Question

Convergent or divergent

Integral from 0 to infinity arctan(x)/sqrt(x) dx

Answer 1

i want to use the comparison test, but i dont know which fraction i can use

Answer 2

limit comparison ?

Answer 3

this looks sorta like a integral test to me

Answer 4

well if you want to see if the sum converges or diverges i would do comparison to 1/sqrt(x) right as arctan(x) stays within y=pi/2 and y=-pi/2 right? and as the sum 1/sqrt(x) diverges by the p-test you can say that it diverges too. Something like that. Would have to actually work it out to know if what Im saying is correct, but yeah

Answer 5

he! i cant use this... because arctanx/sqrt(x) is < 1/sqrt(x)

Answer 6

it needs to be > because its divergent.. :s

Answer 7

ok so dont use the comparison test, but try to use the limit comparison test (i.e. take the limit to inf of (arctanx/sqrtx)/(1/sqrt(x))). This gives you lim_x^inf of arctanx which is pi/2 right? so they behave similarly. Thus as sum of 1/sqrtx goes to inf so does arctanx/sqrtx voila: divergence

Answer 8

now if this isnt the right way to do it I have no idea how to help :p

Answer 9

also please just check out my question as you said you understood the Lagrangian and give any tips if you have them Much appreciated :)