Integrating the determinant of the inverse of a Toeplitz matrix

Original post where i shared my solution.

For convenience we write $a= e^{-\pi m}$ and our matrix $H(m)$ reduces to $H(m)=\left(\begin{array}{cccc} a &a ^2 & a^3 & a^4 \\ a^2 & a & a^2 & a^3 \\ a^3 & a^2 & a & a^2 \\ a^4& a^3 & a^2 & a\end{array}\right)$ We note that $H(m)$ is a $4\times 4$ $\text{Square Matrix}$ .So to determine the $\det(H(m))$ we can exploit the general formula for $n\times n$ Square matrix as $\Delta =\sum_{i=1}^{n}\sum_{j=1}^{n} (-1)^{i+j} a_{ij} M_{ij}=\sum_{i,j}^{n} (-1)^{i+j}a_{ij}M_{ij}$ Now for $1\leq k\leq 4$ $a^k= e^{-k\pi m} \neq 0$ which implies $H(m)$ doesn't have any $0$ entries and taking any row and columns the determinant doesn't change justified by $\textbf{Laplace expansion}$ . We take $j=1$ and then $\Delta = \sum_{i,1}^{4} (-1)^{i+1} a_{i1}M_{i1}= a_{11}M_{11}-a_{21}M_{21} +a_{31}M_{31} -a_{41}M_{41}\cdots(1)$ here $M_{ij}$ is the corresponding minors so we calculate the minors as $M_{11} =\left| \begin{array}{ccc} a & a^2 & a^3 \\ a^2 & a &a^2 \\ a^3 & a^2 & a\end{array}\right| =a\left|\begin{array}{cc} a & a^2 \\ a^2 & a \end{array}\right|- a^2\left|\begin{array}{cc} a^2 & a^2 \\ a^3 & a \end{array}\right|+a^3\left|\begin{array}{cc} a^2 & a \\ a^3 & a^2 \end{array}\right|= a^3-2a^5+a^7$ here calculating the minors like that of $M_{11}$ we can obtained that $M_{21}=a^4-2a^6+a^8$ and $M_{31}=M_{41}=0$ . Plugging the obtained values in $1$ we have $\Delta = a_{11}M_{11} -a_{21}M_{21} =a(a^3-2a^5+a^7) -a^2(a^4-2a^6+a^8)$ and hence $\det(H(m)) = a^4(1-3a^2+3a^4-a^6)$ . since $H(m)$ is an $\text{invertible Matrix}$ $\Rightarrow H(m)\cdot H(m)^{-1} =I_4 \Rightarrow \det(H(m))\cdot H(m)^{-1} =1\Rightarrow \det(H(m)= \frac{1}{\det({H(m)}^{-1})}=a^4(1-3a^2+3a^4-a^6)$ where $I_4$ is identity matrix hence

$\begin{aligned} \int_{0}^{\infty}e^{-4\pi m}(1-3e^{-2\pi m}+3e^{-4\pi m} -e^{-6\pi m})dm \\=\frac{1}{4\pi}-\frac{3}{6\pi}+\frac{3}{8\pi}-\frac{1}{10\pi}=\frac{1}{40\pi}\end{aligned}$ Therefore required $\beta =40$ .

The given integral is $\dfrac{1}{π}\int_0^1 {a^3(1-a^2)^3da}$ , where $a=e^{-πm}$ . The value of the integral is $\dfrac{1}{40π}$ . Therefore $β=\boxed {40}$