How do you show whether the improper integral #int (79 x^2/(9 + x^6)) dx# converges or diverges from negative infinity to infinity?

Question

Charles Anctil · Accepted Answer

undefined To determine the convergence or divergence of the improper integral $ \int_{-\infty}^{\infty} \frac{79x^2}{9 + x^6} \, dx $, we'll use the comparison test.

1. Consider the behavior of $ \frac{79x^2}{9 + x^6} $ as $ x $ approaches $ \pm \infty $.
   
   As $ x $ approaches $ \pm \infty $, the term $ x^6 $ dominates, and the function behaves like $ \frac{79x^2}{x^6} = \frac{79}{x^4} $.

2. Since the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, the function approaches zero as $ x $ approaches $ \pm \infty $.

3. We can compare $ \frac{79x^2}{9 + x^6} $ with $ \frac{79}{x^4} $ for large $ x $ values.

As $ x $ approaches $ \pm \infty $:
   $$ \frac{79x^2}{9 + x^6} < \frac{79}{x^4} $$

4. The integral $ \int_{-\infty}^{\infty} \frac{79}{x^4} \, dx $ can be evaluated as a p-series with $ p = 4 $, and it converges since $ p > 1 $.

5. By the Comparison Test, since $ \frac{79x^2}{9 + x^6} < \frac{79}{x^4} $ and $ \int_{-\infty}^{\infty} \frac{79}{x^4} \, dx $ converges, $ \int_{-\infty}^{\infty} \frac{79x^2}{9 + x^6} \, dx $ converges as well.

Emma Aldridge · Answer

I would integrate by trigonometric substitution, then check that the limit exists.

We can take out a constant factor, so #int_-oo^oo (79x^2/(9 + x^6)) dx# converges if and only if #int_-oo^oo (x^2/(9 + x^6)) dx# converges. #int (x^2/(9 + x^6)) dx = 1/9tan^-1(x^3/3)# As #xrarroo#, we have #tan^-1(x^3/3) rarr pi/2# (and as #xrarr-oo#, we have #tan^-1(x^3/3) rarr -pi/2#) so both #int_-oo^0 (x^2/(9 + x^6)) dx# and #int_0^oo (x^2/(9 + x^6)) dx# converge. Therefore, #int_-oo^oo (x^2/(9 + x^6)) dx# converges.