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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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    The identity $\log(t/n)=\log(t)-\log(n)$ seems to settle the issue.2012-01-14

1 Answers 1