Question

Let\[L(x)=\int_{1}^{x}\frac{1}{t}dt.\]How can I show that\[L(ab)=L(a)+(b)?\]

Answer 1

The key to this, would be to notice that the integral of 1/t is the natural log, and that property holds with natural log.

Answer 2

I only want to use integral properties. Otherwise, this problem would be trivial.

Answer 3

perhaps you can start by showing that
\[\int_1^{2x}\frac{dt}{t}=\int_1^2\frac{dt}{t}+\int_1^x\frac{dt}{t}\]

Answer 4

take the derivatives, show that they are the same, therefore the functions must differ by a constant, find the constant

Answer 5

first solve for the integral in the LHD so that will give you
\[L(x)= \int\limits_{1}^{x} 1/t dt =\ln\left| x \right|-\ln\left| 1 \right|=\ln\left| x \right|\]
so since:
\[ L(x)=\ln \left| x \right|\]
\[ L(ab)=\ln\left| ab \right|=\ln\left| a\right|\left| b \right|=\ln\left| a \right|+\ln\left| b \right|=L(a)+L(b)\]
therefore, L(ab)=L(a)+L(b)
notice i used a property of logs

Answer 6

in my example
\[\frac{d}{dx}\int_1^{2x}\frac{dt}{t}=\frac{1}{x}\]
\[\frac{d}{dx}\int_1^x\frac{dt}{t}=\frac{1}{x}\] therefore
\[\int_1^{2x}\frac{dt}{t}=\int_1^{x}\frac{dt}{t}+C\]

Answer 7

and you can find C easily

Answer 8

\[ \int_{1}^{ab} 1/x dx = \int_{1}^{a}1/x dx + \int_{a}^{ab}1/x dx \]

let x = at, when x=a, t=1, and when x =ab, t=b, dx = a dt
\[ \int_{a}^{ab} 1/x dx = \int_{1}^{b} adt/at = \int_{1}^{b} dt/t \]

We have, \[ \int_{1}^{a}1/x dx + \int_{1}^{b}1/t dt = \int_{1}^{a}1/x dx + \int_{1}^{b}1/x dx \]

Answer 9

lol

Answer 10

or if you like @pre-algebra :
\[ L(ab)=\int\limits_{1}^{ab} 1/t dt=\ln \left| ab \right|-\ln\left| 1 \right|=\ln\left| ab \right|=\ln\left| a \right|\left| b \right|=\ln\left| a \right|+\ln\left| b \right|\]
\[=\ln\left| a \right|-0+\ln\left| b \right|-0=\ln\left| a \right|-\ln\left| 1 \right|+\ln\left| b \right|-\ln\left| 1 \right|=\int\limits_{1}^{a}1/tdt+\int\limits_{1}^{b}1/tdt\]
\[=L(a)+L(b)\]
so L(ab)=L(a)+L(b)

Answer 11

nvm

Answer 12

@satellite73, in this example, is it possible to find \(C\), or is that just a generalization?

Answer 13

just an example.

Answer 14

\[f(x)=\int_1^{x}\frac{dt}{t}\]
\[f(ax)=\int_1^{ax}\frac{dt}{t}\]
\[f'(ax)=f'(x)\] is an easy check. therefore
\[f(ax)=f(x)+C\] now to find C, we know that
\[f(1)=0\] so
\[f(a)=f(a\times 1)=f(1)+C=C\] hence \(C=f(a)\) and thefore \[f(ax)=f(a)+f(x)\]

Answer 15

Just for the record, the way I ended up doing this was by substitution.

For this example, we have that\[L(ab)=\int_{1}^{a}\frac{dt}{t}+\int_{a}^{ab}\frac{dt}{t}.\]If we let \(t=au\), then \(dt=adu\), and\[\int_{a}^{ab}\frac{dt}{t}=\int_{1}^{b}\frac{adu}{au}=\int_{1}^{b}\frac{du}{u}.\]Hence,\[L(ab)=\int_{1}^{a}\frac{dt}{t}+\int_{1}^{b}\frac{du}{u}=L(a)+L(b).\]