How Does U-Substitution Prove the Equality of Two Logarithmic Integrals?

[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

Thanks for your help
Todd

 

[HallsofIvy]

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

That's not a very useful substitution, is it?

if x=1
t=xy u=y(1)=yt=x u=1

but x is not 1.

or
u=1/t and du/ln [t] = dt

No, if u= 1/t, then du= -dt/t²

if x=1
t=xy u=1/y
t=x y=1

Thanks for your help
Todd

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.