Integral involving binomial expression of an exponential 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.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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
  $Incomplete\ Beta\ Function$. 

