MHB Solid of revolution vs. area below 1/x

SweatingBear
Messages
119
Reaction score
0
How come the area below the graph of $$\frac 1x$$ between $$[1, \infty)$$ does not exist, but the solid of revolution below the same graph in that same interval does exist? I do not see the logic.
 
Physics news on Phys.org
sweatingbear said:
How come the area below the graph of $$\frac 1x$$ between $$[1, \infty)$$ does not exist, but the solid of revolution below the same graph in that same interval does exist? I do not see the logic.

The calculus demonstrates that the area... $\displaystyle A = \int_{1}^{\infty} \frac{dx}{x} = \infty$ (1)

...and the volume of its the solid of revolution is finite and is... $\displaystyle V= \pi\ \int_{1}^{\infty} \frac{dx}{x^{2}} = \pi$ (2)

It may seem incredible but this 'paradox' was discovered by Pappo di Alessandria in the fourth century after Christ [!]...Kind regards $\chi$ $\sigma$
 
Thank you both for your replies.
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...
Back
Top