how do you solve the integral of 1/[(square root of x)(lnx)] from 2 to infinity?

i did the p- integral theorem with 1/square root of x and got it to be a divergent integral. however i was told this was the wrong way and that i should do it by integration by parts. but i can't figure it out by that method. please help. thanks.

1 answer

You can proceed as follows. Substitute:

x = e^t. Then the integral becomes:

Integral from ln(2) to infinity of

e^(t/2)/t dt

We can get rid of the factor 2 in the exponential by putting y = t/2. The integral becomes:

Integral from 1/2 ln(2) to infinity of

e^(y)/y dy

For positive y we have

e^(y) > 1

It follows from this that the integral is larger than

Integral from 1/2 ln(2) to infinity of

1/y dy

but this is already divergent, so the integral diverges.