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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 $c$

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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...

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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 $y$.

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