How do you test the improper integral #int x^(-1/2) dx# from #[1,oo)# and evaluate if possible?

Question

Paisley Johnson · Accepted Answer

#int_1^oo x^(-1/2)dx = oo#

Based on the integral test the convergence of the integral:

#int_1^oo x^(-1/2)dx#

is equivalent to the convergence of the series:

#sum_(n=1)^oo 1/n^(1/2)#

which we can immediately see is divergent based on the #p#-series criteria.

In fact, the integrand function is defined and continuous in #[1,oo)#, so:

#int_1^oo x^(-1/2)dx = lim_(t->oo) int_1^t x^(-1/2)dx#

#int_1^oo x^(-1/2)dx = lim_(t->oo) [x^(1/2)/(1/2)]_1^t#

#int_1^oo x^(-1/2)dx = lim_(t->oo) 2sqrt(t)-2 = oo#

Adam Adkins · Answer

undefined To test the improper integral $ \int_1^\infty x^{-1/2} \, dx $, you can check for convergence or divergence using the limit comparison test.

Given the integral $ \int_1^\infty x^{-1/2} \, dx $, set up a comparison integral with a function $ g(x) $ that is easier to integrate and whose convergence or divergence is known.

One common choice is to compare with $ g(x) = x^{-1/2} $, which is also the integrand function.

Apply the limit comparison test by computing the following limit:

$$ \lim_{x \to \infty} \frac{f(x)}{g(x)} $$

If the limit is finite and positive, then both integrals either converge or diverge. If the limit is zero or infinite, then the integral $ \int_1^\infty x^{-1/2} \, dx $ behaves the same as the comparison function.

In this case, since $ f(x) = x^{-1/2} $ and $ g(x) = x^{-1/2} $, the limit simplifies to 1, which is finite and positive.

Therefore, the integral $ \int_1^\infty x^{-1/2} \, dx $ converges.

To evaluate the integral, you can integrate $ x^{-1/2} $ from 1 to $ \infty $:

$$ \int_1^\infty x^{-1/2} \, dx = \lim_{b \to \infty} \int_1^b x^{-1/2} \, dx $$

$$ = \lim_{b \to \infty} \left[2x^{1/2}\right]_1^b $$

$$ = \lim_{b \to \infty} \left(2\sqrt{b} - 2\right) $$

Since $ \lim_{b \to \infty} \sqrt{b} = \infty $, the limit becomes $ \infty - 2 $.

So, the value of the integral is $ \infty $.