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Consider:$$\int {t^n e^{t}}\ \mathrm dt$$ is there any closed formula for this? W|A gave me this but I don't know what is Gamma function: $$\int {t^n e^t\ \mathrm dt} = (-t)^{-n}\ t^n\ \Gamma(n+1, -t)+ \text{constant}$$

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  • $\begingroup$ When I type integrate t^n Exp[t] it returns an answer along with the text "$\Gamma(a,x)$ is the incomplete gamma function"; if you hover over this text it will provide you with links to Documentation, Properties, and Definition. Here special functions are unavoidable. $\endgroup$
    – anon
    Commented Aug 20, 2014 at 11:42
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    $\begingroup$ Without knowledge of the incomplete Gamma function, you can integrate by parts $n$ times, with which you can derive a formula by induction. In the case of the integral from $0$ to $\infty$ (the complete Gamma function), this is used to show that $\Gamma(n+1)=n!$. $\endgroup$
    – Ian
    Commented Aug 20, 2014 at 11:43
  • $\begingroup$ apply partial integration n times to get rid of the polynomial and exponential combination. good luck. $\endgroup$
    – user170727
    Commented Aug 20, 2014 at 11:47
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    $\begingroup$ Your problems may originate from the fact that W|A doesn't know that $n$ is a positive integer. Try this with $n=2$, $n=3$ et cetera. $\endgroup$ Commented Aug 20, 2014 at 11:54

2 Answers 2

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Hint: Assuming $n$ is a positive integer, there is a closed formula including a sum $$\int t^n e^t \mathrm{d}t = n!e^t\sum_{k=0}^n (-1)^{n-k}\frac{t^k}{k!} + C $$ I got it by using the formulas (1) and (2) for the incomplete Gamma function. Once you have it, it can be verified by taking the derivative or you can try proving it with mathematical induction.

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It is easily proved by induction and integration by parts that $$ \int {t^n e^{t}}\,dt = p_n(t) e^t $$ where $p_n$ is a polynomial of degree $n$ and $$ p_{n+1}(t) = t^{n+1}-(n+1)p_n(t) $$ This will give you the formula mentioned by gammatester.

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