This is an integral problem. What is the area of (ln x)/x^2 from one to positive infinity?

Question

Also, could you provide specific steps to your answer? I would appreciate it.

Answer 1

∫ (ln x)/x^2 [1, ∞]

Rewrite it so it's not an improper integral
lim ∫ (ln x)/x^2 [1, n]
n->∞

Integrate by parts
u = ln(x)
du = 1/x
dv = 1/x^2
v = -1/x

∫ (ln x)/x^2 = -ln(x)/x + ∫ 1/x^2
∫ (ln x)/x^2 = -ln(x)/x - 1/x

evaluating form 1 to n
[-ln(n)/n - 1/n] - [ln(1)/1 - 1/1]
[-ln(n)/n - 1/n] - [0 - 1/1]
-ln(n)/n - 1/n + 1

Letting n go to infinity
lim -[ln(n)-1]/n + 1
n->∞

lim -[ln(n)-1]/n + lim 1
n->∞

yields an an indeterminate form [∞/∞] so you can apply L'hopitals to the -[ln(n)-1]/n portion

differentiating the top and bottom you get
-1/x / 1

Taking the limit you get
0/1

leaving just 0 +1
∫ (ln x)/x^2 [1, ∞] converges to 1

Answer 2

That integration by parts stuff is great, but in my never-ending quest for shortcuts, I'd try a change of variables, namely y = ln x.

x = e^y

dy/dx = 1/x.

Then your integral looks a lot like the integral of e(-y)dy, where y varies from 1 to 0.

And that looks a lot like -(-1) = 1.

Answer 3

∫ln(x)/x^2 dx

let ln x = y : e^y = x

1/x dx = dy

∫ln(x)/x^2 dx = ∫(ln x/x) (1/x) dx

∫y/e^y dy

∫y e^-y dy

integrating by parts

u = y : du = dy

dv = e^-y : v = - e^-y

∫y e^-y dy = uv - ∫v du

=> - ye^-y - ∫-e^-y dy

=> -y e^-y - e^-y

=> -lnx (1/x) - (1/x)

limits between 1 to +∞

assuming that 1/∞ = 0

[0 + 0)] -[0 - 1]

=> 1

Answer 4

∫(ln x)/x^2 dx, x from one to infinity
= -∫(ln x) d((1/x), x from one to infinity
= -lnx / x + ∫1/x^2 dx, x from one to infinity
= 1