Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

?

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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