Show the equality of two expressions

[grepecs]

Homework Statement

Show that

\sum_{k=0}^{N-1}e^{\gamma \tau k}\int_{0}^{\tau}F'(k\tau+s)ds

can be written as

\int_{0}^{t}e^{\gamma t'}F'(t')dt'

Homework Equations

1. t=N\tau

2. \int_{0}^{\tau}F'(k\tau+s)ds has the same statistical properties for each interval of length \tau, and is statistically independent with respect to k.

The Attempt at a Solution

I barely know where to start. As a first step, I'm thinking that perhaps "same statistical properties" means that the integral is the same regardless of k, so that

\int_{0}^{\tau}F'(k\tau+s)ds=\int_{0}^{\tau}F'(s)ds,

i.e., k=0. Is this correct?

 

[mfb]

Without the exponential, the first expression would just be a piecewise definition of the second integral, and a substitution would transform them into each other (can you see how? Hint: which argument values are used in F in the first, second, ... integral?). I guess you need the second relevant equation to get the same result including the exponential, but it could be still interesting to make that substitution.

 

[grepecs]

Without the exponential, the first expression would just be a piecewise definition of the second integral, and a substitution would transform them into each other (can you see how? Hint: which argument values are used in F in the first, second, ... integral?). I guess you need the second relevant equation to get the same result including the exponential, but it could be still interesting to make that substitution.

Thanks. That gives me

\sum_{k=0}^{N-1}e^{\gamma \tau k}\int_{k\tau}^{(k+1)\tau}F'(t')dt'.

What's left now is to move the exponential into the integrand, but I'm not sure how that can be justified.

 

[mfb]

That I don't know. It does not work for general functions F', but it works if you take the limit N -> infinity (with finite t), and it might work for some special F' even with finite N.