# 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 $M^er^r\neq C^hr^is^tm^as!$. \\\\

# A Picturesque Reduction of An Integral Determinant Equation 1

In this post we discuss example solutions of the following integral determinant equation that was derived on my Determinant Integrals page:

(1) …… $\displaystyle \left(\int_{\Omega}\det(A)\mathrm{d}x\right)\left(\int_{\Omega}\det(A^{-1})\mathrm{d}x\right)=1$

Let $\Omega \subset \mathbb{R}^m$ be a bounded open set. Given an invertible matrix $A\in\mathcal{M}_n(\Omega,\textbf{r})$ (see the Determinant Integral page for a description of the set $\mathcal{M}_n(\Omega,\textbf{r})$ if needed) we know that its determinant is integrable and

$\displaystyle \text{det}(A)=\frac{1}{\text{det}(A^{-1})}\text{ }\text{ a.e in }\Omega.$

Consequently, if equation (1) holds we have

$\displaystyle \left(\int_{\Omega}\text{det}(A)\text{ }\mathrm{d}x\right)\left(\int_{\Omega}\frac{1}{\text{det}(A)}\text{ }\mathrm{d}x\right)=1.$

This leads us to consider examples of bounded open sets $\Omega$ and scalar functions $f$ which satisfy

(2) …… $\displaystyle \left(\int_{\Omega}f\text{ }\mathrm{d}x\right)\left(\int_{\Omega}\frac{1}{f}\text{ }\mathrm{d}x\right)=1.$

Clearly, equation (2) is equation (1) in the case when $A$ is a $1\times 1$ matrix, so we’ve massively simplified the problem presented by equation (1). Given an integer $n\geq 2$, notice equation (1) on its own does not have a unique $n\times n$ matrix solution $A$ defined a.e in $\Omega$ if there exists a function $f$ satisfying equation (2). With a function $f$ known to satisfy equation (2), we could construct an uncountable number of matrices $A$ for which $\text{det}(A)=f$ a.e in $\Omega$. For example, consider triangular matrices $A$ whose main diagonals take the form $\text{diag}(1,\cdots,1,f,1,\cdots,1)\in\mathbb{R}^n$. We then have full freedom to determine the possibly non-degenerate off-diagonal entries of $A$, according to whether the matrix is upper or lower triangular.

We now present an example solution pair $(f,\Omega)$ to equation (2). Denote by $I[f,\Omega]$ the left-hand side of equation (2):

$\displaystyle I[f,\Omega]:= \left(\int_{\Omega}f\text{ }\mathrm{d}x\right)\left(\int_{\Omega}\frac{1}{f}\text{ }\mathrm{d}x\right).$

Let $f(x)=\cos(\alpha x)$ be defined over $\Omega=(a,b)$ with $\alpha \neq 0$ and $a, b\in\mathbb{R}$ such that $a\neq b$. I’ve allowed simply for the condition $a\neq b$ as opposed to the natural condition $a because equation (2) holds even if the limits of integration are interchanged, which allows for the case $a>b$. On the other hand, equation (2) can’t hold if $a=b$ or else we have $0=1$. Nonetheless, for suitable $a,b\in\mathbb{R}$ there holds

$\displaystyle I[f,\Omega]=\frac{1}{\alpha^2}\left(\sin(\alpha b)-\sin(\alpha a)\right)\log\left|\frac{\sec(\alpha b)+\tan(\alpha b)}{\sec(\alpha a)+\tan(\alpha a)}\right|.$

If we let

$z_{\alpha}(a,b)=\left(\sin(\alpha b)-\sin(\alpha a)\right)\log\left|\frac{\sec(\alpha b)+\tan(\alpha b)}{\sec(\alpha a)+\tan(\alpha a)}\right|$

equation (2) reads $z_{\alpha}(a,b)=\alpha^2$. Below we plot this equation implicitly in Python over $(a,b)\in\left[0,\frac{\pi}{2}\right]^2$ for $\alpha=1,2,3,4,5,$ and $9$.

In Figure 1 all curves of a given colour constitute the solution locus corresponding to one choice of $\alpha$. We describe this correspondence below:

• $\alpha=1\to$ Light Blue curves
• $\alpha=2\to$ Orange curves
• $\alpha=3 \to$ Red curves
• $\alpha=4\to$ Blue curves
• $\alpha=5\to$ Magenta curves
• $\alpha=9\to$ Green curves

In essence, for fixed $\alpha$ that generate curves in the plane of positive length, there are infinitely many choices of $\Omega$ that ensure equation (2) holds. In Figure 2 below we extend the plot to $[0,4\pi]^2$ which gives a picturesque pattern that would make for a neat wallpaper!

Between figures 1 and 2 we observe closed curves of decreasing diameter for $\alpha=1,2,3,4,$ and $9$, whereas for $\alpha=5$ we see small horizontal/vertical spikes which do not cross the red curves although they appear to be touching prior to zooming in. In Figure 2 we see a pattern repeating for any given coloured curve corresponding to a choice of $\alpha$. This is expected as $z_{\alpha}(a,b)$ is periodic with period $2\pi$ in both its arguments when $\alpha$ is an integer. It would be interesting then to see what family curves we observe when $\alpha$ is not an integer or irrational.

In this post I present my argument which proves integral no. 5 under the Radical Integrals section of The Integral Corner. The result in question reads as follows.

Radical Integral 5. Let $P$ and $Q$ real satisfy $0< P\leq Q$. Consider the function

$\displaystyle a(x):=1+e\gamma(x+1,1)$

for $x\geq 0$ where $\gamma(u,v)$ is the incomplete gamma function with integral representation

$\displaystyle \gamma(u,v)=\int_0^{v}t^{u-1}e^{-t}\mathrm{d}t.$

Then,

$\displaystyle \int_P^Q\Lambda(x)\log \sqrt[x+1]{a(x)^{x+2}\sqrt[x+2]{a(x)\sqrt[x+3]{a(x)\sqrt[x+4]{a(x)\sqrt[x+5]{\cdots}}}}}\text{ }\mathrm{d}x=\frac{1}{4e}\log\left(H(P,Q)\right)$

with

$\displaystyle \Lambda(x):=\int_0^{1}t^{x}e^{-t}\log(t)\text{ }\mathrm{d}t$

and

$\displaystyle H(P,Q):=\frac{a(Q)^{2a(Q)^2}}{a(P)^{2a(P)^2}}\exp(a(P)^2-a(Q)^2).$

Proof. If we let

$\displaystyle L(x):=\sqrt[x+1]{a(x)^{x+2}\sqrt[x+2]{a(x)\sqrt[x+3]{a(x)\sqrt[x+4]{a(x)\sqrt[x+5]{\cdots}}}}} \text{ }\text{ }(x\geq 0)$

we have for each $x>0$

$\displaystyle L(x)=a(x)^{1+\frac{1}{x+1}+\frac{1}{(x+1)(x+2)}+\frac{1}{(x+1)(x+2)(x+3)}+\cdots}.$

This can be further simplified to

$\displaystyle L(x)=a(x)^{1+e\gamma(x+1,1)}=a(x)^{a(x)}$

after establishing

$\displaystyle \sum_{k=0}^{\infty}\prod_{j=1}^{k+1}\frac{1}{x+j}=e\gamma(x+1,1)$

for $x\geq 0$. To see this, note that the “lower” incomplete gamma function $\gamma(s,z)$ is a holomorphic function with singularities at points $(s,z)$ where $z=0$ or $s$ is a non-positive integer (check out the Incomplete gamma function Wikipedia page). Moreover, it admits the representation

$\displaystyle \gamma(s,z)=z^s\Gamma(s)e^{-z}\sum_{k=0}^{\infty}\frac{z^k}{\Gamma(s+k+1)}.$

Setting $s=x+1$ and $z=1$, we find for each $x>0$

$\displaystyle \gamma(x+1,1)=e^{-1}\Gamma(x+1)\sum_{k=0}^{\infty}\frac{1}{\Gamma(x+k+2)}=e^{-1}\Gamma(x+1)\left(\frac{1}{\Gamma(x+2)}+\frac{1}{\Gamma(x+3)}+\cdots\right)$

Rearranging, we get

$\displaystyle \sum_{k=0}^{\infty}\frac{\Gamma(x+1)}{\Gamma(x+k+2)}=e\gamma(x+1,1).$

But notice that, formally,

$\prod_{j=1}^{k+1}\frac{1}{x+j}\equiv\frac{\Gamma(x+1)}{\Gamma(x+k+2)}.$

As such, we arrive at the desired identity for the infinite sum, justifying the identity $L(x)=a(x)^{a(x)}$ for $x>0$. Writing out $a$ as

$\displaystyle a(x)=1+e\int_0^{1}t^xe^{-t}\mathrm{d}t \text{ }(x\geq 0)$

we see that $a$ is differentiable over $(0,\infty)$ with derivative given by

$\displaystyle \frac{da}{dx}=e\int_0^1t^xe^{-t}\log(t)\mathrm{d}t=e\Lambda(x).$

Consequently, $a$ is monotone decreasing over $(0,\infty)$. Therefore, our proposed integral for given $0 can be evaluated as follows.

$\displaystyle \int_P^Q\Lambda(x)\log{L(x)}\mathrm{d}x=e^{-1}\int_P^Qa(x)\log(a(x))\frac{da}{dx}\mathrm{d}x=e^{-1}\int_{a(P)}^{a(Q)}v\log(v)\mathrm{d}v=e^{-1}\left[\frac{v^2}{2}\log(v)-\frac{1}{4}v^2\right]_{a(P)}^{a(Q)} =\frac{1}{4e}\left[\log\left((v^{2v^2}\exp(-v^{2})\right)\right]_{a(P)}^{a(Q)}.$

Simplification leads to the stated result:

$\displaystyle \int_P^Q\Lambda(x)\log\sqrt[x+1]{a(x)^{x+2}\sqrt[x+2]{a(x)\sqrt[x+3]{a(x)\sqrt[x+4]{a(x)\sqrt[x+5]{\cdots}}}}}\text{ }\mathrm{d}x=\frac{1}{4e}\log\left(H(P,Q)\right)$

with

$\displaystyle \Lambda(x):=\int_0^{1}t^{x}e^{-t}\log(t)\text{ }\mathrm{d}t$

and

$\displaystyle H(P,Q):=\frac{a(Q)^{2a(Q)^2}}{a(P)^{2a(P)^2}}\exp(a(P)^2-a(Q)^2).$