# Find all $p\in \mathbb{R}$ such that improper integral $\int\limits_1^\infty e^{-x}x^p$ converges. [closed]

$p\in \mathbb{R}$ such that $\int\limits_1^\infty e^{-x}x^p$ converges. I only have that $\frac{1}{e^x x^p}\leq \frac{1}{x^2}$ which converges.

## closed as off-topic by Did, Leucippus, Daniel W. Farlow, Henrik, ShaileshAug 24 '16 at 0:09

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• Hint: the $e^{-x}$ term decreases so quickly that it dominates $x^p$ for any $p$, so it should converge for any $p$. – florence Aug 23 '16 at 11:29

$Hint:$ Notice that you can use induction to prove convergence for all $p \in \mathbb{N}$ (try integration by parts). Then, using basic inequalities this can be extended for any $p \in \mathbb{R}$.

Hope this helps.

If $p\le0$, then for $x\ge 1$ we have $x^p\le 1$ and hence $$0<\int_1^{\infty}e^{-x}x^pdx\le \int_1^{\infty}e^{-x} dx = \frac{1}{e}.$$

Let now $p> 0$, then intergration by parts yields $$\int_1^{\infty}e^{-x}x^pdx = (-e^{-x}x^p)\big|_{x=1}^{x=+\infty}+p\int_1^{\infty}e^{-x}x^{p-1}dx = 1/e+p\int_1^{\infty}e^{-x}x^{p-1}dx.$$ Essentially, we managed to split the integral in two: a constant part and apart with $p$ replaced by $p-1$. By iterating this step $k$ times we will arrive to the form $$C+p(p-1)\cdot\ldots\cdot(p-k-1)\int_1^{\infty}e^{-x}x^{p-k}dx$$ where $C$ is some finite constant.

If $k=\lceil p\rceil$ (smalles integer superior to $p$), then the latter intergral converges.

Therefore, the final answr is that the integral is finite for all $p$.

Use asymptotic analysis: $\;\mathrm e^{-x}=o(x^\alpha),\quad x\to\infty\;$ for all $\alpha \in \mathbf R$. In particular, $\;\mathrm e^{-x}=o\biggl(\dfrac1{x^{p+2}}\biggr)$, so $$\mathrm e^{-x}x^p=o\biggl(\dfrac1{x^{p+2}}\biggr)x^p=o\biggl(\dfrac1{x^2}\biggr),$$ so it converges.

We can evaluate the integral using the upper incomplete Gamma function $$\Gamma(a,x) = \int\limits_{x}^{\infty} \mathrm{e}^{-z} z^{a-1} \mathrm{d} z$$ Thus $$\int\limits_{1}^{\infty} \mathrm{e}^{-x} x^{p} \mathrm{d} x = \Gamma(1+p,1)$$ As can be seen here, the upper incomplete Gamma function, $\Gamma(a,x)$, is an entire function for all $a$ when $x \ne 0$. Thus the integral converges for all values of $p$.