Question

Using L(X) = int [1,X] dt/t, show -L(X) = L(1/X). Dr. Jerison wrote a comment that int[a,b]= -int[b,a], but he was careful to warn that that was a notation, not a part of any definition. Am I remembering correctly?

Answer 1

\[-L(X) \]

Answer 2

0

Answer 3

\[L(x) = \int\limits_{1}^{x}1/t*dt\]
\[ = [\ln t] from 1 \to x = \ln x - \ln 1 = \ln x\] (as ln 1=0 because e^0=1)
Now -L(x)
= \[-\ln(x) = \ln (x^-1)\] (Using logarithmic properties)
=\[\ln(1/x)\] which is equal to L(1/x)

And hence -L(x)=L(1/x)
[Majorly due to the properties of logarithms. You had to realize that the integral of 1/t is ln t which is a logarithmic function]

Answer 4

From int [1, x] dt/t = L(X); previously proved product rule, L(1)= L( X* 1/X) = L(X)+ L(1/X) = L(1) = 0; L(X) + L(1/X) = 0 => -L(X) = L(1/X).