# Basic Formula

We shall now prove that polynomials can be integrated using the following Basic Formula: If $v(s)=s^p$, then with $x(0)=0$, then we have

• $x(t)= \frac{t^{p+1}}{p+1} = \int_0^ts^p ds,\quad p = 0,1,...$

To prove this, we note that for $p=0$ we have

• $t= \int_0^t 1 ds$,

which is the same as

• $\frac{d}{dt}t = \frac{dt}{dt} = 1$.

To see this we note that if $x(t)=t$, then $dx(t)=x(t+dt)-x(t)=t+dt -t=dt$ and thus $\frac{dx}{dt}=1$.

For $p=1$ we are supposed to prove that

• $\frac{t^{2}}{2} = \int_0^ts ds$,

which is the same as $\frac{d}{dt}t^2 = 2t$. To see this we note that if $x(t)=t^2$, then

• $dx(t)=x(t+dt)-x(t)=(t+dt)(t+dt) -t^2=t^2+2tdt+dtdt-t^2=2tdt-dtdt$.

If now $dt$ is small then we can argue that $dtdt$ is so small that it can be neglected, and thus $\frac{d}{dt}t^2=2t$.

For $p=2$ we are supposed to prove that

• $\frac{t^{3}}{3} = \int_0^ts^2 ds$,

which is the same as

• $\frac{d}{dt}t^3 = 3t^2$.

To see this, we note that if $x(t)=t^3$, then

• $dx(t)=x(t+dt)-x(t)=(t+dt)(t+dt)(t+dt)-t^2$
• $=t^3+3t^2dt+3tdtdt+dtdtdt-t^3=3t^2dt+3tdtdt +dtdtdt$.

If now $dt$ is small then we can argue that $3tdtdt$ and $dtdtdt$ are so small that they can be neglected, and thus $\frac{d}{dt}t^3=3t^2$. Similarly, one can show the Basic Formula:

• $\frac{d}{dt}t^p=pt^{p-1} \quad\mbox{for }p=1,2,3...$

In MTS we show that his formula also holds for negative exponents

• $\frac{d}{dt}t^p=pt^{p-1}\quad\mbox{for }t>0, \quad p=-1,-2,-3,....$

# The Logarithm

We shall also discover that the case $p=0$ is special, and gives rise to the logarithm $\log(t)$
as the solution of

• $\dot u(t)=t^{-1}=\frac{1}{t}, \quad\mbox{for }t>0,\quad u(1)=0$,

that is

• $\log(t)=\int_1^t\frac{1}{s}\, ds, \quad\mbox{for }t>0$.