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


\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}


\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!. \\\\

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