# 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.

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Is this homework? –  Aryabhata Sep 8 '10 at 14:04
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

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: $$\int_0^a t^{b-1}e^{-t}dt=a^be^{-a}\sum_0^{\infty}\frac{a^n}{b(b+1)\ldots(b+n)}$$

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


\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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