Differentiate an Integral w.r.t. it's Limit in This Special Case.

Question

Can someone explain how to differentiate when there is a something like a chain function?
My problem is this:

$s(t) = \int_{t}^{\infty} (q(u) - \alpha s(u) )e^{-i(u-t)}du.$

differentiated w.r.t. $t$ becomes something like:

$\dot{s} = -q(t) + (i+\alpha)s(t)$.

What confuses me is where the $i$ within the $i+\alpha$ factor comes from.

Here, $u$ is the general time variable and $t$ represents specific points in time, while $i , \alpha >0$.

I don't know how to cope with the function within the same function. Additionally I am not sure if the integral limit gets replaced as well by $u$ in the case of $s(u)$.

I also tried to solve the integral before differentiating (e.g. with integration by parts) but that turned out in a big mess and only gave me $0 = 0$.

Help highly appreciated.

Answer 1

Okay, even though I found a lot for "Differentiation under the integral sign" I disregarded my findings and did not thought it would apply here.

But I was wrong! Thanks to the post of SDiv

The solution to this is straight forward applying the formula from Wikipedia: Differentiation Under The Integral Sign.

Just insert what is there into:

$\frac{\text{d}}{\text{d}x}\left( \int_{a(x)}^{b(x)}f(x,t)\text{d}t \right ) = f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int_{a(x)}^{b(x)}f_x(x,t)\text{d}t.$

Thereby we have:

$\frac{d}{dt} \int_{t}^{B \rightarrow \infty} f(t,u)du = f(t,B) \cdot \frac{dB}{dt} - f(t,t) \cdot \frac{dt}{dt} + \int_{t}^{B}\left(\frac{d}{dt} f(t,u)du\right)$

$\frac{d}{dt} \int_{t}^{B \rightarrow \infty} f(t,u)du =(1) - (2) + (3)$

$(1) = (q(B)-\alpha s(B)) e^{-i(B-t)} \cdot 0 = 0$
$(2) = (q(t)-\alpha s(t)) e^{-i(t-t)} \cdot 1 = q(t)-\alpha s(t)$
$(3) = \int_{t}^{B}\left(\frac{d}{dt} f(t,u)du\right) = i \cdot \int_{t}^{B}f(t,u)du = i s(t)$

Together this results in:

$\frac{d}{dt} \int_{t}^{B \rightarrow \infty} f(t,u)du = \dot{s} = - \left(q(t) - \alpha s(t)\right) + is(t) = -q(t) + (i + \alpha)s(t)$