Deciding whether this integral is convergent or divergent: $\int_1^\infty \frac{e^{\sqrt {x}}}{\sqrt{x}}\,\mathrm dx$

Question

Decide if the following integral is convergent or divergent. If it is convergent evaluate the integral. $$\int_1^{\infty} \frac{e^{-\sqrt {x}}}{\sqrt{x}}\,\mathrm dx$$

I evaluated the integral, but I could not decide via comparison test whether the integral is convergent or divergent.

Answer 1

$$e^{\sqrt{x}}=\sum_{n=0}^{\infty}\frac{x^{0.5n}}{n!}$$ $$\implies e^{\sqrt{x}}>x$$ $$\therefore \sqrt{x}e^{\sqrt{x}}>x^{1.5}$$ $$\frac{1}{\sqrt{x}e^{\sqrt{x}}}<x^{-1.5}$$ Since $\int_1^{\infty} x^{-1.5} dx$ is convergent then the $\int_1^{\infty} \frac{1}{\sqrt{x}e^{\sqrt{x}}}$ is convergent.

Answer 2

Hint. Try the transformation: $y=\sqrt{x}$.

Answer 3

Let $u=\sqrt{x}$ Then, $2du=\frac{dx}{\sqrt{x}}$. $\sqrt{1} = 1$, and $\sqrt{\infty}$ should be $\infty$ when we take limits. So, the integral becomes $2\int_1^{\infty} e^{-u} du = -2e^{-u}|_1^{\infty} = \frac{2}{e}$, again, when we use limits. Thus, the integral converges as our value is finite.