P-test for Improper integral

Question

I am trying to see if this integral converges or not, so we have $$\int_{1}^{\infty}\frac{1}{x^{1+\frac{1}{x}}}$$ Using P-Test I know if $1+\frac{1}{x}>1$, then this integral converges, but could the $x$ be negative so $1+\frac{1}{x}<1$?? then the integral diverges? Thanks

You have defined your integral over the region $x>1$ so no it cannot be negative — Henry Lee, Dec 10 '20 at 18:44

score 4 · Answer 1 · answered Dec 10 '20 at 18:40

4

The integral has form $\int_{1}^{\infty} f(x) dx$, where $\frac{1}{f(x)} = x \cdot x^{\frac{1}{x}} = x \cdot e^{\frac{\ln x}{x}} = x \cdot e^{\bar{o}(1)} \sim x$. Hence $f(x) \sim \frac1{x}$. It's well known that $\int_{1}^{\infty} \frac1{x} dx$ diverges. So the intergal $\int_{1}^{\infty} f(x) dx$ diverges also.

answered Dec 10 '20 at 18:40

Botnakov N.

5,660

In fact, for $x\ge 1$ $$\frac1{x^{1+1/x}}\ge \frac1{e^{1/e}x}$$ – Mark Viola Dec 10 '20 at 19:04

P-test for Improper integral

1 Answers1