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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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    $\begingroup$ The identity $\log(t/n)=\log(t)-\log(n)$ seems to settle the issue. $\endgroup$ – Did Jan 14 '12 at 23:33
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You have $$ \begin{eqnarray} \int_{0}^{\infty}\frac{\ln(t)e^{-nt}}{\sqrt{t}} dt &=& \frac{1}{\sqrt{n}}\int_{0}^{\infty}\frac{\ln(t/n)e^{-t}}{\sqrt{t}}dt \\ &=& \frac{1}{\sqrt{n}}\int_{0}^{\infty}\frac{\ln(t)e^{-t}}{\sqrt{t}}dt - \frac{\ln(n)}{\sqrt{n}}\int_{0}^{\infty}\frac{e^{-t}}{\sqrt{t}}dt \\ &=&\frac{-\sqrt{\pi}(\gamma + \ln(4))-\ln(n)\sqrt{\pi}}{\sqrt{n}} \\ &=& -\sqrt{\frac{\pi}{n}}\left(\gamma+\ln(4n)\right), \end{eqnarray} $$ where $\gamma=0.5772...$ is the Euler (or Euler-Mascheroni) constant.

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  • $\begingroup$ I know that $ \int_{0}^{\infty} \ln(t)e^{-t} \mathrm dt=-\gamma $ but I don't understand why $ \int_{0}^{\infty} \frac{\ln(t)e^{-t}}{\sqrt{t}} \mathrm dt= 2 \int_{0}^{\infty} \ln(t)e^{-t^2} \mathrm dt=-\sqrt{\pi}(\gamma+\ln(4)) $. $\endgroup$ – Chon Jan 15 '12 at 9:06

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