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?$