To celebrate the new year 2021, I’ve created an integral for the occasion which reads as follows.
2021 New Years Integral. Let and be real-valued continuous functions defined in such that
for all . Furthermore, let be functions given by
where are constants. If is a function on defined by
then the integral
Proof. Our first step to evaluating the integral is to simplify the nested radical appearing in the integrand. For this, we employ Theorem 6 found on my Nested Radicals page.
Theorem 6. Let and be periodic sequences of real numbers with periods such that , and for all . If . Then
for , with and .
This result can be derived using the idea behind the proof of Theorem 2 found on my Nested Radicals page. For this it suffices to consider the pair of sequences and given by
- for , for with , and
- , and for .
Then one simplifies the resulting expressions to arrive at Theorem 6.
For our current problem, we let and be periodic sequences of functions defined on where
It is clear that, for each , the sequence fits the assumptions of the theorem with . But for we need to check that for each since we already know that and for all . From the definition of and the assumed bounds on we find
for all . Noting that and , we find in the context of the theorem that and , giving
for . Let’s rewrite and in terms of the functions and . By periodicity of we find
Similarly, we get
With these, we can simplify the expression for the function given by
From our application of Theorem 6, we get for each
Recalling the definition of the function , our integrand thus takes the form
Using the expression for we can simplify the coefficient of as follows: for each
Let denote the integral that we are seeking to evaluate. With the above, our problem is now simplified massively: evaluate
Apply Fubini’s theorem to find that
The integrals in the above products are all of the form
which evaluates to whenever . Consequently,
The conclusion then follows once we note that