Find the constants ## A ## and ## B ##

[Math100]

Homework Statement

Define the logarithmic integral $li(x)$ by $li(x)=\int_{2}^{x}\frac{dt}{\log t}$, for $x>2$. Prove that $li(x)=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}+A$ and $li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}+B$, for some constants $A$ and $B$ that you should determine.

Relevant Equations

None.

Proof:

Observe that
\begin{align*}
&li(x)=\int_{2}^{x}\frac{dt}{\log t}\\
&=[\frac{t}{\log t}]_{2}^{x}-\int_{2}^{x}t(\frac{1}{\log t})'dt\\
&=\frac{x}{\log x}-\frac{2}{\log 2}-\int_{2}^{x}t(-\frac{1}{t\log^2 t})dt\\
&=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2}\\
&=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}+A\\
\end{align*}
where $A=-\frac{2}{\log 2}$.
This implies $li(x)=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2}$.
Since
\begin{align*}
&\int_{2}^{x}\frac{dt}{\log^2 t}=[\frac{t}{\log^2 t}]_{2}^{x}+2\int_{2}^{x}\frac{dt}{\log^3 t}\\
&=\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}-\frac{2}{\log^2 2},\\
\end{align*}
it follows that $li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}-\frac{2}{\log 2}-\frac{2}{\log^2 2}$.
Thus $li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}-(\frac{2}{\log 2}+\frac{2}{\log^2 2})$, where $B=-\frac{2}{\log 2}-\frac{2}{\log^2 2}$.
Therefore, $li(x)=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}+A$ and
$li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}+B$
where $A=-\frac{2}{\log 2}$ and $B=-\frac{2}{\log 2}-\frac{2}{\log^2 2}$.

 

[fresh_42]

Homework Statement:: Define the logarithmic integral $li(x)$ by $li(x)=\int_{2}^{x}\frac{dt}{\log t}$, for $x>2$. Prove that $li(x)=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}+A$ and $li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}+B$, for some constants $A$ and $B$ that you should determine.
Relevant Equations:: None.

Proof:

Observe that
\begin{align*}
&li(x)=\int_{2}^{x}\frac{dt}{\log t}\\
&=[\frac{t}{\log t}]_{2}^{x}-\int_{2}^{x}t(\frac{1}{\log t})'dt\\
&=\frac{x}{\log x}-\frac{2}{\log 2}-\int_{2}^{x}t(-\frac{1}{t\log^2 t})dt\\
&=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2}\\
&=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}+A\\
\end{align*}
where $A=-\frac{2}{\log 2}$.
This implies $li(x)=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2}$.
Since
\begin{align*}
&\int_{2}^{x}\frac{dt}{\log^2 t}=[\frac{t}{\log^2 t}]_{2}^{x}+2\int_{2}^{x}\frac{dt}{\log^3 t}\\
&=\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}-\frac{2}{\log^2 2},\\
\end{align*}
it follows that $li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}-\frac{2}{\log 2}-\frac{2}{\log^2 2}$.
Thus $li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}-(\frac{2}{\log 2}+\frac{2}{\log^2 2})$, where $B=-\frac{2}{\log 2}-\frac{2}{\log^2 2}$.
Therefore, $li(x)=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}+A$ and
$li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}+B$
where $A=-\frac{2}{\log 2}$ and $B=-\frac{2}{\log 2}-\frac{2}{\log^2 2}$.

Looks good. The formula for integration by parts is
$$
\int_a^b u'(x)v(x)\,dx=\left[u(x)v(x)\right]_a^b - \int_a^b u(x)v'(x)\,dx
$$
and you applied it to $u'(x)=1$ twice. If you quote the formula under "relevant equations" and then simply note $u'=1$ then it is a bit easier to read.

A way to remember the formula is to note that it is at its core the Leibniz rule of differentiation:
\begin{align*}
(u\cdot v )'&=u'v +uv'\\
u'v&=(u\cdot v )' - uv'\\
\int u'v &=\int [(u\cdot v )' -uv']\\
\int u'v &=uv-\int uv'
\end{align*}

The quotient rule of differentiation is also simply the Leibniz rule:
$$
\left(\dfrac{u}{v}\right)'=(u\cdot v^{-1})'=u'v^{-1}+u\left(v^{-1}\right)'=\dfrac{u'v}{v^2}-\dfrac{uv'}{v^2}=\dfrac{u'v-uv'}{v^2}
$$

In other parts of mathematics, the Leibniz rule takes the form of the Jacobi-identity, or the defining equation for derivations. All these formulas are simply the Leibniz rule $(u\cdot v )'=u'v +uv'$ which is all one has to remember.