Definite Integral of 1/x from 0 to 1 problem

[TylerH]

Given that 1/x is symetric across y=x, why can't we say \int^1_0 1/x - x dx= \int^\infty_1 1/x + x dx? Geometrically, it makes sense, but ln(0) is clearly undefined.

 

[matiasmorant]

from 1 to infinity the integral is unboundedly large. ln(x) tends to -infinity as x tends to 0, so the other integral is also unboundedly large. I think there's nothing wrong with saying the areas are equal

 

[TylerH]

Yeah... I should have waited to post. I solved it and got x=0, but I'm not sure that means anything...

\int^1_0 (1/x - x)dx = \int^\infty_1 dx/x + \int^1_0 xdx
\right[lnx\left]^1_0 - x^2/2 = \right[lnx\left]^\infty_1 + x^2/2
ln1 - \lim_{x \to 0+}lnx = \lim_{x \to \infty}lnx + ln1 + x^2
-\lim_{x \to 0+}lnx = \lim_{x \to \infty}lnx + x^2
\lim_{x \to \infty}lnx + \lim_{x \to 0+}lnx + x^2 = 0
\lim_{x \to \infty}(lnx + ln1/x)+x^2=0
\lim_{x \to \infty}\frac{x}{x}+x^2=0
and so on...

What is the geometric interpretation of x=0?