How do you test the improper integral #int x^(-1/3) dx# from #(-oo, oo)# and evaluate if possible?

Question

Ezra Adams · Accepted Answer

It does not converge.

For any #n >= -1#, #x^n >= 1/x# and #int_0^oo 1/x dx# diverges.

Adam Adkins · Answer

undefined To test the improper integral $ \int_{-\infty}^{\infty} x^{-1/3} \, dx $ for convergence or divergence, we use the limit comparison test. We compare it to the integral $ \int_{1}^{\infty} x^{-1/3} \, dx $ since the function is not defined at $ x = 0 $.

To evaluate the integral $ \int_{1}^{\infty} x^{-1/3} \, dx $, we find its antiderivative and then evaluate it at the limits of integration. The antiderivative of $ x^{-1/3} $ is $ 3x^{2/3} $.

Now, we evaluate the antiderivative at the limits of integration:

$$ \lim_{a \to \infty} 3a^{2/3} - \lim_{b \to 1} 3b^{2/3} $$

As $ a $ approaches infinity, $ 3a^{2/3} $ approaches infinity, and as $ b $ approaches 1, $ 3b^{2/3} $ approaches 3.

Therefore, the integral $ \int_{1}^{\infty} x^{-1/3} \, dx $ diverges.

Since $ \int_{-\infty}^{\infty} x^{-1/3} \, dx $ can't converge if the part from $ 1 $ to $ \infty $ diverges, the entire integral $ \int_{-\infty}^{\infty} x^{-1/3} \, dx $ also diverges.