evaluating $\lim_{\tau\rightarrow \infty} \int_{-c/\tau}^0 e^{-\beta x}(\tau x+c)^{(\lambda/\tau) -1}dx$

Question

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.

Answer 1

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 $\tau \to \infty$.

Answer 2

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}