# Logarithm defined as integral

1. Apr 25, 2006

### fstam2

Confused, but tried it this way:

Use u-substitution to show that (for y a positive number and x>0)

$$\int_{x}^{xy} \frac{1}{t} dt = \int_{1}^{y} \frac{1}{t} dt$$

so, u=t and du=dt
if x=1
t=xy u=y(1)=y
t=x u=1

or
u=1/t and du/ln [t] = dt
if x=1
t=xy u=1/y
t=x y=1

Todd

2. Apr 25, 2006

### HallsofIvy

That's not a very useful substitution, is it?
but x is not 1.

No, if u= 1/t, then du= -dt/t2
Again, you cannot just say "if x= 1"- it's not, it's a variable. Also you haven't used those substitutions- you haven't put them into either integral.

Look at the upper limits on each integral. On one it is xy, on the other it is just y. To show that the two integrals are equal, you need to change one into the other by some substitution. Okay,xy/x= y so we need to divide by x. Try u= t/x in the left integral only.