How do you use the integral test to determine whether #int lnx/(xsqrtx)# converges or diverges from #[3,oo)#?

Question

Cameron Alexander · Accepted Answer

The integral converges to #2/sqrt3(ln3+2)#.

We won't use the "integral test," since that test uses integrals to determine whether or not certain series converge or diverge. We will, however, integrate the given function and evaluate it at its infinite limit and #3# and determine whether or not the integral converges or diverges. #intlnx/(xsqrtx)dx=intx^(-3/2)lnxdx# Let's use integration by parts. Let: #{(u=lnx" "=>" "du=1/xdx),(dv=x^(-3/2)dx" "=>" "v=-2x^(-1/2)):}# So the (presently unbounded) integral becomes: #=-2x^(-1/2)lnx-int-2x^(-1/2)(1/xdx)# #=-(2lnx)/sqrtx+2intx^(-3/2)dx# As we've already determined: #=-(2lnx)/sqrtx-4/sqrtx# #=-(2(lnx+2))/sqrtx# Thus, the improper integral is: #int_3^oolnx/(xsqrtx)dx=-2[(lnx+2)/sqrtx]_3^oo# #=-2lim_(xrarroo)(lnx+2)/sqrtx-(-2)(ln3+2)/sqrt3# The limit is #0#, since a polynomial function like #sqrtx# will always grow a logarithmic function like #lnx# as #xrarroo#. #=2/sqrt3(ln3+2)# The integral converges.

Adam Adkins · Answer

undefined To determine the convergence or divergence of the integral ∫(ln(x) / (x√x)) from [3, ∞), you can use the integral test. First, verify that the function f(x) = ln(x) / (x√x) is continuous, positive, and decreasing for x ≥ 3. Then, evaluate the improper integral ∫(ln(x) / (x√x)) from 3 to ∞. If the integral converges, the series also converges. If the integral diverges, the series also diverges.