- #1
Math100
- 781
- 220
- 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) ##.
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) ##.