How can I find the first term of the series expansion of
$$ \int_{0}^{\infty} \frac{\ln(t)}{\sqrt{t}}e^{-nt} \mathrm dt,n\rightarrow\infty ?$$
Or:
As $$ \int_{0}^{\infty} \frac{\ln(t)}{\sqrt{t}}e^{-nt} \mathrm dt = \frac{1}{\sqrt{n}}\int_{0}^{\infty} \frac{\ln(\frac{t}{n})}{\sqrt{t}}e^{-t} \mathrm dt $$
What is $$ \lim_{n\rightarrow\infty} \int_{0}^{\infty} \frac{\ln(\frac{t}{n})}{\sqrt{t}}e^{-t} \mathrm dt?$$
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