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.

Let $c>0$ and $r >0$ and consider the integral for $n \geq 2$

$$ I_{r,n} (c) =\int_{1}^{\infty} \frac{\exp\left( - (y-c)^n\right)}{y^r} \mathrm{d} y$$

How do I show $I_{r,n} (c) < \infty$?

I am not sure if this is even true without making additional assumptions on $r,n$ or $c$

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

For $y\ge 1$, $r>0$ $$0\le { \exp( -(y-c)^n)\over y^r}\le \exp( -(y-c)^n).$$ And for $y>c+1$ $$ \exp( -(y-c)^n)\le \exp(-(y-c)).$$ Since $\int_{c+1}^\infty \exp(-(y-c))\,dy$ is convergent, the original integral can be shown to be converge by using a comparison test.

As Robert Israel points out though, the integral converges for $n>0$ regardless of the values of $r$ and $c$. This would take a bit more work...

share|improve this answer
add comment

It's true for any $n > 0$ without restrictions on $r$ and $c$. The point is that the exponential goes to $0$ as $y \to \infty$ faster than any power of $y$.

share|improve this answer
add comment

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.