Posted by kathryn on Tuesday, August 25, 2009 at 7:52pm.
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.
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