An Integral for the Occasion: Merry Christmas!

This post is dedicated to a proof of the following result.

Christmas Determinant Integral. Let $C,M,e,r,i,t,m,a,s,h,y>0$ be given constants such that $M^er^r\neq C^hr^is^tm^as!$ . Then, the determinant integral

$\displaystyle I:=\det\int_{0}^{\infty}\exp\begin{pmatrix}\log\left(\frac{\log^2\left(1+\frac{y}{x^2}\right)}{4\pi\log(2)}\right)-C^hr^is^tm^as!w & M^eC^hr^iw \\-s^tm^as!r^rw &\log\left(\frac{\log^2\left(1+\frac{y}{x^2}\right)}{4\pi\log(2)}\right)+M^er^rw\end{pmatrix}\mathrm{d}x$

where

$\displaystyle w:=\frac{\log\left(M^er^r\right)-\log\left(C^hr^is^tm^as!\right)}{M^er^r-C^hr^is^tm^as!},$

evaluates to

$\displaystyle I=\frac{M^er^ry}{C^hr^is^tm^as!}.$

Proof. Let $A$ be a 2×2 matrix

$\displaystyle A=\begin{pmatrix} \tilde{a} & \tilde{b} \\ \tilde{c} & \tilde{d} \end{pmatrix}$

that is diagonalisable, so that there exists an inveritable matrix $P$

$\displaystyle P=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$

and a diagonal matrix $D$

$\displaystyle D=\begin{pmatrix} g & 0 \\ 0 & f \end{pmatrix}$

satisfying $A=PDP^{-1}$ (where of course $ad\neq bc$). As such, we have

$\displaystyle A=\frac{1}{ad-bc} \begin{pmatrix} adg-bcf & abf-abg \\ cdg-cdf & adf-bcg \end{pmatrix}.$

With the above decomposition, it can be shown that the exponential of $A$ reads

$\displaystyle \exp(A)=P\begin{pmatrix}\exp(g) & 0 \\ 0 & \exp(f) \end{pmatrix} P^{-1}$

or

$\displaystyle \exp(A)=\frac{1}{\det(P)}\begin{pmatrix} ad\exp(g)-bc\exp(f) & ab\exp(f)-ab\exp(g) \\cd\exp(g)-cd\exp(f) & ad\exp(f)-bc\exp(g) \end{pmatrix}.$

Assuming that $a,b,c,d$ are constants, $\Omega\subset\mathbb{R}^n$ is given, and that $\exp(g)$ and $\exp(f)$ are integrable with respect to Lebesgue measure over $\Omega$, we have

$\displaystyle T(\exp(A))=\frac{1}{\det(P)}\begin{pmatrix} \int_{\Omega}(ad\exp(g)-bc\exp(f))\mathrm{d}x & \int_{\Omega}(ab\exp(f)-ab\exp(g))\mathrm{d}x \\ \int_{\Omega}(cd\exp(g)-cd\exp(f))\mathrm{d}x & \int_{\Omega}(ad\exp(f)-bc\exp(g))\mathrm{d}x \end{pmatrix}.$

Here, I’ve let $T(\exp(A))$ denote the integration of the matrix $\exp(A)$ entry-wise. This operation is discussed on my Determinant Integrals page.

Now, assume $a,b,c,d> 0$, and suppose $f:=\log\left(\frac{ad}{bc}\exp(g)\right)$. Then,

$\displaystyle T(\exp(A))=\frac{1}{\det(P)}\begin{pmatrix} 0 & \int_{\Omega}(ab\exp(f)-ab\exp(g))\mathrm{d}x \\ \int_{\Omega}(cd\exp(g)-cd\exp(f))\mathrm{d}x & \int_{\Omega}(ad\exp(f)-bc\exp(g))\mathrm{d}x\end{pmatrix}$

which implies

$\displaystyle \det(T(\exp(A)))=\frac{abcd}{(ad-bc)^2}\left(\int_{\Omega}(\exp(f)-\exp(g))\mathrm{d}x\right)^2$

or rather

(1)……. $\displaystyle \det(T(\exp(A)))=\frac{ad}{bc}\left(\int_{\Omega}\exp(g)\mathrm{d}x\right)^2.$

With the choice of $f$ made above, it can be shown that

$\displaystyle A=\begin{pmatrix} g-bc\frac{\log(ad)-\log(bc)}{ad-bc} & ab\frac{\log(ad)-\log(bc)}{ad-bc} \\ -cd\frac{\log(ad)-\log(bc)}{ad-bc} & g+ad\frac{\log(ad)-\log(bc)}{ad-bc} \end{pmatrix}.$

Setting $\Omega=(0,\infty)\subset\mathbb{R}$ and

$\displaystyle g:=\log\left(\frac{\log^2\left(1+\frac{y}{x^2}\right)}{4\pi\log(2)}\right), \text{ }x>0$

where $y>0$ is an arbitrary constant, it follows from (1) that

$\displaystyle \det(T(\exp(A)))=\frac{ad}{bc}\left(\int_{\Omega}\exp(g)\mathrm{d}x\right)^2=\frac{ad}{bc}\left(\int_{\Omega}\frac{\log^2\left(1+\frac{y}{x^2}\right)}{4\pi\log(2)} \mathrm{d}x\right)^2=\frac{ady}{bc}.$

Here, I have used the fact that

$\displaystyle \int_{0}^{\infty}\log^2\left(1+\frac{q}{x^2}\right)\mathrm{d}x=4\pi\sqrt{q}\log(2)$

holds for all $q>0$. This follows by using integration by parts and a trigonometric substitution to reduce the integral to the problem of evaluating $\int_0^{\pi/2}\frac{x}{\tan{x}}\mathrm{d}x$. Wolfram Alpha indicates that this latter integral is equal to $\frac{\pi}{2}\log{2}.$

We now have that $A$ takes the form

$\displaystyle A(x)=\begin{pmatrix} \log\left(\frac{\log^2\left(1+\frac{y}{x^2}\right)}{4\pi\log(2)}\right) -bc\frac{\log(ad)-\log(bc)}{ad-bc} & ab\frac{\log(ad)-\log(bc)}{ad-bc} \\ -cd\frac{\log(ad)-\log(bc)}{ad-bc} & \log\left(\frac{\log^2\left(1+\frac{y}{x^2}\right)}{4\pi\log(2)}\right)+ad\frac{\log(ad)-\log(bc)}{ad-bc} \end{pmatrix},$

for $x>0$, with

$\displaystyle \det(T(\exp(A)))=\frac{ady}{bc},\text{ }y>0.$

Setting $w$ to be the real number given by

$\displaystyle w:= \frac{\log(ad)-\log(bc)}{ad-bc},$

the proposed result follows by setting $a=M^e,$ $d=r^r$, $b=C^hr^i,$ and $c=s^tm^as!$, where $C,M,e,r,i,t,m,a,s,h>0$ are arbitrary constants such that $M^er^r\neq C^hr^is^tm^as!$. \\\\