`int_2^oo 1/(x(lnx)^p) dx = int_2^oo 1/(x((lnx)^2)^(p/2)) dx`

Using the logarithmic rule `lnx^a = alnx`

`int_2^oo 1/(x(lnx)^p) = int_2^oo 1/(x(p/2)(lnx)^2) dx`

Now `int 1/(x(lnx)^2) dx = -1/lnx + C`

So

`int_2^oo 1/(x(lnp)^p) dx = -2/(plnx)|_2^oo = -2/p(lim_(x->oo) (1/lnx) - 1/ln2)`

`= -2/p(0 - 1/ln2) = 2/(pln2)`

**The integral converges for p not equal to 0**

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