How can you be sure that an integral does not exist, if it has no indefinite integral?

Question

Say you have the integral $\displaystyle\int_1^\infty{\frac{1}{x^{1+\frac{1}{x}}}}\;\mathrm{d}x$

This integral cannot be completed. Not that it goes to infinity, but it physically just cannot be completed. How can you realize this if you encounter it? How can you prove it?

Answer 1

@KaviRamaMurthy and @player2326 have answered this question. The comparison test can be employed to solve this question.

Additional references:

How to see this improper integral diverges?

Checking whther the integral $\int_1^∞ \frac{1}{x^{\frac{1}{x}+1}} dx$ convergent

Answer 2

$$I=\int_{1}^{\infty}\frac{dx}{x^p}=\left .\frac{x^{1-p}}{1-p}\right|_{1}^{\infty}=\frac{1}{p-1}, ~if~ p>1,$$ because $0^{1-p}=0$ if only $p>1$, otherwise it is infinite. Hence the integral converges when $p>1$.