$p\in \mathbb{R}$ 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, Shailesh Aug 24 '16 at 0:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Leucippus, Daniel W. Farlow, Henrik, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 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 \begin{equation} \Gamma(a,x) = \int\limits_{x}^{\infty} \mathrm{e}^{-z} z^{a-1} \mathrm{d} z \end{equation} Thus \begin{equation} \int\limits_{1}^{\infty} \mathrm{e}^{-x} x^{p} \mathrm{d} x = \Gamma(1+p,1) \end{equation} 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 $p$.

Not the answer you're looking for? Browse other questions tagged or ask your own question.