-1
$\begingroup$

Calculate the derivative of the integral expression: $$\frac{d}{dt}\int_0^tce^{\delta(t-s)}f(s)ds$$ where $c$ and $\delta$ are positive constants.

  • 0
    https://en.wikipedia.org/wiki/Leibniz_integral_rule#Formal_statement – 2017-02-06

2 Answers 2

0

It's good to know differentiate under the integral sign, but in this case we can do this way:

$$\frac {d}{dt} \int_{0}^{t} ce^{\delta (t-s)}f (s)\mathrm {d}s=$$

$$=c\frac {d}{dt}\left(e^{\delta t}\int_0^te^{-\delta s}f(s)\mathrm ds\right)=$$

$$=c\frac {d}{dt}e^{\delta t}\int_0^te^{-\delta s}f(s)\mathrm ds+ce^{\delta t}\frac {d}{dt}\int_0^te^{-\delta s}f(s)\mathrm ds=$$

$$=\delta ce^{\delta t}\int_0^te^{-\delta s}f(s)\mathrm ds+ce^{\delta t}e^{-\delta t}f(t)=\delta \int_0^tce^{\delta(t- s)}f(s)\mathrm ds+cf(t)$$

Procedure not very different from the one following the general statment, indeed.

  • 0
    Thank you! your argument is precise, but you used the fact $e^{(t-s)}=e^te^{-s}$. In the general case $$\frac{d}{dt}\int_0^tf(t,s)ds$$ is it possible obtain the same result? – 2017-02-06
  • 0
    Yes. In fact, using the Leibniz rule you get directly the second term of the result. $\frac {d}{dt} \int_{0}^{t} ce^{\delta (t-s)}f (s)\mathrm {d}s=\int_0^t\frac {\partial}{\partial t}ce^{\delta (t- s)}f(s)\mathrm ds+ce^{\delta (t- t)}f(t)-0=\delta \int_0^tce^{\delta(t- s)}f(s)\mathrm ds+cf(t)$ – 2017-02-06
  • 0
    Very nice! Thank you for your suport. – 2017-02-07
1

Just use the concept of differentiation under the integral sign to get $$\frac {d}{dt} \int_{0}^{t} ce^{\delta (t-s)}f (s)\mathrm {d}s $$ $$=ce^{\delta (t-t)}f (t)\frac {d}{dt} - 0 = ce^0 f (t) = cf (t) $$ Hope it helps.

  • 0
    We need have in mynd that the variable $t$ is an "end point" of the interval of integration, and this do not let us use defferentiation under the integral sign. – 2017-02-06