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<b 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.

Figure 1: Loci described by z_{\alpha}=\alpha^2 over \left[0,\pi/2\right]^2

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!

Figure 2: Loci described by z_{\alpha}=\alpha^2 over \left[0,4\pi\right]^2

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.

Radical Integrals: No. 5

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<P\leq Q 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).