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

  • Thread starter Math100
  • Start date
In summary, the proof shows that ## li(x) \sim \frac{x}{\log x} ## and the Prime Number Theorem can be expressed as ##\pi(x) \sim li(x)##. However, it should be noted that calculations with infinity as a number are not valid and should be disregarded.
  • #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) ##.
 
Physics news on Phys.org
  • #2
Math100 said:
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,
 
  • Like
Likes Math100
  • #3
Wikipedia has a nice image about the difference between the curves:

PrimeNumberTheorem2.png


For more details, see:
https://www.physicsforums.com/insights/the-history-and-importance-of-the-riemann-hypothesis/
 
  • Informative
Likes Math100
  • #4
fresh_42 said:
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.
 
  • Like
Likes fresh_42

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

1. What does the notation li(x) mean in this equation?

The notation li(x) represents the logarithmic integral function, which is defined as the integral of 1/ln(t) from 2 to x. In simpler terms, it is a function that approximates the number of prime numbers less than or equal to x.

2. How is the equation li(x) ~ x/log(x) derived?

The approximation of li(x) ~ x/log(x) is a result of the Prime Number Theorem, which states that the number of prime numbers less than or equal to x is approximately equal to x/log(x). This theorem was proven by mathematicians in the late 19th and early 20th centuries.

3. Is the approximation li(x) ~ x/log(x) accurate for all values of x?

No, the approximation is only accurate for large values of x. As x approaches infinity, the ratio between li(x) and x/log(x) gets closer and closer to 1. However, for smaller values of x, the approximation may not be as accurate.

4. How is this equation useful in mathematics?

The equation li(x) ~ x/log(x) is useful in understanding the distribution of prime numbers. It allows us to estimate the number of prime numbers within a given range and can be used in various mathematical proofs and calculations involving prime numbers.

5. Are there any other approximations for the number of prime numbers?

Yes, there are other approximations for the number of prime numbers, such as the Gauss's prime number theorem and the Chebyshev's theorem. However, the approximation li(x) ~ x/log(x) is one of the most commonly used and studied approximations in mathematics.

Similar threads

Replies
1
Views
1K
Replies
6
Views
876
Replies
7
Views
1K
Replies
14
Views
1K
Replies
25
Views
1K
Replies
7
Views
1K
Replies
7
Views
6K
Back
Top