Find Laplace transform for this function "$g$" $$g_n(t)=\begin{cases}\frac{(1-e^{-t})^n}{t^n}&:t>0,\\0&:t\le0.\end{cases}$$ Then Take advantage of it to calculate the following integration "$I$" $$I_n=\int_0^\infty\frac{(1-e^{-t})^n}{t^n}\,dt,n\geq2$$
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Show us your steps so that we can help you find the answer. – imranfat Feb 04 '14 at 16:53
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Perhaps $\hat{g}{n+1}(s) = \int_0^s \hat{g}{n}(z)dz$? – copper.hat Feb 04 '14 at 17:39
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Related. – metamorphy Dec 01 '20 at 18:42
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I guess one way to do this is to differentiate the integral $n$ times, which means that you have to integrate the result $n$ times back.
$$G_n(s) = \int_0^{\infty} dt \left ( \frac{1-e^{-t}}{t}\right )^n e^{-s t}$$
$$\begin{align}(-1)^n \frac{d^n}{ds^n} G_n(s) &= \int_0^{\infty} dt \left ( 1-e^{-t} \right )^n e^{-s t}\\ &= \sum_{k=0}^n (-1)^k \binom{n}{k}\int_0^{\infty} dt \, e^{-(s+k) t}\\ &=\sum_{k=0}^n \frac{(-1)^k}{s+k} \binom{n}{k}\end{align}$$
For example,
$$\frac{d^2}{ds^2}G_2(s) = \frac1{s} - \frac{2}{s+1} + \frac1{s+2}$$ $$\frac{d}{ds} G_2(s) = \log{s} - 2 \log{(s+1)} + \log{(s+2)} + A$$ $$\begin{align}G_2(s) &= s \log{s}-s - 2 (s+1) \log{(s+1)} + 2 (s+1) + (s+2) \log{(s+2)} - (s+2) + A s + B\\ &= s \log{s}- 2 (s+1) \log{(s+1)}+(s+2) \log{(s+2)}+A s+B\end{align}$$
We get $A$ and $B$ by considering the limit as $s\to\infty$; in this limit, both of these constants drop out. Thus,
$$G_2(s) = s \log{s}- 2 (s+1) \log{(s+1)}+(s+2) \log{(s+2)}$$
The integral $I_2 = G_2(0) = 2 \log{2}$.
Higher values of $n$ may be done similarly.
Ron Gordon
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The integral (before the $\large k$-sum) is $\displaystyle{\large{\rm B}\left(s,n + 1\right) = n!,{\Gamma\left(s\right) \over \Gamma\left(s + n + 1\right)}}$. – Felix Marin Feb 04 '14 at 19:13
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@FelixMarin: I know, but I did not want to end up integrating that $n$ times. – Ron Gordon Feb 04 '14 at 20:01
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@RonGordon How I can get a simplified formula of Gn(s) not the derivatives of Gn(s). – omran Feb 06 '14 at 01:30