CSvax:Pucc-H:Pucc-I:ags@pur-ee.UUCP (09/12/83)
Here is a different "proof" that 0=1 for you calculus fans. Take the
basic integration-by-parts formula:
{integral} u dv = uv - {integral} v du
and let u = 1 / log x, v = log x.
Then du = - (1/x) * (1/log x)**2 dx
and dv = dx / x.
Substituting:
{integral} dx / (x log x) = 1 + {integral} dx / (x log x)
Cancelling like terms:
0 = 1.
Dave Seaman
pur-ee!pucc-I!agsecn-ec:ecn-pc:ecn-ed:vu@pur-ee.UUCP (09/13/83)
Unimaginable. How about:
{integral} dx = 1 + {integral} dx ????????
DON'T YOU KNOW that there is a constant of integration ??????????
Hao-Nhien Vu (pur-ee!vu)