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I have spent some time to solve this equation and I don't know how to solve it: $$\int_0^t e^{-rs}dk_s ,$$ where r is constant. I was thinking the result would be $e^{-rs}k_s $. But should the exponential term affect too?

Ultimately, I want to take the derivative of the above expression with respect to t. Seems like the answer is $e^{-rt}dk_t$. Could anyone please explain how to arrive at the answer? Thanks!

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  • $\begingroup$ I suppose you know what $dk_s$ means. If not, sit down and recall the definition and how to obtain $ds$ from $dk_s$. If you know this, you can easily solve the integral. $\endgroup$ – user526015 Feb 19 at 10:34
  • $\begingroup$ Hi, thanks for the reply. Unfortunately, there is no definition about $dk_s$, so I can't find a way to restate $dk_s$ in terms of $ds$. $\endgroup$ – U_ng Feb 19 at 10:56
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I will write $k(s)$ instead of $k_s$. Using "physicists notation", $$\frac{dk(s)}{ds}=k'(s) \Longleftrightarrow dk(s)=k'(s)ds.$$ So you have to calculate $$\int_0^t e^{-rs}dk(s)=\int_0^te^{-rs}k'(s)ds.$$ Using the Fundamental Theorem of Calculus, this is $F(t)-F(0)$ where $F$ is a primitive function for the integrand. Thus, differentiating w.r.t. $t$ should give back the integrand, i.e. $e^{-rt}k'(t).$

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  • $\begingroup$ Thank you so much for the help $\endgroup$ – U_ng Feb 19 at 11:12
  • $\begingroup$ You're welcome. $\endgroup$ – user526015 Feb 19 at 11:12

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