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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 $$n!\int_x^\infty \frac{e^{-y}}{y^{n+1}}dy <n!\int_x^\infty e^{-y} dy$$


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For a fixed constant x, we have y>x so 1/y < 1/x. Raise both sides to (n+1)th power, multiply by $e^{-y}$ and integrate with respect to y. The inequality follows. – Shahab May 6 '12 at 15:57
You can change $y$ to $x$ if you put in the correct inequality. Then note that $x$ is a constant, so can be taken outside the integral sign. This is the opposite way round from your question - you can't take $y$ outside the integral sign. – Mark Bennet May 6 '12 at 16:01
Thanks for your help, understood it better! – Jonathan May 6 '12 at 16:45
up vote 1 down vote accepted

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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Thanks @DavidMitra , may I ask is my 2nd inequality valid? (2nd part of the question). – Jonathan May 6 '12 at 16:15
@Jonathan It's certainly true for $x\ge 1$, since then for $y>x$, you'd have ${e^{-y}\over y^{n+1}}<e^{-y}$. For $x<1$, it's not necessarily true, I think. Here is a Worlfram computation with $n=1$, $x=.1$. – David Mitra May 6 '12 at 16:26
Thanks again for your help, most appreciated! – Jonathan May 6 '12 at 16:52

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