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.
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
-(1/2)[ln x]/x^2 +1/(4x^2)
wait, are you doing the convergent problem or the partial fraction problem?
I am just doing the integral of (ln x)dx / x^3
In other words I think
integral of ln x dx/x^3= (-1/2)(ln x)/x^2 + 1/(4x^2) + constant
but check my arithmetic !!
To determine the convergence of the given integral, you can apply the limit comparison test. Here's how you can proceed:
1. Start by setting up the integral as you did: the definite integral from 1 to infinity of (ln(x)/(x^3)).
2. Apply the limit comparison test by choosing a simpler reference function. In this case, a standard choice is 1/x^2. So, let's consider the integral of (ln(x)/(x^2)) from 1 to infinity. This integral is easier to evaluate.
3. Compute the reference integral: ∫(ln(x)/(x^2)) dx. To do this, you can use integration by parts. Let u = ln(x) and dv = (1/x^2), then differentiate u and integrate dv to find du and v, respectively. Applying the formula for integration by parts, you'll obtain ∫(ln(x)/(x^2)) dx = -ln(x)/x + ∫(1/x^3) dx.
4. Evaluate the reference integral: ∫(ln(x)/(x^2)) dx = -ln(x)/x - 1/(2x^2).
5. Apply the limit comparison test by taking the limit as R approaches infinity of the ratio of the original integral to the reference integral:
lim (R → ∞) [∫(ln(x)/(x^3)) from 1 to R] / [∫(ln(x)/(x^2)) from 1 to R]
= lim (R → ∞) [∫(ln(x)/(x^3)) from 1 to R] / [-ln(R)/R - 1/(2R^2)]
6. Simplify the expression: Using L'Hôpital's rule, you can take the derivative of the numerator and denominator with respect to R and evaluate the limit again. After applying L'Hôpital's rule once, the limit simplifies to:
= lim (R → ∞) [(1/R) / (-1/R^2 - 1/(2R^3))]
= -2
7. Analyzing the limit, we find that it is a finite non-zero value (-2). Since this limit is finite and non-zero, it implies that both integrals converge or diverge together.
Therefore, since the integral of (ln(x)/(x^2)) converges, the original integral ∫(ln(x)/(x^3)) from 1 to infinity also converges.