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I want to compute

$$\frac{d}{dt}\left[ \int_t^{\tau}e^{-r(s-t)}(-pc) \,ds\right]$$ First, I know that I can just solve the integral and take the derivative:

$$\frac{d}{dt}\left[ \int_t^{\tau}e^{-r(s-t)}(-pc) \,ds\right]=\frac{d}{dt}\left[ \frac{-pc}{r}\left(1-e^{-r(\tau-t)}\right)\right]$$ $$=pc\cdot e^{-r(\tau-t)}$$

however, I want to solve it "straight away" using the Fundamental Theorem of Calculus. However, I must be doing something wrong because I obtain the following:

$$\frac{d}{dt}\left[ \int_t^{\tau}e^{-r(s-t)}(-pc) \,ds\right]=-e^{-r(t-t)}(-pc)-pc\int_t^{\tau} \frac{d}{dt}e^{-r(s-t)}ds$$ $$=pc-pc\cdot r\int_t^{\tau} e^{-r(s-t)}ds$$ $$=pc+pc [e^{-r(\tau-t)}-1]$$ $$=pc\cdot e^{-r(\tau-t)}$$

Comment: Initially, I did an algebra mistake and wasn't getting the same result for the second equation. Now it seems to be right. The magic of typing in latex code :). Already wrote the question though. So tell me if the way I compute this second derivative is correct according to you.

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    You are using the wrong formula $$\frac{d}{dt}\int_{a(t)}^{b(t)}u(t,s)ds=\int_{a(t)}^{b(t)}\frac{\partial u}{\partial t}(t,s)ds$$ instead of $$\frac{d}{dt}\int_{a(t)}^{b(t)}u(t,s)ds=\int_{a(t)}^{b(t)}\frac{\partial u}{\partial t}(t,s)ds+b'(t)u(t,b(t))-a'(t)u(t,a(t))$$ In your case, $$\frac{d}{dt}\int_t^\tau u(t,s)ds=\int_t^\tau\frac{\partial u}{\partial t}(t,s)ds-u(t,t)$$ – 2017-02-14

1 Answers 1

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$\frac{d}{dt} \int_t^\tau e^{-r(s-t)}(-pc)ds$

$= \frac{d}{dt}\left[- \int_\tau^t e^{-r(s-t)}(-pc)ds\right]$

$= \frac{d}{dt} \int_\tau^t pc \cdot e^{rt} \cdot e^{-rs}ds $

$= pc \cdot \frac{d}{dt} \left [ e^{rt}\cdot \int_\tau^t f(s)ds\right], $

where $f(s) = e^{-rs}$. Try using the fundamental theorem now.

This doesn't really make it much easier, because you'll still have to evaluate the integral when using the product rule.

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    I don't think this is right. Note that $f(s)$ actually depends on $t$! – 2017-02-14
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    I realized that so I updated my answer. You need to use the product rule too – 2017-02-14