Improper Integration?

Question

I'm having a little bit of trouble doing this question I wonder if anyone could help me.

The integration between e to infinity of (lnx / x) dx . I know its divergent but I can't figure out how to integrate it. Thanks!!!

Answer 1

substitute lnx=t
dx/x=dt
so integrating tdt it is t^2/2

=>(lnx)^2/2
apply the limits

Answer 2

I ln(x)/x dx .We put ln(x)=z so 1/x dx =dz Sose integral becomes Iz*dz = z^2/2 ) = (ln(x))^2/2

At e the value is 1/2 .As x tends to infinity the function tends to infinity .So the integral is divergent

I means integral

Answer 3

integration between e to infinity of (lnx / x) dx
= integration between e to infinity of lnx dlnx
= ln(lnx) between e to infinity
= ln(ln infinity)
= infinity

Answer 4

if you put u=lnx then du 1/x dx

and you obtain int of udu = u^2/2

so the primitive of your fonction is (lnx)^2 +cte

and the integrale is I = (ln (infinite) )^2 - (ln e)^2

ln e =1 and ln infinite =1

so result infinite

Answer 5

make this the essential from 3 to b of one million/x^(3/2) dx, then we are able to take the minimize as b is going to infinity. ?a million/x^(3/2)dx = ?x^-(3/2)dx = -2x^-(a million/2) = -2/?x eval from 3 to b = -2/?b + 2/?3 shall we take the minimize as b-->infinity - 2/?infinity +2/?3 0 + 2?3 = 2/?3 wish this permits

Answer 6

i hope we have to substitute the value of 'e' as 2.7182.. then we might get the ans..