Expert: Alon Mandes - 12/10/2010

Question
I have a problem which should be fairly simple, but I wind
up stuck at an early point.

The problem states

Make an appropriate u-substitution to show that:

Int(a,ab) [(1/t) dt] = Int(1,b) [(1/t) dt]

and,

Use this equality to obtain another proof of ln (ab) = ln a
+ ln b

I think I can do the second part if I can just do the first
part.

My first idea is to use the u-substitution u=t/a

And indeed this link:

http://myyn.org/m/article/natural-logarithm2/

Tells me to do this.

But when I get that I get
u=t/a
t=ab, u=b
and
t=a, u=1

So far so good.

But I also get du = dt/a
or a(du) =dt

So that the equation would be
                 a
Int [1,b] ------------------------ du
                 u
which would give me an integral of [a ln b]
and not
             1
Int [1,b] ------------------------ du
             u

Which is what I need, and what the link I posted above seems
to imply.

There is obviously something I am missing, but I don’t see
what it is. Can you help?


Answer
Let's perform the substitution : t=av . Thus,
dt=adv AND when t=a then v=1 , AND where t=ab --> v=b . Hence,

ab       1
∫dt/t = ∫dv/v . Now we perform substitution v=t . We get :
a      b

ab       b      b  
∫dt/t = ∫dv/v = ∫dt/t . Q.E.D
a      1       1

Now, let's calculate this integral :

ab        
∫dt/t = Ln(t) {form a to ab } = Ln(ab)-Ln(a).
a

b         
∫dt/t = Ln(t) {from 1 to b } = Ln(b)-Ln(1)=Ln(b)-0=Ln(b).
1

So, we get : L(ab)-Ln(a)=Ln(b) --> Ln(ab)=Ln(a)+Ln(b) . Q.E.D

Alon.