An Integral for the Occasion: Happy New Year!

To celebrate the new year 2021, I’ve created an integral for the occasion which reads as follows.

2021 New Years Integral. Let \alpha,\beta, and \gamma be real-valued continuous functions defined in \mathbb{R}^3 such that

\displaystyle 1<\gamma(x)<\frac{1+\sqrt{5}}{2},

\displaystyle 1<\alpha(x)<\frac{\gamma(x)}{1+\gamma(x)-\gamma(x)^2},

and

\displaystyle \beta(x):=\gamma(x)+\frac{1}{\alpha(x)}-\frac{1}{\gamma(x)},

for all x\in\mathbb{R}^3. Furthermore, let f,g:\mathbb{R}^3\to\mathbb{R} be functions given by

\displaystyle f(x):=\pi^{-\frac{3}{2}}\exp\left(-\frac{x_1^2}{H^2}-\frac{x_2^2}{a^2p^2}-\frac{x_3^2}{p^2y^2}\right),

and

\displaystyle g(x):=\pi^{-\frac{3}{2}}\exp\left(-\frac{x_1^2}{4}-\frac{x_2^2}{2019!^2}-\frac{x_3^2}{2041210^2}\right),

where H,a,p,y>0 are constants. If G is a function on \mathbb{R}^3 defined by

\displaystyle G(x):=\frac{(\alpha(x)\beta(x)\gamma(x))^2-1}{\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1}\text{ }\text{ \emph{for all} }x\in\mathbb{R}^3,

then the integral

\displaystyle \int_{\mathbb{R}^3}G(x)\log\sqrt[\alpha(x)]{\exp\left({\frac{f(x)}{\gamma(x)}}\right)\sqrt[\beta(x)]{\exp({g(x)})\sqrt[\gamma(x)]{\exp\left({\frac{f(x)}{\gamma(x)}}\right) \sqrt[\alpha(x)]{\exp({g(x)})\sqrt[\beta(x)]{\cdots}}}}}\text{ }\mathrm{d}x

evaluates to

Happy+2021!

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 (a_n)_{n\in\mathbb{Z}_{\geq 1}} and (b_n)_{n\in\mathbb{Z}_{\geq 1}} be periodic sequences of real numbers with periods T_a,T_b\in \mathbb{Z}_{\geq 1} such that a_n>0, a_n=a_{n+T_a} and b_n=b_{n+T_b} for all n\geq 1. If |b_1|,|b_2|,\cdots |b_{T_b-1}|,|b_{T_b}|>1. Then

\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}=\prod_{n=1}^{T_a}a_n^{S(n)}

where

\displaystyle S(n)=\frac{(b_1\cdots b_{T_b})^{A}}{(b_1\cdots b_{T_b})^{A}-1}\sum_{k=0}^{B-1}\prod_{j=1}^{T_ak+n}b_j^{-1}

for n\in\mathbb{Z}_{\geq 1}, with A=\frac{l.c.m(T_a,T_b)}{T_b} and B=\frac{l.c.m(T_a,T_b)}{T_a}.

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 (\tilde{a}_n) and (\tilde{b}_n) given by

  • \tilde{a}_n:=a_n for 1\leq n \leq T, \tilde{b}_n:=b_n for 1\leq n\leq T with T=l.c.m(T_a,T_b), and
  • \tilde{a}_n=\tilde{a}_{n+T}, and \tilde{b}_n=\tilde{b}_{n+T} for n\geq 1.

Then one simplifies the resulting expressions to arrive at Theorem 6.

For our current problem, we let (a_n)=(a_n(x)) and (b_n)=(b_n(x)) be periodic sequences of functions defined on \mathbb{R}^3 where

\displaystyle a_n=a_{n+2},\text{ }\text{ }b_{n}=b_{n+3}\text{ }\forall n\geq 1

with

\displaystyle a_1:=\exp\left({\frac{f(x)}{\gamma(x)}}\right),\text{ }a_2:=\exp(g(x)),

\displaystyle b_1:=\alpha(x),\text{ }b_2:=\beta(x),\text{ }\text{ and }\text{ }b_3:=\gamma(x).

It is clear that, for each x\in\mathbb{R}^3, the sequence (a_n(x)) fits the assumptions of the theorem with T_a=2. But for (b_n) we need to check that |b_2|=|\beta(x)|>1 for each x\in\mathbb{R}^3 since we already know that |b_1|=\alpha(x)>1 and |b_3|=\gamma(x)>1 for all x\in\mathbb{R}^3. From the definition of \beta and the assumed bounds on \alpha we find

\displaystyle \beta(x)=\gamma(x)+\frac{1}{\alpha(x)}-\frac{1}{\gamma(x)}>\gamma(x)+\frac{1+\gamma(x)-\gamma(x)^2}{\gamma(x)}-\frac{1}{\gamma(x)}=1

for all x\in\mathbb{R}^3. Noting that T_a=2 and T_b=3, we find in the context of the theorem that A=2 and B=3, giving

\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}=\prod_{n=1}^{2}a_n^{S(n)}

where

\displaystyle S(n)=\frac{(b_1b_2 b_{3})^2}{(b_1b_2 b_{3})^{2}-1}\sum_{k=0}^{2}\prod_{j=1}^{2k+n}b_j^{-1}

for n\in\{1,2\}. Let’s rewrite S(1) and S(2) in terms of the functions \alpha,\beta and \gamma. By periodicity of (b_n) we find

\displaystyle S(1)=\frac{(b_1b_2b_3)^2}{(b_1b_2b_3)^{2}-1}\sum_{k=0}^{2}\prod_{j=1}^{2k+1}b_j^{-1}=\frac{(b_1b_2b_3)^2}{(b_1b_2b_3)^{2}-1}\left(b_1^{-1}+b_1^{-1}b_2^{-1}b_3^{-1}+b_1^{-2}b_2^{-2}b_3^{-1}\right)

or

\displaystyle S(1)=\frac{\gamma(x)\left(\alpha(x)\beta(x)^2\gamma(x)+\alpha(x)\beta(x)+1\right)}{(\alpha(x)\beta(x)\gamma(x))^2-1},\text{ }\forall x\in\mathbb{R}^3.

Similarly, we get

\displaystyle S(2)=\frac{(b_1b_2 b_{3})^2}{(b_1b_2 b_{3})^{2}-1}\sum_{k=0}^{2}\prod_{j=1}^{2k+2}b_j^{-1}=\frac{\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1}{(\alpha(x)\beta(x)\gamma(x))^2-1},

for all x\in\mathbb{R}^3.

With these, we can simplify the expression for the function l:\mathbb{R}^3\to\mathbb{R} given by

\displaystyle l(x):=\log\sqrt[\alpha(x)]{\exp\left({\frac{f(x)}{\gamma(x)}}\right)\sqrt[\beta(x)]{\exp({g(x)})\sqrt[\gamma(x)]{\exp\left({\frac{f(x)}{\gamma(x)}}\right) \sqrt[\alpha(x)]{\exp({g(x)})\sqrt[\beta(x)]{\cdots}}}}}.

From our application of Theorem 6, we get for each x\in\mathbb{R}^3

\displaystyle l(x)=\log \prod_{n=1}^{2}a_n^{S(n)}=S(1)\log\left(\exp\left({\frac{f(x)}{\gamma(x)}}\right) \right)+S(2)\log( \exp({g(x)}))

or

\displaystyle l(x)=\frac{ (\alpha(x)\beta(x)^2\gamma(x)+\alpha(x)\beta(x)+1)f(x)+(\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1)g(x)}{(\alpha(x)\beta(x)\gamma(x))^2-1}.

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

\displaystyle G(x)l(x)=\frac{ (\alpha(x)\beta(x)^2\gamma(x)+\alpha(x)\beta(x)+1)f(x)+(\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1)g(x)}{\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1}

or

\displaystyle G(x)l(x)=\left(\frac{\alpha(x)\beta(x)^2\gamma(x)+\alpha(x)\beta(x)+1}{\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1}\right)f(x)+g(x)\text{ }\forall x\in \mathbb{R}^3.

Using the expression for \beta we can simplify the coefficient of f as follows: for each x\in\mathbb{R}^3

\displaystyle \frac{\alpha(x)\beta(x)^2\gamma(x)+\alpha(x)\beta(x)+1}{\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1}=\frac{\beta(x)^2\gamma(x)+\beta(x)+\frac{1}{\alpha(x)}}{\beta(x)\gamma(x)^2+\frac{\beta(x)\gamma(x)}{\alpha(x)}+\frac{1}{\alpha(x)}}

\displaystyle = \frac{\beta(x)^2\gamma(x)+\beta(x)+\beta(x)+\frac{1}{\gamma(x)}-\gamma(x)}{\beta(x)\gamma(x)^2+\beta(x)\gamma(x)\left(\beta(x)+\frac{1}{\gamma(x)}-\gamma(x)\right)+ \beta(x)+\frac{1}{\gamma(x)}-\gamma(x) }

\displaystyle =\frac{\beta(x)^2\gamma(x)+2\beta(x)+\frac{1}{\gamma(x)}-\gamma(x)}{\beta(x)^2\gamma(x)+ 2\beta(x)+\frac{1}{\gamma(x)}-\gamma(x) }=1.

Let I denote the integral that we are seeking to evaluate. With the above, our problem is now simplified massively: evaluate

\displaystyle I=\int_{\mathbb{R}^3}(f(x)+g(x))\mathrm{d}x

with

\displaystyle f(x)=\pi^{-\frac{3}{2}}\exp\left(-\frac{x_1^2}{H^2}-\frac{x_2^2}{a^2p^2}-\frac{x_3^2}{p^2y^2}\right),

and

\displaystyle g(x)=\pi^{-\frac{3}{2}}\exp\left(-\frac{x_1^2}{4}-\frac{x_2^2}{2019!^2}-\frac{x_3^2}{2041210^2}\right).

Apply Fubini’s theorem to find that

\displaystyle \int_{\mathbb{R}^3}f(x)\mathrm{d}x=\pi^{-\frac{3}{2}}\left(\int_{\mathbb{R}}e^{-\frac{x_1^2}{H^2}}\mathrm{d}x_1\right)\left(\int_{\mathbb{R}}e^{-\frac{x_2^2}{(ap^2)}}\mathrm{d}x_2 \right)\left(\int_{\mathbb{R}}e^{-\frac{x_3^2}{(py)^2}}\mathrm{d}x_3\right),

and

\displaystyle \int_{\mathbb{R}^3}g(x)\mathrm{d}x=\pi^{-\frac{3}{2}}\left(\int_{\mathbb{R}}e^{-\frac{x_1^2}{4}}\mathrm{d}x_1\right)\left(\int_{\mathbb{R}}e^{-\frac{x_2^2}{2019!^2}}\mathrm{d}x_2 \right)\left(\int_{\mathbb{R}}e^{-\frac{x_3^2}{2041210^2}}\mathrm{d}x_3\right).

The integrals in the above products are all of the form

\displaystyle \int_{\mathbb{R}}e^{-\frac{x^2}{w^2}}\mathrm{d}x

which evaluates to \sqrt{\pi}w whenever w>0. Consequently,

\displaystyle \int_{\mathbb{R}^3}f(x)\mathrm{d}x=Happy

while

\displaystyle \int_{\mathbb{R}^3}g(x)\mathrm{d}x=2\times 2019!\times 2041210.

The conclusion then follows once we note that

\displaystyle 2\times 2019!\times 2041210=2!(2021-2)!\times 2041210=2021!\frac{2041210}{\begin{pmatrix} 2021 \\ 2 \end{pmatrix}}

and

\displaystyle \begin{pmatrix} 2021 \\ 2 \end{pmatrix}=2041210,

giving

\displaystyle \int_{\mathbb{R}^3}g(x)\mathrm{d}x=2021!

\\\\