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Let $f$ be a smooth function and $\alpha >0$, i would like to know how to calculate the following expression: $$ A:=\dfrac{d}{dt}\left[\int_0^t f(s) \sin(\alpha(t-s))ds\right] $$ I followed a very simple method but i think that it is wrong. I assumed that $$ \int_0^t f(s) \sin(\alpha(t-s))ds=G(t)-G(0) $$ where $$G(s)=\int g(s) ds$$ and $$g(s)=f(s)\sin(\alpha(t-s))$$ Then we have $$ A= \frac{d}{dt}\left(G(t)-G(0)\right)= g(t)=0$$ Question:

Did i make any mistakes ?

How to proceed to calculate this type of integral ?

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Note that $t$ is also present in the integrand, so you need to apply the chain rule: $$ \frac{d}{dt}\int_0^t g(t,s)ds=g(t,t)+\int_0^t ds \frac{d}{dt}g(t,s)\ . $$ In your case, $g(t,t)$ is indeed equal to zero, but the second chunk is not...

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  • $\begingroup$ I don't how you have used the chain rule here exactly $\endgroup$ – hamza boulahia Dec 20 '17 at 20:16
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Write $F(u,t)=\int_0^t f(s) \sin(\alpha(u-s))ds$ and $G(t)=(t,t)$ compute the differential of $F\circ G$

$dF={\partial F\over{\partial u}}du+{\partial F\over{\partial t}}dt$

$=\int_0^t\alpha f(s)cos(\alpha(u-s)du+f(t)sin(\alpha(u-t)dt$

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  • $\begingroup$ i don't understand why you changed the $t$ in the assignment by $u$ ? i think by doing that you have changed the whole integral expression into another one $\endgroup$ – hamza boulahia Dec 20 '17 at 20:09
  • $\begingroup$ because I use the composition of two differentiable functions, $F$ has two variables and $G$ one this enables to prove the formula of Pierpalo Vivo. $\endgroup$ – Tsemo Aristide Dec 20 '17 at 20:12
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$$ A:=\frac{d}{dt}\left[\int_0^t f(s) \sin(\alpha(t-s))ds\right] =$$ $$=\frac{d}{dt}\left[\sin(\alpha{t})\int_0^t f(s)\cos(\alpha{s}) ds-\cos(\alpha{t})\int_0^t f(s)\sin(\alpha{s}) ds\right]=$$ $$=\alpha\cos{\alpha{t}}\int_0^t f(s)\cos(\alpha{s}) ds+\alpha\sin{\alpha{t}}\int_0^t f(s)\sin(\alpha{s}) ds=$$ $$\alpha\int_{0}^{t}f(s)\cos{(\alpha(t-s))}ds$$

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