I am trying to calculate the following integral :
$$I(\lambda,\alpha)=\int_{\lambda}^1 \mathrm{d}\tau \frac{1-\tau^\alpha}{1-\tau}\exp(-k \tau)$$
where $\lambda<1$, $k$ is a positive constant and $\alpha$ is a large integer.
I was thinking of doing the substitution $y=\alpha (1-\tau)$, in which case the integral becomes :
$$e^{-k}\int_0^{\alpha(1-\lambda)}\frac{\mathrm{d}y}{y}\left(1-\left(1-\frac{y}{\alpha}\right)^\alpha\right)\exp{\left(\frac{ky}{\alpha}\right)} $$
and I expect at some point to say that for large $\alpha$, then $(1-\frac{y}{\alpha})^\alpha\simeq e^{-y}$, but this is false when $y$ is close to $\alpha(1-\lambda)$.
Is there any way to properly control the error in this assumption and still get an asymptotic equivalent for $I(\lambda,\alpha)$ as $\alpha\to \infty$ ?
Thanks in advance.