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In my notes it was said $\begin{eqnarray*} n!\int_x^\infty \frac{e^{-y}}{y^{n+1}} \, dy &<& \frac{n!}{x^{n+1}}\int_x^\infty e^{-y} \, dy \\ &=& \frac{n!e^{-x}}{x^{n+1}}\end{eqnarray*}$

How did they get from the first line to the second line? Can I just pull out the $y^{n+1}$ term and change it to $x^{n+1}$?

Also is it the case that $n!\int_x^\infty \frac{e^{-y}}{y^{n+1}}dy

?

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    Thanks for your help, understood it better!2012-05-06

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For your first question:

They are using the fact that if $f(x)< g(x)$ on $(a,\infty)$, then $\int_a^\infty f(x)\,dx<\int_a^\infty g(x)\, dx$. This is a standard comparison test for improper integrals.

Here, we have $y^{n+1}>x^{n+1}$ for $y$ in the interval $(x,\infty)$ (note, then, that $y> x$); so for $y$ in the interval $(x,\infty)$, we have ${e^{-y}\over y^{n+1}} <{e^{-y}\over x^{n+1}}$. Thus $\int_x^\infty {e^{-y}\over y^{n+1}} \,dy<\int_x^\infty {e^{-y}\over x^{n+1}}\, dy$.

Finally, since the integration is with respect to $y$, the term $1\over x^{n+1}$ is a constant as far as the integration is concerned and can be factored out of the integral sign.


Though it would lead to the correct result, you shouldn't think of pulling $y^{n+1}$ out first, since you are integrating with respect to $y$. You can change it to $x^{n+1}$ first, introducing an inequality, and then pull it out.

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    @Jonathan It's certainly true for $x\ge 1$, since then for y>x, you'd have {e^{-y}\over y^{n+1}}. For x<1, it's not necessarily true, I think. Here is a [Worlfram computation](http://www.wolframalpha.com/input/?i=%28integrate+e^%28-y%29%2F+y%2C+y%3D.1..infty+%29-%28integrate+e^%28-y%29+%2C+y%3D.1..infty+%29) with $n=1$, $x=.1$.2012-05-06