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 tha*t

*and*

*for all* . *Furthermore, let* *be functions given by *

*and *

*where* *are constants. If* *is a function on* *defined by*

*then the integral*

*evaluates to*

*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*

*where*

*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

with

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

where

for . Let’s rewrite and in terms of the functions and . By periodicity of we find

or

Similarly, we get

for all

With these, we can simplify the expression for the function given by

From our application of Theorem 6, we get for each

or

Recalling the definition of the function , our integrand thus takes the form

or

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

with

and

Apply Fubini’s theorem to find that

and

The integrals in the above products are all of the form

which evaluates to whenever . Consequently,

while

The conclusion then follows once we note that

and

giving

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