfrac{dy}{dt} = 0$.

Edit: The reason I don't feel comfortable using the Fundamental Theorem of Calculus is since the lower bound is not constant. Instead it is negative infinity.

Thank you for all the responses.

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  • $\begingroup$ Welcome! Here's a MathJax tutorial. $\endgroup$
    – k170
    Commented Sep 1, 2014 at 17:25
  • $\begingroup$ I edited your question in order to do it more clear. Tell me if is what you wanted to say. $\endgroup$
    – rlartiga
    Commented Sep 1, 2014 at 17:37

1 Answer 1

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Don't forget that you have $t$ also inside the integrand. You can deal with this in several different ways.

The simplest way here is to write $$\int_{-\infty}^t(t-\tau)u(\tau)d\tau = t\int_{-\infty}^tu(\tau)d\tau - \int_{-\infty}^t\tau u(\tau)d\tau$$

Now by taking the derivative using the product rule you get

$$\frac{d}{dt}\left(\int_{-\infty}^t(t-\tau)u(\tau)d\tau\right) = \int_{-\infty}^tu(\tau)d\tau + t \frac{d}{dt}\left(\int_{-\infty}^tu(\tau)d\tau\right) - \frac{d}{dt}\left(\int_{-\infty}^t\tau u(\tau)d\tau\right)$$

and use the fundamental theorem of calculus (see also Andre's answer here).

The other way to do it is to use Leibniz rule which says that

$$\frac{d}{dt} \int_a^t f(t,\tau) d\tau = f(t,t) + \int_a^t \frac{\partial }{\partial t}f(t,\tau) d\tau$$

where for your case $f(t,\tau) = (t-\tau)u(\tau) \to \frac{\partial}{\partial t} f(t,\tau) = u(\tau)$. In either case, if you do it correctly, you should end up with $\int_{-\infty}^tu(\tau)d\tau$ as the final answer.

${\bf Edit:}$

As a previous answer that is now removed tried to say: it does not matter that its $-\infty$ in the lower limit of the integral, you can still use the fundamental theorem of calculus. To see this write (for any $a$) $$\int_{-\infty}^t g(x) dx = \int_{-\infty}^a g(x) dx + \int_{a}^t g(x) dx$$

and now take the derivative to get

$$\frac{d}{dt}\left(\int_{-\infty}^t g(x) dx\right) = g(t)$$

since the derivative of the first term (which is a constant) vanishes.

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