Question

at http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegralsCompTest.aspx in the second example it says that int (upper limit inf & lower limit 3) 1/x dx diverges by the fact that the power of the denominator is > 1 but then immediately after this statement in this example it says "We’ve got a larger function that is divergent" which contradicts that it diverges, so is it like a printing mistake (typo) or am I wrong to understand it?

Answer 1

anyone please help me :(

Answer 2

The statement says *probably* will converge. 1/x does not converge (but it's close)

Answer 3

in the scond example?

Answer 4

so he's looking for a function that is larger than 1/x^p that does converge.

Answer 5

diverges by the fact that
\[p\leq 1\]

Answer 6

Yes
if the integrand goes to zero faster than 1/x then the integral will probably converge

Answer 7

1/x is the harmonic series, and it diverges very very slowly

Answer 8

so 1/x+e^x<1/x<1/e^x

Answer 9

so according to this 1/e^x is a greater function that others

Answer 10

sorry i am lost

Answer 11

1/e^x converges so it is < 1/x

Answer 12

okkkkkkk

Answer 13

and 1/(x+e^x) < 1/e^x so it also converges.

Answer 14

but then 1/x also converges so by the comparison test it is fine then it says in the example that "We’ve got a larger function that is divergent" (just copy and find ctrl+f on that website

Answer 15

no, 1/x diverges

Answer 16

how?

Answer 17

okkkkkkkkkkkkkk
got it if p<= 1 then diverges

Answer 18

got it

Answer 19

thanks a tonne bro

Answer 20

It's not really too complicated.

Answer 21

just confusing.

Answer 22

i was not reading it properly i was reading it as converging :}

Answer 23

1/x was famous back in the day. All kinds of arguments about if it converged or not.