# 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

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