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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.

Chappers
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Someone1348
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  • A closed form solution exists for this integral, but it involves a sum, I'm guessing that is not what you want though right? – Gregory May 23 '17 at 17:54
  • is $\lambda>0$ ? – tired May 23 '17 at 18:01
  • i didn't make this a full answer because it lacks some rigor which i don't have the time to fill in – tired May 23 '17 at 22:41
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    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

    $$ g_k(n)=\frac{1}{n}e^{-k}-\frac{k}{n}g_k(n-1)=\frac{1}{n}e^{-k}-\frac{k}{n(n-1)}e^{-k}+\frac{k^2}{n(n-1)}e^{-k}g_k(n-2) $$

     – tired May 24 '17 at 00:09
  • 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 $I_0(n)=H_n\sim\log(n)+\gamma \rightarrow C=0$)

     – tired May 24 '17 at 00:09
  • Thanks a lot ! I did know that I could write the fraction $\frac{\tau^\alpha-1}{\tau-1}$ as a sum, but then I had a sum of differences of incomplete gamma functions, and then I couldn't really get much insight into the asymptotics. – Someone1348 May 24 '17 at 12:50
  • @Someone1348 you are welcome. Interestingly this approach also works for non-integer $n$ in case you are interested. – tired May 28 '17 at 23:17

1 Answers1

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$$I(\lambda,\alpha)=\int_{\lambda}^1 \frac{1-\tau^\alpha}{1-\tau}\,e^{-k \tau}\,d\tau$$

The antiderivative is not too difficult and $$I(\lambda,\alpha)=\frac{1-\lambda }{k^2}\Bigg ( (k (1-\lambda )-1)\,e^{-k \lambda }+\frac{\Gamma (\alpha+2,k \lambda )-k \Gamma (\alpha+1,k \lambda )}{k^a} \Bigg)$$ The asymptotic is not very pleasant. According to Mathematica, it could write $$I(\lambda,\alpha)=\frac{(\lambda -1) k^{-\alpha -2} e^{-\alpha -k \lambda }}{288 \alpha }\,(A-B)$$ with $$A=288 e^{\alpha } k^{\alpha } \left(\alpha +k^2 (\lambda -1) \lambda ^{\alpha +1}+\alpha k (\lambda -1)\right)$$ $$B=\sqrt{2 \pi } \alpha ^{\alpha +\frac{1}{2}} \left(288 \alpha ^2+312 \alpha -24 (12 \alpha +1) k+25\right) e^{k \lambda }$$

Trying for $k=2$, $\lambda=\frac 12$ and $\alpha=123$ the above gives $\color{red}{1.7419052}90\times 10^{169}$

Claude Leibovici
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