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Let $\alpha>0$. Then please prove that: $$\lim_{x\rightarrow+\infty}\left(\int_0^{+\infty}t^{-\alpha t+x}dt\right)\left[\sqrt{\frac{2\pi}{e^\alpha}} x^{1/2\alpha} \exp\left(\frac{\alpha}{e}x^{\frac{1}{\alpha}}\right)\right]^{-1}=1,$$ where $\exp(x):=e^x$.

(This question is asked by an undergraduate student in grade 1 and it is an exercise in a Chinese book. This question passed through a lot of people and then to me, so I do not know the name of that book.)

DATO
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    Very challenging question indeed. Can you edit your question to include where you found this equation? I think I understand the $\sim$ sign. Please mention if you've done or seen similar problems. Consider using Approach0 to search for similar problems and see if you can adapt these techniques for your problem. – Sarvesh Ravichandran Iyer Jan 01 '22 at 09:46
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    If we write $f(t)=(-\alpha t+x)\log t$, then the Laplace's method suggests that the integral is asymptotically $$\sqrt{\frac{2\pi}{|f''(t_0)|}} e^{f(t_0)} \sim \sqrt{\frac{2\pi}{x}} t_0 e^{\alpha t_0(\log t_0)^2}, $$ where $t_0=\frac{x}{\alpha W(ex/\alpha)}$ is the unique critical point of $f$ on $(0, \infty)$ and $W$ is the Lambert W-function. But numerical simulations sugest that your asymptotic formula is likely not true. – Sangchul Lee Jan 01 '22 at 12:33

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