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
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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