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I'm trying to determine for which values of $q\in\mathbb{R}$ this improper integral converges and diverges. I have discrepancy in my solutions, so there appears to be something wrong, or some subtle nuance I'm unaware of. $$\int_{1}^\infty x^q e^{x^{q+1}} dx$$ Let $u=x^{q+1}$, so $\frac{du}{dx}=(q+1)e^q$. Then,

$$\frac{1}{q+1}\int_{1}^\infty e^u du=\frac{1}{q+1} \lim_{L\rightarrow \infty}\int_{1}^L e^u du= \frac{1}{q+1}\lim_{L\rightarrow \infty} [e^u]_{1}^L$$

Now if I sub $L$ and $1$ in, I get divergence to $\infty$ if $q>-1$ and divergence to $-\infty$ if $q<-1$. However, if I sub $x^{q+1}$ back in for $u$, I get,

$$\frac{1}{q+1}\lim_{L\rightarrow \infty} [e^{x^{q+1}}]_{1}^L = \frac{1}{q+1} (\lim_{L\rightarrow \infty} (e^{L^{q+1}}) - e)$$

Hence, the integral diverges to $\infty$ for $q\geq-1$ and converges to $\frac{1-e}{q+1}$ if $q<-1$.

Why do the solutions change if I sub $x^{q+1}$ back in? Which solution set is correct and why?

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    $\begingroup$ Are you sure that $u\to+\infty$ (your upper integration bound) as $x\to+\infty$ when $q<-1$? $\endgroup$ – A.Γ. May 20 '18 at 18:14
  • $\begingroup$ @A.Γ. Thanks, that answers my question :) $\endgroup$ – HumptyDumpty May 20 '18 at 18:17
  • $\begingroup$ @A.Γ. Could you please convert your comment into an answer so this question can be removed from the "Unanswered" queue? Seems like it was exactly the kind of advice OP was looking for. $\endgroup$ – Robert Howard Dec 1 '18 at 2:07

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