## Solution of a Simple Exponential Equation via the Lambert W Function

How to solve

x = \mathrm{exp}\left(\frac{2}{x}\right)

was posed at Mathematics Stack Exchange.

To solve it, we use the Lambert W function.

1 = \frac{1}{x} \mathrm{exp}\left(\frac{2}{x}\right)

2 = \frac{2}{x} \mathrm{exp}\left(\frac{2}{x}\right)

\mathrm{W}(2) = \frac{2}{x}

x = \frac{2}{\mathrm{W}(2)} \approx 2.34575

Where $$\mathrm{W}(z)$$ is the Lambert W function.

A good tutorial can be found here, where we see that for $$a=b\mathrm{e}^{b}$$, we have $$b=\mathrm{W}(a)$$.

## Evaluation of Definite Integrals Via the Cauchy-Schlomilch Transformation

The Cauchy-Schlomilch transformation is one of many so called master formulas for evaluating definite integrals. The idea is that such formulas can be used directly on a variety of integrals and also indirectly by various additional transformations such as differentiating with respect to certain parameters, changes in variables, etc. Here, I present the basic Cauchy-Schlomilch transformation as it appeared in the references listed below. Note that in [1] there is a corollary to this result, but I will not present it here.

For $$a,c \gt 0$$

\int\limits_{0}^{\infty} f\Big[\left(cx – \frac{a}{x} \right)^{2} \Big] \mathrm{d} x = \frac{1}{c} \int\limits_{0}^{\infty} f(u^{2}) \mathrm{d}u
\tag{1}
\label{eq:cst1-1}

Proof: Let $$y=a/(cx)$$, so that \eqref{eq:cst1-1} becomes
\begin{align}
I &= \int\limits_{0}^{\infty} f\Big[\left(cx – \frac{a}{x} \right)^{2} \Big] \mathrm{d} x \\
&= -\frac{a}{c} \int\limits_{0}^{\infty} \frac{1}{y^{2}} f\Big[\left(\frac{a}{y}-cy \right)^{2} \Big] \mathrm{d} y \\
&= \frac{a}{c} \int\limits_{0}^{\infty} \frac{1}{x^{2}} f\Big[\left(cx – \frac{a}{x} \right)^{2} \Big] \mathrm{d} x
\tag{2}
\label{eq:cst1-2}
\end{align}

Combining equations \eqref{eq:cst1-1} and \eqref{eq:cst1-2} yields

2I =
\frac{1}{c} \int\limits_{0}^{\infty} \frac{1}{x^{2}} f\Big[\left(cx – \frac{a}{x} \right)^{2} \Big] \left(c+\frac{a}{x^{2}} \right) \mathrm{d} x
\tag{3}
\label{eq:cst1-3}

Now make the substitution $$u=cx-(a/x)$$

2I = \frac{1}{c} \int\limits_{-\infty}^{\infty} f(u^{2}) \mathrm{d}u = \frac{2}{c} \int\limits_{0}^{\infty} f(u^{2}) \mathrm{d}u

Solving for I yields our desired result.

Notice that $$a$$ vanishes.

## Example 1

\begin{align}
\int\limits_{0}^{\infty} \mathrm{exp}\left(-\Big[ax-\frac{b}{x}\Big]^{2n}\right) \mathrm{d}x
&= \int\limits_{0}^{\infty} \mathrm{exp}\left(-\Big[(ax-\frac{b}{x})^{n}\Big]^{2}\right) \mathrm{d}x \\
&= \frac{1}{a} \int\limits_{0}^{\infty} \mathrm{exp}\left(-[y^{n}]^{2}\right) \mathrm{d}y \\
&= \frac{1}{2an} \Gamma\left(\frac{1}{2n}\right)
\end{align}

## Example 2

\int\limits_{0}^{\infty} \mathrm{exp}\left(-\alpha x^{2} – \frac{\beta}{x^{2}}\right) \mathrm{d}x

Notice that if we expand the expression inside of the integrand in equation \eqref{eq:cst1-1} we have

\left(cx – \frac{a}{x} \right)^{2} = c^{2} x^{2} – 2ca + \frac{a^{2}}{x^{2}}

Rearrangement yields

-c^{2} x^{2} – \frac{a^{2}}{x^{2}} = -2ca \,- \left(cx – \frac{a}{x} \right)^{2}

Thus, $$c^{2} = \alpha$$, $$a^{2} = \beta$$, and

-\alpha x^{2} – \frac{\beta}{x^{2}} = -2\sqrt{\alpha \beta} \,- \left(\sqrt{\alpha} x \,- \frac{\sqrt{\beta}}{x}\right)^{2}

Now we apply the Cauchy-Schlomilch transformation to obtain our final result
\begin{align}
\int\limits_{0}^{\infty} \mathrm{exp}\left(-\alpha x^{2} – \frac{\beta}{x^{2}}\right) \mathrm{d}x
&= \mathrm{e}^{-2\sqrt{\alpha \beta}} \int\limits_{0}^{\infty} \mathrm{exp}\left(-\Big[\sqrt{\alpha} x – \frac{\sqrt{\beta}}{x}\Big]^{2}\right) \mathrm{d}x \\
&= \frac{\mathrm{e}^{-2\sqrt{\alpha \beta}}}{\sqrt{\alpha}} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}} \mathrm{d}y \\
&= \frac{\mathrm{e}^{-2\sqrt{\alpha \beta}}}{\sqrt{\alpha}} \frac{\sqrt{\pi}}{2} \lim_{y \to \infty} \mathrm{erf}(y) \\
&= \frac{1}{2} \sqrt{\frac{\pi}{\alpha}} \mathrm{e}^{-2\sqrt{\alpha \beta}}
\end{align}
Note that this integral was evaluated with a different method in Example 8.4.1 here.

References

1. T. Amdeberhan, M. L. Glasser, M. C. Jones, V. H. Moll, R. Posey, D. Varela. The Cauchy-Schlomilch Transformation. 2010.
2. G. Boros, V. Moll. Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals. 2004.
3. M. L. Glasser. A remarkable property of definite integrals. Math. Comp., 40(40):561–563, 1983.

## Generalized Gould-Hopper Polynomials by Emil Horozov

Classical orthogonal polynomial systems of Jacobi, Hermite, Laguerre and Bessel have the property that the polynomials of each system are eigenfunctions of a second order ordinary differential operator. According to a famous theorem by Bochner they are the only systems with this property. In a recent paper [22] we extended the class of polynomial systems having both the property to be eigenfunctions of a differential operator (now without restriction on the order) and the 3-term recurrence relation, responsible for the orthogonality is relaxed to a d+2-term relation with $$d \in \mathbb{N}$$. We found such systems also when instead of a differential operator we consider a difference one. Our polynomial systems came with many properties. In the present paper we continue to study a class of these systems in the spirit of the classical orthogonal polynomials. Their most important properties, are hypergeometric representations that we find for them. From the hypergeometric representations we derive generating functions and in some cases we find Mehler-Heine type formulas.

The paper is available here.

## Integrals of products of Hurwitz zeta functions via Feynman parametrization and two double sums of Riemann zeta functions by M. A. Shpot, R. B. Paris

We consider two integrals over $$x \in [0,1]$$ involving products of the function $$\zeta_{1}(a,x) \equiv \zeta(a,x) − x^{−a}$$, where $$\zeta(a,x)$$ is the Hurwitz zeta function, given by

\int\limits_{0}^{1} \zeta_{1}(a,x) \zeta_{1}(b,x) \mathrm{d}x

and

\int\limits_{0}^{1} \zeta_{1}(a,x) \zeta_{1}(b,1-x) \mathrm{d}x

when $$\Re (a,b) \gt 1$$. These integrals have been investigated recently in [23]; here we provide an alternative derivation by application of Feynman parametrization. We also discuss a moment integral and the evaluation of two doubly infinite sums containing the Riemann zeta function $$\zeta(x)$$ and two free parameters a and b. The limiting forms of these sums when $$a+b$$ takes on integer values are considered.

The paper is available here.

## Evaluate the Integral $$\int_{0}^{\infty} x^{s-1} \cos(2\pi ax) \mathrm{d}x$$

How to evaluate

\int\limits_{0}^{\infty} x^{s-1} \cos(2\pi ax) \mathrm{d}x
\tag{1}
\label{eq:mtc-1}

was part of a question posed at Mathematics Stack Exchange. Note that this is the Mellin transform of the indicated cosine function.

The original answer that I provided required some rather questionable steps regarding the limits of integration, so here I provide another solution that avoids such difficulties.

In Volume 2 of Higher Transcendental Functions (Bateman Manuscript), Section 9.10, Equation 1 we have a generalization of the fresnel integrals attributed to Bohmer:
\begin{align}
\mathrm{C}(x,a) &= \int\limits_{x}^{\infty} z^{a-1} \cos(z) \mathrm{d}z \\
&= \frac{1}{2} \Big[\mathrm{e}^{i\pi a/2} \Gamma(a,-ix) + \mathrm{e}^{-i\pi a/2} \Gamma(a,ix)\Big]
\end{align}

Thus

\mathrm{C}(0,a) = \int\limits_{0}^{\infty} z^{a-1} \cos(z) \mathrm{d}z
= \Gamma(a) \cos\left(\frac{\pi}{2}a\right)

For our integral, let $$z=2\pi ax$$:

\int\limits_{0}^{\infty} x^{s-1} \cos(2\pi ax) \mathrm{d}x
= (2\pi a)^{-s} \int\limits_{0}^{\infty} z^{s-1} \cos(z) \mathrm{d}z
= (2\pi a)^{-s} \Gamma(s) \cos\left(\frac{\pi}{2}s\right)

## Evaluate the Integral $$\int_{0}^{\pi /2} \frac{x}{\sin(x)} \mathrm{d}x$$

The proof of

\int\limits_{0}^{\pi /2} \frac{x}{\sin(x)} \mathrm{d}x = 2\mathrm{G}
\tag{1}
\label{eq:xosinx-1}

where G is Catalan’s constant was part of a question at Mathematics Stack Exchange. The question could be answered without explicitly evaluating this integral, so I decided to evaluate it.

We begin by expanding the denominator in exponential functions and evaluating the following indefinite integral
\begin{align}
\int \frac{1}{\mathrm{e}^{ix}-\mathrm{e}^{-ix}} \mathrm{d}x &= -\int \frac{\mathrm{e}^{ix}}{1-\mathrm{e}^{i2x}} \mathrm{d}x \\
&= -\int \frac{\mathrm{e}^{ix}}{(1+\mathrm{e}^{ix})(1-\mathrm{e}^{ix})} \mathrm{d}x \\
&= -\frac{1}{2} \int \Big[\frac{\mathrm{e}^{ix}}{1+\mathrm{e}^{ix}} + \frac{\mathrm{e}^{ix}}{1-\mathrm{e}^{ix}} \Big] \mathrm{d}x \\
&= \frac{i}{2} \big[\ln(1+\mathrm{e}^{ix}) – \ln(1-\mathrm{e}^{ix})\big] + \mathrm{const}
\tag{2}
\label{eq:xosinx-2}
\end{align}

Now we evaluate the following integral by parts

\int \frac{x}{\mathrm{e}^{ix}-\mathrm{e}^{-ix}} \mathrm{d}x = ab \,- \int b \,\mathrm{d}a
\tag{3}
\label{eq:xosinx-3}

where $$a = x ,\, \mathrm{d}a = \mathrm{d}x ,\, \mathrm{d}b = \mathrm{d}x/(\mathrm{e}^{ix}-\mathrm{e}^{-ix}) ,\, b =$$ equation \eqref{eq:xosinx-2}.

The integral on the right in equation \eqref{eq:xosinx-3} is
\begin{align}
\int \ln(1+\mathrm{e}^{ix}) \mathrm{d}x &= i \int \frac{\ln(u)}{1-u} \mathrm{d}u \\
&= -i \int \frac{\ln(1-y)}{y} \mathrm{d}y \\
&= i \mathrm{Li}_{2}(y) \\
& = i \mathrm{Li}_{2}(-\mathrm{e}^{ix})
\tag{4}
\label{eq:xosinx-4}
\end{align}
where we used the following substitutions in succession: $$u = 1 + \mathrm{e}^{ix} ,\, y = 1-u$$. $$\mathrm{Li}_{2}(z)$$ is the dilogarithm.

Likewise, we have

\int \ln(1-\mathrm{e}^{ix}) \mathrm{d}x = i \mathrm{Li}_{2}(\mathrm{e}^{ix})
\tag{5}
\label{eq:xosinx-5}

Putting all of the pieces together, we obtain the desired result
\begin{align}
\int\limits_{0}^{\pi /2} \frac{x}{\sin(x)} \mathrm{d}x &= i2 \int\limits_{0}^{\pi /2} \frac{x}{\mathrm{e}^{ix}-\mathrm{e}^{-ix}} \mathrm{d}x \\
&= x\big[\ln(1-\mathrm{e}^{ix}) – \ln(1+\mathrm{e}^{ix}) \big] + i\Big[\mathrm{Li}_{2}(-\mathrm{e}^{ix}) – \mathrm{Li}_{2}(\mathrm{e}^{ix}) \Big] \Big|_{0}^{\pi /2} \\
&= \frac{\pi}{2} \big[\ln(1-i) – \ln(1+i) \big] + i\Big[\mathrm{Li}_{2}(-i) – \mathrm{Li}_{2}(i)\Big] – 0 -i\Big[\mathrm{Li}_{2}(-1) – \mathrm{Li}_{2}(1)\Big] \\
&= \frac{\pi}{2} \Big[\frac{\ln(2)}{2} -i\frac{\pi}{4} – \frac{\ln(2)}{2} – i\frac{\pi}{4}\Big] + i\Big[-\frac{\pi^{2}}{48}-i\mathrm{G}+\frac{\pi^{2}}{48}-i\mathrm{G} \Big] – i\Big[-\frac{\pi^{2}}{12}-\frac{\pi^{2}}{6} \Big] \\
&= 2\mathrm{G}
\end{align}

## Evaluate the Integral $$\int_{-\infty}^{\infty} \frac{\sin(ax)}{\mathrm{e}^{x/2}-\mathrm{e}^{-x/2}}\mathrm{d}x$$

How to prove

I=\int\limits_{-\infty}^{\infty} \frac{\sin(ax)}{\mathrm{e}^{x/2}-\mathrm{e}^{-x/2}}\mathrm{d}x = \pi \tanh(\pi a)
\tag{1}
\label{eq:sinsinh-1}

was a question on Mathematics Stack Exchange. I will provide a solution similar to one at MSE but more fully worked out. Also, I will show an interesting relation between the digamma function and the tangent function due to another solution at MSE.

\begin{align}
I &= -i \int\limits_{0}^{\infty} \frac{\mathrm{e}^{iax}-\mathrm{e}^{-iax}}{\mathrm{e}^{x/2}-\mathrm{e}^{-x/2}}\mathrm{d}x \\
&= -i \int\limits_{0}^{\infty} \mathrm{e}^{-x/2} \frac{\mathrm{e}^{iax}-\mathrm{e}^{-iax}}{1-\mathrm{e}^{-x}}\mathrm{d}x \\
&= -i \int\limits_{0}^{\infty} \mathrm{e}^{-x/2} \left( \mathrm{e}^{iax}-\mathrm{e}^{-iax} \right) \sum\limits_{n=0}^{\infty} \mathrm{e}^{-nx}\mathrm{d}x \\
&= -i \int\limits_{0}^{\infty} \Big[\sum\limits_{n=0}^{\infty} \mathrm{e}^{-x(1/2+n-ia)} \,-\, \sum\limits_{n=0}^{\infty} \mathrm{e}^{-x(1/2+n+ia)} \Big] \mathrm{d}x \\
\tag{2}
\label{eq:sinsinh-2}
&= -i \Big[\sum\limits_{n=0}^{\infty} \frac{1}{1/2+n-ia} \,-\, \sum\limits_{n=0}^{\infty} \frac{1}{1/2+n+ia}\Big] \\
&= 2a \sum\limits_{n=0}^{\infty} \frac{1}{(n+1/2)^{2}+a^{2}}
\end{align}

The Mittag-Leffler expansion of the hyperbolic tangent function is

\tanh(z) = 2z \sum\limits_{n=0}^{\infty} \frac{1}{n^{2}(\pi/2)^{2}+z^{2}}
\tag{3}
\label{eq:sinsinh-3}

For $$z=a\pi$$ we have our final result

\int\limits_{-\infty}^{\infty} \frac{\sin(ax)}{\mathrm{e}^{x/2}-\mathrm{e}^{-x/2}}\mathrm{d}x = 2a \sum\limits_{n=0}^{\infty} \frac{1}{(n+1/2)^{2}+a^{2}} = \pi \tanh(\pi a)

Now, let us examine equation \eqref{eq:sinsinh-2} and use it to recover a definition of the tangent in terms of the digamma function.

Using a series definition of the digamma function:

\psi(z) = -\gamma + \sum\limits_{n=0}^{\infty} \left( \frac{1}{n+1} – \frac{1}{n+z} \right)
\tag{4}
\label{eq:sinsinh-4}

we have

\psi\left(\frac{1}{2}+ia\right) = -\gamma + \sum\limits_{n=0}^{\infty} \left( \frac{1}{n+1} – \frac{1}{(n+1/2)+ia} \right)
\tag{5a}
\label{eq:sinsinh-5a}

and

\psi\left(\frac{1}{2}-ia\right) = -\gamma + \sum\limits_{n=0}^{\infty} \left( \frac{1}{n+1} – \frac{1}{(n+1/2)-ia} \right)
\tag{5b}
\label{eq:sinsinh-5b}

Subtracting equation \eqref{eq:sinsinh-5b} from \eqref{eq:sinsinh-5a} and multiplying the result by $$-i$$ yields equation \eqref{eq:sinsinh-2} which is equal to equation \eqref{eq:sinsinh-1}.

Making the substitution $$z=\frac{1}{2}+ia$$ yields
\begin{align}
-i\big[\psi(z)-\psi(1-z)\big] &= \pi \tanh\Big[-i\pi\left(z-\frac{1}{2}\right)\Big] \\
&= -i \pi \tan\Big[\pi\left(z-\frac{1}{2}\right)\Big] \\
&= -i \pi \cot(\pi z)
\end{align}

Rearrangement yields

\psi(z)-\psi(1-z) = \frac{-\pi}{\tan(\pi z)}

an expression which can be found here.

## Josef Meixner: his life and his polynomials by Paul L. Butzer, Tom H. Koornwinder

This paper starts with a biographical sketch of the life of Josef Meixner. Then his motivations to work on orthogonal polynomials and special functions are reviewed. Meixner’s 1934 paper introducing the Meixner and Meixner-Pollaczek polynomials is discussed in detail. Truksa’s forgotten 1931 paper, which already contains the Meixner polynomials, is mentioned. The paper ends with a survey of the reception of Meixner’s 1934 paper.

The entire paper is available here.

## Integrate $$\int_{0}^{\infty} \mathrm{e}^{x^{2}} (1 + \mathrm{e}^{x^{2}})^{-2} \mathrm{d}x$$

How to evaluate this integral was a question at Mathematics Stack Exchange.

We begin by considering:

\begin{align}
\left(1 + a\mathrm{e}^{x^{2}}\right)^{-1}
&= \frac{1}{a}\mathrm{e}^{-x^{2}} \frac{1}{1+(a\,\mathrm{exp}(x^{2}))^{-1}} \\
&= \frac{1}{a}\mathrm{e}^{-x^{2}} \sum\limits_{n=0}^{\infty} (-1)^{n} \left( \frac{1}{a}\mathrm{e}^{-x^{2}}\right)^{n} \\
&= \sum\limits_{n=0}^{\infty} (-1)^{n} \frac{1}{a^{n+1}} \mathrm{e}^{-(n+1)x^{2}} \\
\end{align}

Now integrate the exponential function

\int\limits_{0}^{\infty} \mathrm{e}^{-(n+1)x^{2}} \mathrm{d}x = \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{n+1}}

Now we have

\int\limits_{0}^{\infty} \left(1 + a\mathrm{e}^{x^{2}}\right)^{-1} \mathrm{d}x =
\frac{\sqrt{\pi}}{2} \sum\limits_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^{1/2}} \frac{1}{a^{n}}

Taking the derivative of both sides with respect to $$a$$:

\int\limits_{0}^{\infty} \mathrm{e}^{x^{2}} \left(1 + a\mathrm{e}^{x^{2}}\right)^{-2} \mathrm{d}x =
\frac{\sqrt{\pi}}{2} \sum\limits_{n=0}^{\infty} (-1)^{n-1} \frac{1}{n^{-1/2}} \frac{1}{a^{n+1}}

Taking $$\lim_{a \to 1}$$
\begin{align}
\int\limits_{0}^{\infty} \mathrm{e}^{x^{2}} \left(1 + \mathrm{e}^{x^{2}}\right)^{-2} \mathrm{d}x &=
\frac{\sqrt{\pi}}{2} \sum\limits_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n^{-1/2}} \\
&= \frac{\sqrt{\pi}}{2} \eta(-1/2) \\
&\approx 0.336859119
\end{align}

$$\eta(s)$$ is the Dirichlet eta function.

## Integrate $$\int_{0}^{\infty} \frac{x^{a}}{1+x^{2}} \mathrm{d}x$$

How to evaluate this integral was a question at Mathematics Stack Exchange. The first method we present was already answered at MSE but here we fill in the missing steps.

## Method 1

Let $$z=x^2$$
\begin{align}
\int\limits_{0}^{\infty} \frac{x^{a}}{1+x^{2}} \mathrm{d}x
&= \frac{1}{2} \int\limits_{0}^{\infty} z^{(a-1)/2} \frac{\mathrm{d}z}{1+z} \\
\tag{a}
&= \frac{1}{2} \mathrm{B}\left(\frac{a+1}{2}, 1-\frac{a+1}{2} \right) \\
\tag{b}
& = \frac{1}{2} \Gamma\left(\frac{a+1}{2}\right) \Gamma\left(1-\frac{a+1}{2}\right) \\
\tag{c}
&= \frac{\pi}{2\sin(\pi(a+1)/2)}
\end{align}
a. We used the following definition of the beta function

\mathrm{B}(a.b)=\int\limits_{0}^{\infty} \frac{z^{a-1}}{1+z^{a+b}} \mathrm{d}z

b. $$\mathrm{B}(a.b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
c. $$\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}$$

## Method 2

Let

f(z) = \frac{z^{a}}{1+z^{2}}

Using the keyhole contour, we have first order poles at $$\pm i$$, so the residues are

\mathrm{Res}[f(z),i] = \frac{i^{a}}{i2} = \mathrm{e}^{ia\pi/2} \frac{1}{i2}

\mathrm{Res}[f(z),-i] = \frac{(-i)^{a}}{-i2} = -\mathrm{e}^{ia\pi} \mathrm{e}^{ia\pi/2} \frac{1}{i2}

\begin{align}
\oint\limits_{C} f(z) \mathrm{d}z
&= \pi \mathrm{e}^{ia\pi/2} \left(1 – \mathrm{e}^{ia\pi} \right) \\
&= \lim_{\epsilon,R \to 0,\infty} \int\limits_{\epsilon}^{R} f(x) \mathrm{d}x
+ \int\limits_{\Gamma} f(z)\mathrm{d}z
+ \int\limits_{R}^{\epsilon} f(x) \mathrm{d}x
+ \int\limits_{\gamma} f(z)\mathrm{d}z \\
&= \int\limits_{0}^{\infty} \frac{x^{a}}{1+x^{2}} \mathrm{d}x
\,- \int\limits_{0}^{\infty} \mathrm{e}^{ia2\pi} \frac{x^{a}}{1+x^{2}} \mathrm{d}x \\
&= \left(1 – \mathrm{e}^{ia2\pi} \right) \int\limits_{0}^{\infty} \frac{x^{a}}{1+x^{2}} \mathrm{d}x
\end{align}

Thus we have
\begin{align}
\int\limits_{0}^{\infty} \frac{x^{a}}{1+x^{2}} \mathrm{d}x
&= \pi \mathrm{e}^{ia\pi/2} \left(1 – \mathrm{e}^{ia\pi} \right) \frac{1}{\left(1 – \mathrm{e}^{ia2\pi} \right)} \\
&= \frac{\pi \sin(a\pi/2)}{\sin(a\pi)} \\
&= \frac{\pi}{2\cos(a\pi/2)}
\end{align}

Notes:
1. R is the radius of the large circle $$\Gamma$$.
2. $$\epsilon$$ is the radius of the small circle $$\gamma$$.