For the following question:
Prove that $\int_{0}^{1} (-1)^{\lfloor{1994t\rfloor}} (-1)^{\lfloor{1995t\rfloor}} \binom{1993}{\lfloor{1994t\rfloor}}\binom{1994}{\lfloor{1995t\rfloor}} dt=0$
I am having trouble resolving several issues. The integral involves floor functions, and I think to solve it, i am suppose to turn it into some type of alternating sum or an alternating sum of integrals over partition the limit of integration interval [0,1], into $n$ sub-intervals.
There are two exponent terms for floor functions, $\lfloor{1994t\rfloor} \text{ and } \lfloor{1995t\rfloor}$, I am not sure if I should partition the interval [0,1] into $0 \leq \frac{t}{1994}\leq 1, t=0..1993,$ or $0 \leq \frac{t}{1995}\leq 1, t=0..1994.$ My guess for either of the partition is because I know that $gcd(1994,1995)=1$, $1994x+1995y=gcd(1994,1995),$ with $x=y$, I am suppose to do something with $\lfloor{1994t\rfloor}$ and $\lfloor{1994t\rfloor}$, maybe write one of the floor function in terms of the other. I don't know if there is an identity for $\lfloor{pt\rfloor} +\lfloor{(p+1)t\rfloor}$ in terms of a single floor function involving both $pt, (p+1)t$. Also if I let $u=at$, then $du=a dt \text{ and } \frac{du}{a}=dt,$ but what would the value be for $a$.
Also, since the integral involves two $(-1)^{\lfloor{t\rfloor}}$ terms, and I can do something along the following line: since $\lfloor{t\rfloor} \in \mathbb{Z}$, then let $\lfloor{t\rfloor} = n$. If I partition [0,1], $0 \leq t \leq 1$, the intervals between each consecutive integers for $(-1)^{\lfloor{t\rfloor}}$ will be either $1$ or $-1$, depending on whether $n$ is even or odd.
Hence,
$\int_{0}^{1} (-1)^{\lfloor{1994t\rfloor}} (-1)^{\lfloor{1995t\rfloor}} \binom{1993}{\lfloor{1994t\rfloor}}\binom{1994}{\lfloor{1995t\rfloor}} dt = a^{-1}\sum_{u=?}^{?}\int_{u}^{u+1} (-1)^{\lfloor{1994\frac{u}{a}\rfloor}} (-1)^{\lfloor{1995\frac{u}{a}\rfloor}}\binom{1993}{\lfloor{1994\frac{u}{a}\rfloor}}\binom{1994}{\lfloor{1995\frac{u}{a}\rfloor}}du$.
Also, how do I simplify $\binom{1993}{\lfloor{1994\frac{u}{a}\rfloor}}\binom{1994}{\lfloor{1995\frac{u}{a}\rfloor}}?$ Which the product of the two binomial terms are dependent on $t=\frac{u}{a}$.
Lastly, I think the integral will be solved by somehow transforming it into something along the line of an alternating sums of binomial coefficients, similar to $\sum_{j=0}^{n} (-1)^{n+1}j\binom{n}{j}$ It is just an hunch. But I am not sure if it is possible to do so, and if so, how one can go about doing it. Thank you in advance.
Best Answer
If you sub $u=1-t$, you have
$$\int_{u=1}^{u=0}(-1)^{\lfloor{1994-1994u\rfloor}} (-1)^{\lfloor{1995-1995u\rfloor}} \binom{1993}{\lfloor{1994-1994u\rfloor}}\binom{1994}{\lfloor{1995-1995u\rfloor}} (-du)$$
Now note that there are values of $u$ where $\lfloor{1994-1994u\rfloor}$ is an integer, and at such places $\lfloor{1994-1994u\rfloor}\neq1993-\lfloor{1994u\rfloor}$. However those places have measure $0$ and contribute nothing to the integral. Everywhere else, where $\lfloor{1994-1994u\rfloor}$ is not an integer, then $\lfloor{1994-1994u\rfloor}=1993-\lfloor{1994u\rfloor}$.
And there is the analogous statement for $1995$. So you have:
$$-\int_{u=1}^{u=0}(-1)^{1993-\lfloor{1994u\rfloor}} (-1)^{1994-\lfloor{1995u\rfloor}} \binom{1993}{1993-\lfloor{1994u\rfloor}}\binom{1994}{1994-\lfloor{1995u\rfloor}} du$$
And for more basic reasons, this is equal to
$$-\int_{u=0}^{u=1}(-1)^{\lfloor{1994u\rfloor}} (-1)^{\lfloor{1995u\rfloor}} \binom{1993}{\lfloor{1994u\rfloor}}\binom{1994}{\lfloor{1995u\rfloor}} du$$
This is the negative of what we started with, using $u$ as the integration variable instead of $t$. So it all must be $0$.