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

It does not converge.

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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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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