Asked by mika
how do you determine the convergence of :
definite integral from 1--> infinity of lnx/(x^3)?
i set the problem as
lim (R--->infinity) of the integral of lnx/(x^3) from 1--->R, but i can't compute the integral.
definite integral from 1--> infinity of lnx/(x^3)?
i set the problem as
lim (R--->infinity) of the integral of lnx/(x^3) from 1--->R, but i can't compute the integral.
Answers
Answered by
Damon
something *( x^-2) ln x - something else*x^something call it k
say
a (x^-2) ln x - b x^c
differentiate to get
a (x^-2)(x^-1) + a [ ln x] (-2) x^-3 + b c x^(c-1)
clearly -2 a = 1
so
a = -1/2
so - (1/2) x^-3 = b c x^(c-1)
so
c-1 = -3
c = -2
but b c = - a = 1/2
b = -1/4
a (x^-2) ln x - b x^c
is
-(1/2)[ln x]/x^2 +(1/4) x^(1/2)
check my arithmetic!!!
so in the end
say
a (x^-2) ln x - b x^c
differentiate to get
a (x^-2)(x^-1) + a [ ln x] (-2) x^-3 + b c x^(c-1)
clearly -2 a = 1
so
a = -1/2
so - (1/2) x^-3 = b c x^(c-1)
so
c-1 = -3
c = -2
but b c = - a = 1/2
b = -1/4
a (x^-2) ln x - b x^c
is
-(1/2)[ln x]/x^2 +(1/4) x^(1/2)
check my arithmetic!!!
so in the end
Answered by
Damon
-(1/2)[ln x]/x^2 +1/(4x^2)
Answered by
mika
wait, are you doing the convergent problem or the partial fraction problem?
Answered by
Damon
I am just doing the integral of (ln x)dx / x^3
Answered by
Damon
In other words I think
integral of ln x dx/x^3= (-1/2)(ln x)/x^2 + 1/(4x^2) + constant
integral of ln x dx/x^3= (-1/2)(ln x)/x^2 + 1/(4x^2) + constant
Answered by
Damon
but check my arithmetic !!
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