evaluating $\lim_{\tau\rightarrow \infty} \int_{-c/\tau}^0 e^{-\beta x}(\tau x+c)^{(\lambda/\tau) -1}dx$ How do I evaluate:
$$\lim_{\tau\rightarrow \infty} \int_{-c/\tau}^0 e^{-\beta x}(\tau x+c)^{(\lambda/\tau) -1}dx$$ to give $$1/\lambda$$
I suspect I need to convert to gamma function but cannot simplify it to get the answer.
 A: Try the subsitution $s=\beta(x+\frac{c}{\tau})$. Then use: 
\begin{equation}
\int_0^a t^{b-1}e^{-t}dt=a^be^{-a}\sum_0^{\infty}\frac{a^n}{b(b+1)\ldots(b+n)}
\end{equation}
A: A sketch of the proof, as I'm in a hurry - I hope I've done all the calculations correctly:
Set $y = \tau x + c$ to obtain
$$
\frac{1}{\tau} \int_0^c e^{\frac{\beta}{\tau} (c - y)} y^{\lambda / \tau - 1} d y
$$
Now, partial integration yields
$$
\frac{1}{\lambda} c^{\lambda / \tau} + \frac{\beta}{\lambda \tau} \int_0^c e^{\frac{\beta}{\tau} (c - y)} y^{\lambda / \tau} dy
$$
The limit of the first summand is $\frac{1}{\lambda}$, and the integrand in the second part is bounded (maximum is obtained for $y = \frac{\lambda}{\beta}$), so the second summand will vanish when $\tau \to \infty$.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{{\rm e}^{#1}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
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 \newcommand{\ul}[1]{\underline{#1}}%
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\begin{align}
&\lim_{\tau \to \infty}\int_{-c/\tau}^{0}\expo{-\beta x}
\pars{\tau x + c}^{\pars{\lambda/\tau} - 1}\,\dd x
\\[3mm]&=
\lim_{\tau \to \infty}\braces{%
\left.\expo{-\beta x}\,
{\pars{\tau x + c}^{\lambda/\tau} \over \lambda/\tau}\,{1 \over \tau}
\right\vert_{-c/\tau}^{0}
+
{\beta \over \lambda}\int_{-c/\tau}^{0}\expo{-\beta x}
\pars{\tau x + c}^{\lambda/\tau}\,\dd x}
\\[3mm]&=
\lim_{\tau \to \infty}\bracks{%
{c^{\lambda/\tau} \over \lambda}
+
{\beta \over \lambda}
\underbrace{\quad%
\int_{-c/\tau}^{0}\expo{-\beta x}\pars{\tau x + c}^{\lambda/\tau}\,\dd x\quad}
_{\ds{<\ {\expo{\beta c/\tau}c^{\pars{\lambda/\tau} + 1} \over \tau}}}}
\end{align}
The inequality shows that the integral vanishes out in the limit
$\tau \to \infty$. We are left with the first term:
$$\color{#ff0000}{\Large%
\lim_{\tau \to \infty}\int_{-c/\tau}^{0}\expo{-\beta x}
\pars{\tau x + c}^{\pars{\lambda/\tau} - 1}\,\dd x}
=
\lim_{\tau \to \infty}{c^{\lambda/\tau} \over \lambda}
=
\color{#ff0000}{\Large{1 \over \lambda}}
$$
