Let $m(t)$ be an unknown function of $t$. What is the derivative of the integral $\int_0^t m(t-s) \,ds$

Question

I am working on a problem which involves the renewal equation of a renewal process, and I need to find the derivative of the integral as above. One of my classmate told me it is $m(t) - m(0)$ by the fundamental theorem of calculus, but I don't understand how....

Answer 1

That's almost correct. By the Fundamental Theorem of Calculus, we have:

$$\int_0^t m(t-s)\ ds = \int_0^t m(s)\ ds = M(t) - M(0)$$

The reason why $m(t-s)$ can be substituted by $m(s)$ here can be seen by a $u$-substitution, but because intuitively at $s=0$, the start of the interval, we have that $m(t-s) = m(t-0)=m(t)$. However, at the end of the interval, we have that $s = t$ so $m(t-s) = m(t-t) = m(0)$. This is the same as looking at $m(0)$ to $m(t)$, it's just that your original integral is almost going from "right-to-left" in a sense, while the $m(s)$ substitution is going "left-to-right" if you think about integrals as moving from start to end. This may help you see how they're symmetric. You can see how one is just a flipped version of the other, yet the area between the curves and the x-axis is the same in both the regular version and the flipped/symmetric version, so the integrals are the same.

We call $M$ an "antiderivative" of $m$.

However then if we are looking for the derivative,

$$\frac{d}{dt} \int_0^t m(t-s)\ ds = \frac{d}{dt}(M(t) - M(0)) = m(t)$$

We have $\frac{d}{dt}M(t) = m(t)$ because $M$ is an antiderivative of $m$. Furthermore we have that $\frac{d}{dt}M(0) = 0$ since it's a constant. Hence we have that it's $m(t)$.

Answer 2

Hint: by symmetry, $\int_0^t m(t-s)\,ds = \int_0^t m(s)\,ds$. Now it's easy to see how to use the fundamental theorem of calculus. You get $m(t)$, not $m(t)-m(0)$.