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.

  • $\begingroup$ A closed form solution exists for this integral, but it involves a sum, I'm guessing that is not what you want though right? $\endgroup$ – Gregory May 23 '17 at 17:54
  • $\begingroup$ is $\lambda>0$ ? $\endgroup$ – tired May 23 '17 at 18:01
  • $\begingroup$ i didn't make this a full answer because it lacks some rigor which i don't have the time to fill in $\endgroup$ – tired May 23 '17 at 22:41
  • 1
    $\begingroup$ I give an answer to a simplified version (to make things clearer) of the quetsion. Namely i will consider $I_k(n)=I_k(0,n)$ the general case might be easily deduced from it. Let us start with $$ I_k(n)-I_k(n-1)=\int_0^1e^{-kx}\frac{1-x^n-1+x^{n-1}}{1-x}=\int_0^1x^{n-1}e^{-kx}=g_k(n) $$ Now let us apply integration by parts to the right hand side $$ g_k(n)=\frac{1}{n}e^{-k}-\frac{k}{n}g_k(n-1)=\frac{1}{n}e^{-k}-\frac{k}{n(n-1)}e^{-k}+\frac{k^2}{n(n-1)}e^{-k}g_k(n-2) $$ $\endgroup$ – tired May 24 '17 at 0:09
  • $\begingroup$ this means that in the large $n$ limit $g(n)\sim e^{-k}\frac{1}{n}+O(n^{-2})$ and we get the following, simplified recursion relation $$ I_k(n)-I_k(n-1)\sim\frac{1}{n}e^{-k}+O(n^{-2}) $$ or >$$ I_k(n)\sim e^{-k}H_n+C\sim e^{-k}\log(n)+C +e^{-k}\gamma+O(n^{-1}) $$ where $C$ is a $n$ independent contstant (which can be fixed by the boundary condition $I_0(n)=H_n\sim\log(n)+\gamma \rightarrow C=0$) $\endgroup$ – tired May 24 '17 at 0:09

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