# 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

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

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