1
$\begingroup$

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 $c$

2 Answers 2

1

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...

2

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 $y$.