# Show the equality of two expressions

1. Apr 12, 2015

### grepecs

1. The problem statement, all variables and given/known data
Show that

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

can be written as

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

2. Relevant 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 $k$.

3. 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?

2. Apr 12, 2015

### Staff: Mentor

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.

3. Apr 12, 2015

### grepecs

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.

4. Apr 13, 2015

### Staff: Mentor

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.