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.

$\int_{0}^{1} (-1)^{\lfloor{1994t\rfloor}} (-1)^{\lfloor{1995t\rfloor}} \binom{1993}{\lfloor{1994t\rfloor}}\binom{1994}{\lfloor{1995t\rfloor}} dt = a^{-1}\sum_{u=?}^{?}\int_{u}^{u+1} (-1)^{\lfloor{1994\frac{u}{a}\rfloor}} (-1)^{\lfloor{1995\frac{u}{a}\rfloor}}\binom{1993}{\lfloor{1994\frac{u}{a}\rfloor}}\binom{1994}{\lfloor{1995\frac{u}{a}\rfloor}}du$.
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.

  • $\begingroup$ Perhaps you ought to do this first with $2$ and $3$ instead of $1994$ and $1995$ to see if that gives you any insight. $\endgroup$
    – saulspatz
    Commented Apr 28, 2020 at 5:13
  • $\begingroup$ I don't know if it will work, but these sorts of problems often have a hidden symmetry that reveals itself upon the change of variables $u=1-t$. (Without knowing the context in which you found the problem, it's impossible to speculate on what method of solution might be most likely.) $\endgroup$ Commented Apr 28, 2020 at 5:17
  • $\begingroup$ @GregMartin I got the question from an online pdf by David Santos titled: "Number theory of Mathematical Contests". I want to get a better handle on floor and ceiling function in the context of number theory. I was just surprised why in calculus class, first year undergrads were never taught how to integrate floor and ceiling functions, while most of techniques of integration, most can just be happily be forgotten about after the arrival of Maple or Mathematica. $\endgroup$
    – Seth
    Commented Apr 28, 2020 at 5:51

1 Answer 1


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 $0$.

  • $\begingroup$ @SethMai With the right interval partitioning, you can still do this. Your partition would have subintervals on which the function is constant. And the subintervals that are symmetric about $t=0.5$ will have opposite values. But the same symmetry about $t=0.5$ is captured in this argument without bothering to work out the partition. $\endgroup$
    – 2'5 9'2
    Commented Apr 28, 2020 at 7:12
  • $\begingroup$ thank you for your reply. I am not sure why I did not get a notification about it. Your insight make this complicated looking problem seem trivial. $\endgroup$
    – Seth
    Commented Apr 30, 2020 at 7:07

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