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I would like to know the asymptotics of the following sequences of integrals: $$ I_n = \displaystyle { \int _0 ^{+ \infty} \dfrac{t^n}{(t + i)^{n + 1}} \cdot e^{-t} \ dt } $$

Using a deformation of path with Cauchy theorem, I've shown that $$ I_n = \displaystyle { \int _{i \mathbb{R}^+} \dfrac{t^n}{(t + i)^{n + 1}} \cdot e^{-t} \ dt = \int _0 ^{+ \infty} \dfrac{t^n}{(t + 1)^{n + 1}} \cdot e^{-it} \ dt } $$

I therefore have tried using Laplace method, but I have been unable to conclude anything...

Moreover, I didn't manage to use computer software to have some reliable numeric values, because of the oscilating function $t \mapsto e^{-it}$. So, I haven't got any conjecture...

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The saddle point $t^\ast$ is obtained by solving $f'(t)=0$ for $f(t)=-t+n\ln t-(n+1)\ln(t+i)$, for large $n$ we find $f(t^\ast)=-2\sqrt{in}$, so we arrive at the approximation for the integral $I_n\approx\exp(-2\sqrt{in})$.

The plot compares exact (gold) and approximate (blue) absolute values for $n$ up to 200 (multiply horizontal axis labels by 10).

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  • $\begingroup$ Nice cart, to confirm theory! I find two possibilities to the saddle point $t^\star$: $\dfrac{-1 - i \pm \sqrt{2i (2n + 1)}}{2}$. Which one should we consider? $\endgroup$
    – MathTolliob
    Commented Jun 15, 2021 at 18:35
  • $\begingroup$ Do you have a link on with theory and example on Laplace method? I thought it applies on integral $\displaystyle { \int _a ^b g(t) e^{-\lambda f(t)} \ dt } $, with a function $f$ non-depending on the variable $\lambda$? $\endgroup$
    – MathTolliob
    Commented Jun 15, 2021 at 18:35
  • $\begingroup$ It looks to me that the picture shows divergence of the blue and gold plots. $\endgroup$
    – Brendan McKay
    Commented Jun 16, 2021 at 4:09
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    $\begingroup$ Using the saddle node $\dfrac{-1 -i + \sqrt{2i(2n + 1)}}{2}$, I conjecture that $ I_n \underset{n \longrightarrow + \infty}{\sim} \dfrac{\sqrt{\pi}}{n^{\frac{1}{4}}} e^{i \big ( \frac{1}{2} - \frac{\pi}{8} \big )} e^{-2 \sqrt{i n}} $. This has been confirmed by numerical result for $n = 2^k$ for $k \in [\![0;14]\!]$ $\endgroup$
    – MathTolliob
    Commented Jun 16, 2021 at 12:14

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