An Integral for the Occasion: Happy New Year!

To celebrate the new year 2021, I’ve created an integral for the occasion which reads as follows.

2021 New Years Integral. Let \alpha,\beta, and \gamma be real-valued continuous functions defined in \mathbb{R}^3 such that

\displaystyle 1<\gamma(x)<\frac{1+\sqrt{5}}{2},

\displaystyle 1<\alpha(x)<\frac{\gamma(x)}{1+\gamma(x)-\gamma(x)^2},

and

\displaystyle \beta(x):=\gamma(x)+\frac{1}{\alpha(x)}-\frac{1}{\gamma(x)},

for all x\in\mathbb{R}^3. Furthermore, let f,g:\mathbb{R}^3\to\mathbb{R} be functions given by

\displaystyle f(x):=\pi^{-\frac{3}{2}}\exp\left(-\frac{x_1^2}{H^2}-\frac{x_2^2}{a^2p^2}-\frac{x_3^2}{p^2y^2}\right),

and

\displaystyle g(x):=\pi^{-\frac{3}{2}}\exp\left(-\frac{x_1^2}{4}-\frac{x_2^2}{2019!^2}-\frac{x_3^2}{2041210^2}\right),

where H,a,p,y>0 are constants. If G is a function on \mathbb{R}^3 defined by

\displaystyle G(x):=\frac{(\alpha(x)\beta(x)\gamma(x))^2-1}{\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1}\text{ }\text{ \emph{for all} }x\in\mathbb{R}^3,

then the integral

\displaystyle \int_{\mathbb{R}^3}G(x)\log\sqrt[\alpha(x)]{\exp\left({\frac{f(x)}{\gamma(x)}}\right)\sqrt[\beta(x)]{\exp({g(x)})\sqrt[\gamma(x)]{\exp\left({\frac{f(x)}{\gamma(x)}}\right) \sqrt[\alpha(x)]{\exp({g(x)})\sqrt[\beta(x)]{\cdots}}}}}\text{ }\mathrm{d}x

evaluates to

Happy+2021!

Proof. Our first step to evaluating the integral is to simplify the nested radical appearing in the integrand. For this, we employ Theorem 6 found on my Nested Radicals page.
Theorem 6. Let (a_n)_{n\in\mathbb{Z}_{\geq 1}} and (b_n)_{n\in\mathbb{Z}_{\geq 1}} be periodic sequences of real numbers with periods T_a,T_b\in \mathbb{Z}_{\geq 1} such that a_n>0, a_n=a_{n+T_a} and b_n=b_{n+T_b} for all n\geq 1. If |b_1|,|b_2|,\cdots |b_{T_b-1}|,|b_{T_b}|>1. Then

\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}=\prod_{n=1}^{T_a}a_n^{S(n)}

where

\displaystyle S(n)=\frac{(b_1\cdots b_{T_b})^{A}}{(b_1\cdots b_{T_b})^{A}-1}\sum_{k=0}^{B-1}\prod_{j=1}^{T_ak+n}b_j^{-1}

for n\in\mathbb{Z}_{\geq 1}, with A=\frac{l.c.m(T_a,T_b)}{T_b} and B=\frac{l.c.m(T_a,T_b)}{T_a}.

This result can be derived using the idea behind the proof of Theorem 2 found on my Nested Radicals page. For this it suffices to consider the pair of sequences (\tilde{a}_n) and (\tilde{b}_n) given by

  • \tilde{a}_n:=a_n for 1\leq n \leq T, \tilde{b}_n:=b_n for 1\leq n\leq T with T=l.c.m(T_a,T_b), and
  • \tilde{a}_n=\tilde{a}_{n+T}, and \tilde{b}_n=\tilde{b}_{n+T} for n\geq 1.

Then one simplifies the resulting expressions to arrive at Theorem 6.

For our current problem, we let (a_n)=(a_n(x)) and (b_n)=(b_n(x)) be periodic sequences of functions defined on \mathbb{R}^3 where

\displaystyle a_n=a_{n+2},\text{ }\text{ }b_{n}=b_{n+3}\text{ }\forall n\geq 1

with

\displaystyle a_1:=\exp\left({\frac{f(x)}{\gamma(x)}}\right),\text{ }a_2:=\exp(g(x)),

\displaystyle b_1:=\alpha(x),\text{ }b_2:=\beta(x),\text{ }\text{ and }\text{ }b_3:=\gamma(x).

It is clear that, for each x\in\mathbb{R}^3, the sequence (a_n(x)) fits the assumptions of the theorem with T_a=2. But for (b_n) we need to check that |b_2|=|\beta(x)|>1 for each x\in\mathbb{R}^3 since we already know that |b_1|=\alpha(x)>1 and |b_3|=\gamma(x)>1 for all x\in\mathbb{R}^3. From the definition of \beta and the assumed bounds on \alpha we find

\displaystyle \beta(x)=\gamma(x)+\frac{1}{\alpha(x)}-\frac{1}{\gamma(x)}>\gamma(x)+\frac{1+\gamma(x)-\gamma(x)^2}{\gamma(x)}-\frac{1}{\gamma(x)}=1

for all x\in\mathbb{R}^3. Noting that T_a=2 and T_b=3, we find in the context of the theorem that A=2 and B=3, giving

\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}=\prod_{n=1}^{2}a_n^{S(n)}

where

\displaystyle S(n)=\frac{(b_1b_2 b_{3})^2}{(b_1b_2 b_{3})^{2}-1}\sum_{k=0}^{2}\prod_{j=1}^{2k+n}b_j^{-1}

for n\in\{1,2\}. Let’s rewrite S(1) and S(2) in terms of the functions \alpha,\beta and \gamma. By periodicity of (b_n) we find

\displaystyle S(1)=\frac{(b_1b_2b_3)^2}{(b_1b_2b_3)^{2}-1}\sum_{k=0}^{2}\prod_{j=1}^{2k+1}b_j^{-1}=\frac{(b_1b_2b_3)^2}{(b_1b_2b_3)^{2}-1}\left(b_1^{-1}+b_1^{-1}b_2^{-1}b_3^{-1}+b_1^{-2}b_2^{-2}b_3^{-1}\right)

or

\displaystyle S(1)=\frac{\gamma(x)\left(\alpha(x)\beta(x)^2\gamma(x)+\alpha(x)\beta(x)+1\right)}{(\alpha(x)\beta(x)\gamma(x))^2-1},\text{ }\forall x\in\mathbb{R}^3.

Similarly, we get

\displaystyle S(2)=\frac{(b_1b_2 b_{3})^2}{(b_1b_2 b_{3})^{2}-1}\sum_{k=0}^{2}\prod_{j=1}^{2k+2}b_j^{-1}=\frac{\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1}{(\alpha(x)\beta(x)\gamma(x))^2-1},

for all x\in\mathbb{R}^3.

With these, we can simplify the expression for the function l:\mathbb{R}^3\to\mathbb{R} given by

\displaystyle l(x):=\log\sqrt[\alpha(x)]{\exp\left({\frac{f(x)}{\gamma(x)}}\right)\sqrt[\beta(x)]{\exp({g(x)})\sqrt[\gamma(x)]{\exp\left({\frac{f(x)}{\gamma(x)}}\right) \sqrt[\alpha(x)]{\exp({g(x)})\sqrt[\beta(x)]{\cdots}}}}}.

From our application of Theorem 6, we get for each x\in\mathbb{R}^3

\displaystyle l(x)=\log \prod_{n=1}^{2}a_n^{S(n)}=S(1)\log\left(\exp\left({\frac{f(x)}{\gamma(x)}}\right) \right)+S(2)\log( \exp({g(x)}))

or

\displaystyle l(x)=\frac{ (\alpha(x)\beta(x)^2\gamma(x)+\alpha(x)\beta(x)+1)f(x)+(\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1)g(x)}{(\alpha(x)\beta(x)\gamma(x))^2-1}.

Recalling the definition of the function G, our integrand thus takes the form

\displaystyle G(x)l(x)=\frac{ (\alpha(x)\beta(x)^2\gamma(x)+\alpha(x)\beta(x)+1)f(x)+(\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1)g(x)}{\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1}

or

\displaystyle G(x)l(x)=\left(\frac{\alpha(x)\beta(x)^2\gamma(x)+\alpha(x)\beta(x)+1}{\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1}\right)f(x)+g(x)\text{ }\forall x\in \mathbb{R}^3.

Using the expression for \beta we can simplify the coefficient of f as follows: for each x\in\mathbb{R}^3

\displaystyle \frac{\alpha(x)\beta(x)^2\gamma(x)+\alpha(x)\beta(x)+1}{\alpha(x)\beta(x)\gamma(x)^2+\beta(x)\gamma(x)+1}=\frac{\beta(x)^2\gamma(x)+\beta(x)+\frac{1}{\alpha(x)}}{\beta(x)\gamma(x)^2+\frac{\beta(x)\gamma(x)}{\alpha(x)}+\frac{1}{\alpha(x)}}

\displaystyle = \frac{\beta(x)^2\gamma(x)+\beta(x)+\beta(x)+\frac{1}{\gamma(x)}-\gamma(x)}{\beta(x)\gamma(x)^2+\beta(x)\gamma(x)\left(\beta(x)+\frac{1}{\gamma(x)}-\gamma(x)\right)+ \beta(x)+\frac{1}{\gamma(x)}-\gamma(x) }

\displaystyle =\frac{\beta(x)^2\gamma(x)+2\beta(x)+\frac{1}{\gamma(x)}-\gamma(x)}{\beta(x)^2\gamma(x)+ 2\beta(x)+\frac{1}{\gamma(x)}-\gamma(x) }=1.

Let I denote the integral that we are seeking to evaluate. With the above, our problem is now simplified massively: evaluate

\displaystyle I=\int_{\mathbb{R}^3}(f(x)+g(x))\mathrm{d}x

with

\displaystyle f(x)=\pi^{-\frac{3}{2}}\exp\left(-\frac{x_1^2}{H^2}-\frac{x_2^2}{a^2p^2}-\frac{x_3^2}{p^2y^2}\right),

and

\displaystyle g(x)=\pi^{-\frac{3}{2}}\exp\left(-\frac{x_1^2}{4}-\frac{x_2^2}{2019!^2}-\frac{x_3^2}{2041210^2}\right).

Apply Fubini’s theorem to find that

\displaystyle \int_{\mathbb{R}^3}f(x)\mathrm{d}x=\pi^{-\frac{3}{2}}\left(\int_{\mathbb{R}}e^{-\frac{x_1^2}{H^2}}\mathrm{d}x_1\right)\left(\int_{\mathbb{R}}e^{-\frac{x_2^2}{(ap^2)}}\mathrm{d}x_2 \right)\left(\int_{\mathbb{R}}e^{-\frac{x_3^2}{(py)^2}}\mathrm{d}x_3\right),

and

\displaystyle \int_{\mathbb{R}^3}g(x)\mathrm{d}x=\pi^{-\frac{3}{2}}\left(\int_{\mathbb{R}}e^{-\frac{x_1^2}{4}}\mathrm{d}x_1\right)\left(\int_{\mathbb{R}}e^{-\frac{x_2^2}{2019!^2}}\mathrm{d}x_2 \right)\left(\int_{\mathbb{R}}e^{-\frac{x_3^2}{2041210^2}}\mathrm{d}x_3\right).

The integrals in the above products are all of the form

\displaystyle \int_{\mathbb{R}}e^{-\frac{x^2}{w^2}}\mathrm{d}x

which evaluates to \sqrt{\pi}w whenever w>0. Consequently,

\displaystyle \int_{\mathbb{R}^3}f(x)\mathrm{d}x=Happy

while

\displaystyle \int_{\mathbb{R}^3}g(x)\mathrm{d}x=2\times 2019!\times 2041210.

The conclusion then follows once we note that

\displaystyle 2\times 2019!\times 2041210=2!(2021-2)!\times 2041210=2021!\frac{2041210}{\begin{pmatrix} 2021 \\ 2 \end{pmatrix}}

and

\displaystyle \begin{pmatrix} 2021 \\ 2 \end{pmatrix}=2041210,

giving

\displaystyle \int_{\mathbb{R}^3}g(x)\mathrm{d}x=2021!

\\\\


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!. \\\\

New Integral Corner Collection: Complex Analysis to the Rescue!

My latest addition the Integral Corner page is a collection of neat results that can be found using methods from Complex Analysis. Deriving these results by any other method isn’t particularly clear so I’ve called the collection Complex Analysis to the Rescue! Let’s visit the first few results.

The first of the collection asserts that,

1) For a,b,c \in \mathbb{R} satisfying |a|>|b|+|c|>0, and given n \in \mathbb{Z}_{\geq 0}, there hold

\displaystyle \int_0^{2\pi}\frac{\cos(nx)}{a+b\cos(x)+c\sin(x)}dx=2\pi \lambda (a,b,c,n) \cos(n\phi)

and

\displaystyle \int_0^{2\pi}\frac{\sin(nx)}{a+b\cos(x)+c\sin(x)}dx=2\pi \lambda (a,b,c,n) \sin(n\phi)

where

\displaystyle \lambda (a,b,c,n):=\frac{(\text{sgn}(a))^n\left(-|a|+\sqrt{a^2-b^2-c^2}\right)^n}{\sqrt{(a^2-b^2-c^2)(b^2+c^2)^n}}

and

\displaystyle \phi:=\text{arg}(b+ic)\in[-\pi,\pi).

To derive these results one can rewrite each integral as a complex contour integral over the unit circle |z|=1 in \mathbb{C} and employ the Residue Theorem after having checked the positions of poles of the integrand relative to the unit disc (which is where the inequality involving a,b and c comes in).

A similar approach can be used to tackle the second result in my collection which is a generalisation of the above that tells more about the kind of result to expect when integrating quotients of the form

\displaystyle \frac{a_1\cos(nx)+a_2\sin(nx)+a_3}{a_4\cos(mx)+a_5\sin(mx)+a_6}

over one period, provided |a_6|>|a_5|+|a_4|>0 and n,m are nonnegative integers. It reads

2) For any a,b,c,d,e,f\in\mathbb{R} and m,n\in\mathbb{Z}_{\geq 0} such that c>|a|+|b|>0 and m\geq 1,

\displaystyle \int_{0}^{2\pi}\frac{d\cos(nx)+e\sin(nx)+f}{a\cos(mx)+b\sin(mx)+c}\mathrm{d}x=\frac{2\pi }{m\sqrt{c^2-a^2-b^2}}\sum_{k=0}^{m-1}\left(f+\sqrt{d^2+e^2}R\cos\left(\beta-n\alpha_k\right)\right)

where

\displaystyle R:=\left(\frac{c-\sqrt{c^2-a^2-b^2}}{\sqrt{a^2+b^2}}\right)^{\frac{n}{m}},\text{ } \beta:=\text{arg}(d+ie)\in(-\pi,\pi],

and, for k\in\{0,\cdots,m-1\}, we define

\displaystyle \alpha_k:=\frac{\theta+(2k+1)\pi}{m} \text{ with } \theta=\text{arg}(a+bi) \in (-\pi,\pi].

Using the first result above in 1), one can easily tackle the following third item of the collection

3) Given |a|>|b|+|c|>0 where a,b,c\in\mathbb{R} and m\in\mathbb{Z}_{\geq 0}, for all z\in\mathbb{R} there holds

\displaystyle \int_0^{2\pi}\int_0^{2\pi}\frac{(\cos(mx)-\cos(my))(\cos(my)-\cos(mz))}{(a+b\cos(x)+c\sin(x))(a+b\cos(y)+c\sin(y))}\mathrm{d}x\mathrm{d}y=\frac{2\pi^2}{a^2-b^2-c^2}\left(\frac{(b^2+c^2)^m-\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}{\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}\right)

Consequently, for all n\in\mathbb{N} and x_{2n+1}\in\mathbb{R} we have

\displaystyle \int_0^{2\pi}\int_0^{2\pi}\cdots\int_0^{2\pi}\prod_{j=1}^{2n}\frac{\cos(mx_{j})-\cos(mx_{j+1})}{a+b\cos(x_{j})+c\sin(x_{j})}\mathrm{d}x_1\cdots\mathrm{d}x_{2n-1}\mathrm{d}x_{2n}= \frac{2^n\pi^{2n}}{(a^2-b^2-c^2)^n}\left(\frac{(b^2+c^2)^m-\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}{\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}\right)^n.

Notice the first result in 3) does not depend on z\in\mathbb{R}, which allows us to deduce the second result by Fubini’s Theorem and induction on n. The same approach allows one to derive the following 4th result involving an odd number of terms in the product integrand instead.

4) Given |a|>|b|+|c|>0 where a,b,c\in\mathbb{R} and m,n\in\mathbb{Z}_{\geq 0}, for all x_{2n+2}\in\mathbb{R} there holds

\displaystyle \int_0^{2\pi}\int_0^{2\pi}\cdots\int_0^{2\pi}\prod_{j=1}^{2n+1}\frac{\cos(mx_{j})-\cos(mx_{j+1})}{a+b\cos(x_{j})+c\sin(x_{j})}\mathrm{d}x_1\cdots\mathrm{d}x_{2n}\mathrm{d}x_{2n+1}= \Gamma(a,b,c,n,m)\left(\lambda(a,b,c,m)\cos(m\phi)-\lambda(a,b,c,0)\cos\left(mx_{2n+2}\right)\right)

where

\displaystyle \Gamma(a,b,c,n,m):=\frac{2^{n+1}\pi^{2n+1}}{(a^2-b^2-c^2)^n}\left(\frac{(b^2+c^2)^m-\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}{\left(|a|+\sqrt{a^2-b^2-c^2}\right)^{2m}}\right)^n,

\displaystyle \lambda (a,b,c,w):=\frac{(\text{sgn}(a))^w\left(-|a|+\sqrt{a^2-b^2-c^2}\right)^w}{\sqrt{(a^2-b^2-c^2)(b^2+c^2)^w}},

and \phi:=\text{arg}(b+ic)\in[-\pi,\pi).

Other problems I hope to consider involve integrands of the form

\displaystyle \frac{f(a_1\cos(nx)+a_2\sin(nx)+a_3)}{a_4\cos(mx)+a_5\sin(mx)+a_6}

where f is an analytic function locally.

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).

Featured

Welcome!

Below are a few plots of mine for your viewing pleasure. These images depict the evolution of discrete-time dynamical systems that I devised following a coding exercise on Ikeda Maps. While you’re here, feel free to explore

  • The Integral Corner: a place where I collect some of my favourite integrals, most of which I created for fun since August 2017;
  • Nested Radicals: an account of my results on evaluating nested radicals through periodicity. This constitutes a short study that I undertook in July 2020;
  • Determinant Integrals: my introduction to an interplay between matrix determinant and Lebesgue integration. I initiated this work in August 2020 and I look forward to developing it further;
  • Sobolev Spaces: a discussion on fine properties of functions that are ubiquitous in the theory of partial differential equations. Some of my own work on PDE theory will be discussed here.

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