How do I determine if #int_1^oolnx/x^2 dx# converges or diverges?

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Answer 1 · 2017-11-01T21:15:56

The integral is convergent and converges to #1#

To calculate this improper integral, first calculate

#I=int_1^a(lnxdx)/x^2#

Perform this integral by parts

#u'=1/x^2#, #=>#, #u=-1/x#

#v=lnx#, #=>#, #v'=1/x#

#intu'v=uv-intuv'#

Therefore,

#I=int_1^a(lnxdx)/x^2=[-lnx/x]_1^a+int_1^adx/x^2#

#=[-lnx/x]_1^a-[1/x]_1^ a#

#=(-lna/a-0)-(1/a-1)#

#=-lna/a-1/a+1#

Now, calculate the limits

#lim_(a->oo)(-lna/a-1/a+1)=0-0+1=1#

Therefore, the integral is convergent and converges to #1#