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compute $\int_0^1 {{x^{k - 1}}{e^{ - x}}} dx$

This what I did:

By integration by part :

$\int_0^1 {{x^{k - 1}}{e^{ - x}}} dx = \frac{{{e^{ - 1}}}}{k} + \frac{1}{k}\int_0^1 {{x^k}{e^{ - x}}} $

I'am stuck here

Some help would be appreciated

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    $\begingroup$ You probably want to go the other way (decrease the power of $x$, not increase). $\endgroup$ – angryavian Dec 9 '16 at 19:01
  • $\begingroup$ $e^x$ is its own derivative, so there's no way it should disappear upon integration by parts. So you should check / clarify what steps you did to get that expression. $\endgroup$ – Semiclassical Dec 9 '16 at 19:03
  • $\begingroup$ It's just as easy to make the upper limit of the integral a variable instead of 1. $\endgroup$ – marty cohen Dec 9 '16 at 19:05
  • $\begingroup$ related: math.stackexchange.com/q/46469/9464 $\endgroup$ – Jack Dec 9 '16 at 19:12
  • $\begingroup$ $\Gamma\left(k\right) - \Gamma\left(k,1\right)$. $\endgroup$ – Felix Marin Dec 10 '16 at 16:11
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From $$ \int_0^1 {{x^{k - 1}}{e^{ - x}}} dx = \frac{{{e^{ - 1}}}}{k} + \frac{1}{k}\int_0^1 {{x^k}{e^{ - x}}} $$ by dividing both sides by $(k-1)!$, one gets $$ \frac1{(k-1)!}\int_0^1 {{x^{k - 1}}{e^{ - x}}} dx- \frac1{k!}\int_0^1 {{x^{k }}{e^{ - x}}} dx= \frac{{{e^{ - 1}}}}{k!} $$ then summing, one gets, by telescoping terms, $$ \int_0^1 {{e^{ - x}}}dx- \frac1{n!}\int_0^1 {{x^{n }}{e^{-x}}} dx= {{e^{ - 1}}}\sum_{k=1}^n\frac1{k!} $$ or

$$ \frac1{n!}\int_0^1 {{x^{n }}{e^{-x}}} dx=1-e^{-1}\sum_{k=0}^n\frac1{k!},\quad n \ge 0. $$

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    $\begingroup$ Fun fact: This is an illustration of one of the properties of Poisson processes. For a Poisson process with rate $1$, the left-hand side is the probability that the $(n+1)$st arrival is before time $t=1$ (Gamma distribution) and the right-hand side is the probability that there are $\ge n+1$ arrivals in the time interval $[0,1]$ (Poisson distribution). $\endgroup$ – angryavian Dec 9 '16 at 19:15
  • $\begingroup$ @angryavian Nice interpretation. $\endgroup$ – Olivier Oloa Dec 9 '16 at 19:17
  • $\begingroup$ I'm pretty sure $\sum_{k=0}^n\frac1{k!}$ can be turned into a more compact form using the Taylor series of $\exp$ and floor function. $\endgroup$ – Simply Beautiful Art Dec 9 '16 at 19:35
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    $\begingroup$ @SimpleArt Are you thinking about this: $\sum_{k = 0}^{n} \frac {(-1)^k} {k!} = \frac1{n!}\left \lfloor \frac {n!} {e} + \frac {1} {2} \right \rfloor$ ? $\endgroup$ – Olivier Oloa Dec 10 '16 at 21:47
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    $\begingroup$ @SimpleArt Probably this one: $$\sum_{k=0}^n\frac1{k!}=\frac{\lfloor n!e\rfloor}{n!}. $$ $\endgroup$ – Olivier Oloa Dec 22 '16 at 22:40
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Hint: Let $$ a_{k}=\int_{0}^{1}x^{k}e^{-x}dx\text{ for }k\geq0. $$ Then, $$ a_{0}=1-{e}^{-1}. $$ and (by integration by parts) for any integer $k\geq1$, \begin{align*} a_{k} & =-x^{k}e^{-x}\mid_{0}^{1}-\int kx^{k-1}e^{-x}dx\\ & =-e^{-1}-ka_{k-1}. \end{align*} Can you put these things together to find a "closed-form" for $a_k$?

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  • $\begingroup$ If you have trouble with my hint, try using the expression for $a_k$ above to write down an expression for $a_1$. In this expression, $a_0$ should appear on the right-hand side. Now, do the same thing for $a_2$, $a_3$, etc. The idea (induction) should become clear to you by doing this. $\endgroup$ – parsiad Dec 9 '16 at 19:15
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This can be easily done first with this integral $$I(\alpha) = \int_{0}^{1}\exp{(\alpha x)}\,dx=\frac{\exp{(\alpha)}-1}{\alpha}$$ but if we derivate $k-1$ times with respect to $\alpha$ we get yours $$\frac{d^{k-1}}{d\alpha^{k-1}}I(\alpha) = \int_{0}^{1}x^{k-1}\exp{(\alpha x)}\,dx=\frac{d^{k-1}}{d\alpha^{k-1}}\left(\frac{\exp{(\alpha)}-1}{\alpha}\right)\tag1$$

After computing $(1)$ substitute $\alpha=-1$ and you have your desired result

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  • $\begingroup$ This isn't quite right. $\int_0^1e^{ax}dx=\frac1a(e^a-1)$ $\endgroup$ – Simply Beautiful Art Dec 9 '16 at 19:36
  • $\begingroup$ Thank you @SimpleArt, I am used to the usual limits $[-\inf,x]$ $\endgroup$ – HBR Dec 9 '16 at 19:39
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$\displaystyle% \int_{0}^{1}x^{k - 1}\,\mathrm{e}^{-x}\,\mathrm{d}x = \int_{0}^{\infty}x^{k - 1}\,\mathrm{e}^{-x}\,\mathrm{d}x - \int_{1}^{\infty}x^{k - 1}\,\mathrm{e}^{-x}\,\mathrm{d}x = \Gamma\left(k\right) - \Gamma\left(k,1\right)\quad$ with $\displaystyle\quad\Re\left(k\right) > 0$.


$\displaystyle\Gamma\left(k\right)\ \mbox{and}\ \Gamma\left(k,1\right)$ are the Gamma Function and the Incomplete Gamma Function, respectively.

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