Why does this integral diverge?

Question

I am having a hard time solving this one, and it seems so simple. \begin{align*} \int_{1}^{\infty} \dfrac{1}{x^{1+\frac{1}{x}}} dx \end{align*}

I understand that it diverges, because I've tested it. As $x$ approaches infinity, the expression approaches $$\int_{1}^{\infty} \dfrac{1}{x} dx$$

Which we know diverges.

But how can I prove/show that?

Answer 1

Since$$\lim_{x\to\infty}\frac{\frac1{x^{1+1/x}}}{\frac1x}=\lim_{x\to\infty}\frac1{x^{1/x}}=1$$and since$$\int_1^\infty\frac1x\,\mathrm dx$$diverges, your integral diverges too.

Answer 2

Apply the substitution $u = 1/x$, then your integral becomes $\displaystyle \int\limits_{0}^{1} \dfrac{u^u}{u}\, du$. We compute that $\displaystyle \lim\limits_{u \rightarrow 0^+} u^u = 1$.

Let $\epsilon >0$ be small. Then there exists $\delta >0$ such that $u < \delta$ implies $1 - u^u < \epsilon$.

Now $\displaystyle \int\limits_{0}^{\delta} \dfrac{u^u}{u}\, du \geq \int\limits_{0}^{\delta} \dfrac{1-\epsilon}{u}\, du$. But the smaller integral diverges.