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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.

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Is this homework? –  Aryabhata Sep 8 '10 at 14:04
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of course not; I am reading some stuff thats where I ran into this problem. –  Vaolter Sep 8 '10 at 14:15
    
You should use two dollar signs to make the formula be rendered correctly (display style). –  Dario Sep 8 '10 at 20:10

3 Answers 3

up vote 4 down vote accepted

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 $\tau \to \infty$.

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thanx a lot u saved my day! –  Vaolter Sep 8 '10 at 15:43

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}

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thanx a lot u saved my day! –  Vaolter Sep 8 '10 at 15:42

$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{{\rm e}^{#1}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\pp}{{\cal P}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}$

\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:

$$\color{#ff0000}{\Large% \lim_{\tau \to \infty}\int_{-c/\tau}^{0}\expo{-\beta x} \pars{\tau x + c}^{\pars{\lambda/\tau} - 1}\,\dd x} = \lim_{\tau \to \infty}{c^{\lambda/\tau} \over \lambda} = \color{#ff0000}{\Large{1 \over \lambda}} $$

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