int_4^infty 1/(x(ln x)^3)dx=
Substitute u=ln x => du=1/x dx, u_l=ln 4, u_u=ln infty=infty (u_l and u_u denote lower and upper bound respectively).
int_(ln 4)^infty1/u^3 du=-1/(2u^2)|_(ln4)^infty=-1/2(lim_(u to infty)1/u^2-1/(ln 4)^2)=-1/2(0-1/(ln 4)^2)=
1/(2(ln 4)^2)approx0.260171
As we can see the integral converges and its value is 1/(2(ln 4)^2).
The image below shows graph of the function and area under it representing the value of the integral. Looking at the image we can see that the graph approaches -axis (function converges to zero) "very fast". This suggests that the integral should converge to some finite number.
Tuesday, May 22, 2012
Calculus of a Single Variable, Chapter 8, 8.8, Section 8.8, Problem 25
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