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

Question

1465 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-01T13:00:24

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)#.