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I am trying to understand the behavior of the following function w.r.t $b$:

$$ \mathrm{M}\left(b,k\right) = \int_{0}^{\infty}\mathrm{e}^{-kt}\left(2\mathrm{e}^{t} - 1\right)^{b} \,\mathrm{d}t\quad \mbox{where}\ 0 \leq b \leq \frac{k}{2}\ \mbox{and}\ k\ \mbox{is an}\ even\ \mbox{integer}. $$

One way is to expand the $\left(2\mathrm{e}^{t} - 1\right)^{b}$ term ( when $b$ is an integer ) and then integrate after which I end up with a sum involving alternating binomial coefficients along with other terms and I can't really reason about how that sum behaves w.r.t $b$. Is there an approximation for $\mathrm{M}$ which is a simpler closed form expression of $b$ ?.

The reason I am asking is to eventually figure out ( a reasonable approximation would be fine ) where the minima of the following expression lies in the range $0\leq b\leq \frac{k}{2}$:

$\mathrm{M}\left(\exp\left(\frac{k}{\left(k - b\right)\mathrm{M}}\right) - 1\right)$

where $\mathrm{M}$ is a function of $b$ and $k$ as defined in the beginning.

Zaeem Hussain
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1 Answers1

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{M}\pars{b,k} & \equiv \int_{0}^{\infty}\expo{-kt}\pars{2\expo{t} - 1}^{b}\,\dd t = 2^{k}\int_{0}^{\infty}\pars{{1 \over 2}\,\expo{-t}}^{k - b} \pars{1 - {1 \over 2}\expo{-t}}^{b}\,\dd t \end{align}

Set $\ds{x \equiv \expo{-t}/2 \iff t = -\ln\pars{2x}}$

\begin{align} \mrm{M}\pars{b,k} & = 2^{k}\int_{1/2}^{0}x^{k - b}\pars{1 - x}^{b}\pars{-\,{\dd x \over x}} = 2^{k}\int_{0}^{1/2}x^{k - b - 1}\pars{1 - x}^{b}\,\dd x \\[5mm] & = \bbx{2^{k}\,\mrm{B}_{1/2}\pars{k - b,b + 1}} \end{align}

where $\ds{\,\mrm{B}_{z}}$ is the $Incomplete\ Beta\ Function$.

Felix Marin
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