evaluate the integral `int_1^oo (lnx/x) dx`

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evaluate the integral `int_1^oo (lnx/x) dx`

Postby barnamah » Mon 20 June, 2011 12:47

evaluate the integral `int(lnx/x) dx` from x=1 to `x=oo`
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Re: evaluate the integral `int_1^oo (lnx/x) dx`

Postby Integr88 » Mon 20 June, 2011 12:49

this can be written as`int_1^oo (lnx/x) dx`
this is type 1 improper integral which the bounds are between 1 and `oo`
using limit we can solve this:
`lim_(t->oo)int_1^tlnx/xdx`

let u=lnx, du=`1/xdx`
`lim_(t->oo)int_1^tudu` and integrate `lim_(t->oo)[u^2/2]_1^t`
In previous step we set u=ln x so substitute it into our equation and we have `lim_(t->oo)[(lnx)^2/2]_1^t` evaluate with t and 1
`lim_(t->oo)[ln(t)^2/2-ln(1)^2/2]`
`lim_(t->oo)[oo-0]=oo`
`:.` it is `oo` and diverging.
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