How do you determine if the improper integral converges or diverges #int (1 / (u^2 + 3))du # from 0 to infinity?

Question

How do you determine if the improper integral converges or diverges #int  (1 / (u^2 + 3))du # from 0 to infinity?

Ezekiel Alexander · Accepted Answer

The Improper Integral I converges to #pi/(2sqrt3)#.

Let #I=int_0^oo(du)/(u^2+3)#. #:. I=lim_(xrarroo)int_0^x(du)/(u^2+3)#, #=lim_(xrarroo)[1/sqrt3arctan(u/sqrt3)]_0^x#, #=1/sqrt3*lim_(xrarroo)[arctan(x/sqrt3)-arctan0]...........(1)#, #=1/sqrt3{arctan(lim_(xrarroo)x/sqrt3)}.............(2)#, #=1/sqrt3*pi/2#, #=pi/(2sqrt3)#. We note that, #(2)# follows from #(1)#, because, #arctan# is continuous function. Thus, the Improper Integral I converges to #pi/(2sqrt3)#.

Emilia Aguilar · Answer

undefined To determine if the improper integral $\int_{0}^{\infty} \frac{1}{{u^2 + 3}} \, du$ converges or diverges, you can use the limit comparison test. Compare it with a known integral that converges or diverges. In this case, consider comparing it with $\int_{0}^{\infty} \frac{1}{u^2} \, du$. Calculate the limit as $u$ approaches infinity of the ratio of the two integrands. If the limit is finite and non-zero, both integrals converge or diverge together. If it's zero or infinite, they converge or diverge differently. Therefore, calculate:

$$
\lim_{{u \to \infty}} \frac{{\frac{1}{{u^2 + 3}}}}{{\frac{1}{{u^2}}}}
$$

If this limit is finite and non-zero, both integrals behave similarly. If it's zero or infinite, their behaviors differ.