Prove that ## li(x)\sim \frac{x}{\log x} ##.

[Math100]

Homework Statement

Prove that $li(x)\sim \frac{x}{\log x}$, using $li(x)=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2}$. Deduce that the Prime Number Theorem can be expressed in the form $\pi(x)\sim li(x)$.

Relevant Equations

None.

Proof:

Let $f$ and $g$ be functions such that $f$ and $g$ are asymptotically equivalent if
$\lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}=1$.
The Fundamental Theorem of Calculus states that if $F$ is a function defined for all $x$ in $[a, b]$ by
$F(x)=\int_{a}^{x}f(t)dt$, then $F'(x)=f(x)$ where $F$ is an antiderivative of $f$.
Consider $F(x)=\int_{2}^{x}\frac{dt}{\log^2 t}$.
Then $F'(x)=\frac{1}{\log^2 x}$.
Observe that
\begin{align*}
&\frac{d}{dx}(\frac{x}{\log x})=\frac{\log x\frac{dx}{dx}-x\frac{d}{dx}(\log x)}{\log^2 x}\\
&=\frac{\log x-\frac{x}{x}}{\log^2 x}=\frac{\log x-1}{\log^2 x}.\\
\end{align*}
Now we will show that $\lim_{x\rightarrow \infty}\frac{li(x)}{\frac{x}{\log x}}=\lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}$.
Observe that $\lim_{x\rightarrow \infty}(\frac{x}{\log x})=\frac{\infty}{\infty}$.
By L'hopital's rule, we have that
$\lim_{x\rightarrow \infty}\frac{\frac{d}{dx}(x)}{\frac{d}{dx}(\log x)}=\lim_{x\rightarrow \infty}\frac{1}{\frac{1}{x}}=\lim_{x\rightarrow \infty}x=\infty$.
This implies $lim_{x\rightarrow \infty}li(x)=\lim_{x\rightarrow \infty}(\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2})=\infty$,
so $\lim_{x\rightarrow \infty}\frac{li(x)}{\frac{x}{\log x}}=\frac{\infty}{\infty}$.
Applying L'Hopital's rule yields
\begin{align*}
&\lim_{x\rightarrow \infty}\frac{\frac{d}{dx}(li(x))}{\frac{d}{dx}(\frac{x}{\log x})}\\
&=\lim_{x\rightarrow \infty}\frac{\frac{d}{dx}(\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2})}{\frac{d}{dx}(\frac{x}{\log x})}\\
&=\lim_{x\rightarrow \infty}1+\frac{\frac{d}{dx}(\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2})}{\frac{d}{dx}(\frac{x}{\log x})}\\
&=\lim_{x\rightarrow \infty}1+\frac{\frac{1}{\log^2 x}}{\frac{\log x-1}{\log^2 x}}\\
&=\lim_{x\rightarrow \infty}1+\lim_{x\rightarrow \infty}\frac{1}{\log x-1}\\
&=1+0\\
&=1.\\
\end{align*}
Thus $li(x)\sim \frac{x}{\log x}$.
By the Prime Number Theorem, we have that $\pi(x)\sim \frac{x}{\log x}$.
Thus $lim_{x\rightarrow \infty}\frac{\pi(x)}{li(x)}=\lim_{x\rightarrow \infty}(\frac{\pi(x)}{\frac{x}{\log x}}\cdot \frac{\frac{x}{\log x}}{li(x)})=1\cdot \frac{1}{1}=1$.
Therefore, $li(x)\sim \frac{x}{\log x}$ and the Prime Number Theorem can be expressed in the form $\pi(x)\sim li(x)$.

 

[fresh_42]

Homework Statement:: Prove that $li(x)\sim \frac{x}{\log x}$, using $li(x)=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2}$. Deduce that the Prime Number Theorem can be expressed in the form $\pi(x)\sim li(x)$.
Relevant Equations:: None.

Proof:

Let $f$ and $g$ be functions such that $f$ and $g$ are asymptotically equivalent if
$\lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}=1$.
The Fundamental Theorem of Calculus states that if $F$ is a function defined for all $x$ in $[a, b]$ by
$F(x)=\int_{a}^{x}f(t)dt$, then $F'(x)=f(x)$ where $F$ is an antiderivative of $f$.
Consider $F(x)=\int_{2}^{x}\frac{dt}{\log^2 t}$.
Then $F'(x)=\frac{1}{\log^2 x}$.
Observe that
\begin{align*}
&\frac{d}{dx}(\frac{x}{\log x})=\frac{\log x\frac{dx}{dx}-x\frac{d}{dx}(\log x)}{\log^2 x}\\
&=\frac{\log x-\frac{x}{x}}{\log^2 x}=\frac{\log x-1}{\log^2 x}.\\
\end{align*}
Now we will show that $\lim_{x\rightarrow \infty}\frac{li(x)}{\frac{x}{\log x}}=\lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}$.
Observe that $\lim_{x\rightarrow \infty}(\frac{x}{\log x})=\frac{\infty}{\infty}$.
By L'hopital's rule, we have that
$\lim_{x\rightarrow \infty}\frac{\frac{d}{dx}(x)}{\frac{d}{dx}(\log x)}=\lim_{x\rightarrow \infty}\frac{1}{\frac{1}{x}}=\lim_{x\rightarrow \infty}x=\infty$.
This implies $lim_{x\rightarrow \infty}li(x)=\lim_{x\rightarrow \infty}(\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2})=\infty$,
so $\lim_{x\rightarrow \infty}\frac{li(x)}{\frac{x}{\log x}}=\frac{\infty}{\infty}$.
Applying L'Hopital's rule yields
\begin{align*}
&\lim_{x\rightarrow \infty}\frac{\frac{d}{dx}(li(x))}{\frac{d}{dx}(\frac{x}{\log x})}\\
&=\lim_{x\rightarrow \infty}\frac{\frac{d}{dx}(\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2})}{\frac{d}{dx}(\frac{x}{\log x})}\\
&=\lim_{x\rightarrow \infty}1+\frac{\frac{d}{dx}(\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2})}{\frac{d}{dx}(\frac{x}{\log x})}\\
&=\lim_{x\rightarrow \infty}1+\frac{\frac{1}{\log^2 x}}{\frac{\log x-1}{\log^2 x}}\\
&=\lim_{x\rightarrow \infty}1+\lim_{x\rightarrow \infty}\frac{1}{\log x-1}\\
&=1+0\\
&=1.\\
\end{align*}
Thus $li(x)\sim \frac{x}{\log x}$.
By the Prime Number Theorem, we have that $\pi(x)\sim \frac{x}{\log x}$.
Thus $lim_{x\rightarrow \infty}\frac{\pi(x)}{li(x)}=\lim_{x\rightarrow \infty}(\frac{\pi(x)}{\frac{x}{\log x}}\cdot \frac{\frac{x}{\log x}}{li(x)})=1\cdot \frac{1}{1}=1$.
Therefore, $li(x)\sim \frac{x}{\log x}$ and the Prime Number Theorem can be expressed in the form $\pi(x)\sim li(x)$.

You cannot calculate with infinity as if it was a number. You should delete everything between "Now we will show ..." until $\frac{\infty }{\infty }.$

The last calculation (together with the derivative of $x/\log(x)$ from the beginning) is sufficient,

 

[fresh_42]

Wikipedia has a nice image about the difference between the curves:

For more details, see:
https://www.physicsforums.com/insights/the-history-and-importance-of-the-riemann-hypothesis/

 

[Math100]

You cannot calculate with infinity as if it was a number. You should delete everything between "Now we will show ..." until $\frac{\infty }{\infty }.$

The last calculation (together with the derivative of $x/\log(x)$ from the beginning) is sufficient,

Maybe I wrote too much.