Question

Defining e by:
\[\int\limits_{1}^{e}{1\over x} dx = 1\]
show that
\[\int\limits_{1}^{t}{1\over x} dx = ln t\]

Answer 1

In the second integral, we want to change variables such that the limits of integration are no longer 1 to t but 1 to e and we have the same integrand.

That being the case, think about what substitution/change of variable works.

Answer 2

\[x= {tu \over e}\]

when x = t, u =e, but when x =1 u isn't 1

Answer 3

No, it's not. This is a bit tricky. Think of something exponential. Let u be the new variable. Then with u = f(x) we want
f(1) = u = 1
f(t) = u = e

Answer 4

or the other way around x = g(u) where
g(1) = x = 1
g(e) = x = t

Answer 5

hint:
\[ t = e^{\ln t} \]

Answer 6

and 1 = 1^(ln t)

Answer 7

so \[x = e ^{\ln u }\]

Answer 8

No, because when x = t, you have u = e^t

Answer 9

no thats not right

Answer 10

Again, we want g where x= g(u)
1 = x = g(1).
t = x = g(e) Hint: e^(ln t) = t

Answer 11

1 = x = g(1). Hint: 1^(ln t) = 1
t = x = g(e) Hint: e^(ln t) = t

Answer 12

i'm going to sleep on it, I understand what i need to get to ....

Answer 13

ok.

Answer 14

thanks for your help

Answer 15

u = e ^ ln x

Answer 16

No. It will be easier if you try and write as I indicate above x = g(u)

Answer 17

1 = x = g(1). Hint: 1^(ln t) = 1
t = x = g(e) Hint: e^(ln t) = t

And remember that t is a constant.

Answer 18

thanks once again, brain is drained!

Answer 19

ok. I admire your discipline in not asking me the answer ... yet!

Answer 20

i'm not going to!

Answer 21

how about x = u ^ ln t
so when x = 1 u =1, when x = t, u = e

Answer 22

Looks promising. Now, does it work? What is dx and hence the integral
\[ \int_1^t \frac{dx}{x} \]

Answer 23

\[\ln t \int\limits_{1}^{e} {1 \over u} du\]

Answer 24

which gives the answer

Answer 25

blimey that was a hard substitution for A level

Answer 26

Yes. Good.

Answer 27

or maybe i'm just thick

Answer 28

This was a curious question. It it quite frequent to define ln t as that integral.

But I've never seen anyone asked to derive this from a different starting point.

Answer 29

that was fun working it out, well with your help of course

Answer 30

I'm a mathematician and if it makes you feel any better, it took me a couple of minutes to find the substitution as well. Quite unusual.

Answer 31

oh good! makes me feel a bit better

Answer 32

hope yet

Answer 33

right i can sleep easy now
thanks James

Answer 34

have a good weekend.

Answer 35

and you