How do you evaluate #int_1^oo 1/x^x dx# ?

Question

Evelyn Agard · Accepted Answer

By numerical Integration:

#int_1^oo 1/x^x dx ~= 0.70416996# This definite integral requires numerical methods to solve. See article: https://tutor.hix.ai for an introductory explanation if you are not familiar with this. #int_1^oo 1/x^x dx ~= 0.70416996#

Aurora Alvarado · Answer

undefined To evaluate the integral $\int_{1}^{\infty} \frac{1}{x^x} \, dx$, we can use a method called comparison with a known convergent integral. One such integral is the convergent p-series $\int_{1}^{\infty} \frac{1}{x^p} \, dx$, where $p > 1$. Let's compare the given integral to the integral of $1/x^2$. Since $x^x > x^2$ for $x > 1$, we have $\frac{1}{x^x} < \frac{1}{x^2}$. The integral of $\frac{1}{x^2}$ from 1 to infinity is a known convergent integral, given by: $$ \int_{1}^{\infty} \frac{1}{x^2} \, dx = \lim_{t \to \infty} \int_{1}^{t} \frac{1}{x^2} \, dx = \lim_{t \to \infty} \left[-\frac{1}{x}\right]_{1}^{t} = \lim_{t \to \infty} \left(-\frac{1}{t} + 1\right) = 1 $$ Since $\frac{1}{x^x} < \frac{1}{x^2}$ for $x > 1$ and the integral of $\frac{1}{x^2}$ from 1 to infinity converges (equals 1), by the comparison test for integrals, we conclude that the integral $\int_{1}^{\infty} \frac{1}{x^x} \, dx$ also converges.

Charles Arocha · Answer

undefined To evaluate the integral $ \int_1^\infty \frac{1}{x^x} \, dx $, you can use a technique called the comparison test.

First, note that for $ x \geq 1 $, $ x^x \geq x $. So, $ \frac{1}{x^x} \leq \frac{1}{x} $.

Since $ \frac{1}{x} $ is integrable over the interval $ [1, \infty) $, and $ \frac{1}{x^x} $ is less than or equal to $ \frac{1}{x} $ over that interval, we can conclude that $ \frac{1}{x^x} $ is also integrable over the interval $ [1, \infty) $.

Therefore, $ \int_1^\infty \frac{1}{x^x} \, dx $ converges.