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.
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.
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