The Prime Number Theorem says that the counting function for primes, pi(x), is a

Question

The Prime Number Theorem says that the counting function for primes, pi(x), is approximately equal to x/ln(x when x is large. It turns out that pi(x) is even closer to the value of the definite integral dt/ln (t). Show that lim x rightarrow infinity ( dt/ln(t) / (x/ln(x) = 1. This means that dt/ ln(t) and x/ ln(x) are approximately the same when x is large. [ . Use L'HopitaFs rule and the Second Fundamental Theorem of Calculus.] dt/ln(t) = ln(ln(t)) + ln(t) + (ln(t))2 / 2.2! +(ln (t))3 / 3.3! + (ln(t))4 / 4.4!+... Use this series to compute numerically the value of for x = 10, 100. 1000, 104, 106, and 109. Compare the values you get with the values of pi(x) and x/ln(x) given in the table on page 92. Which is closer to pi(X), the integral dt/ln(t) or the function x/ ln(x)? (This problem can be done with a simple calculator, but you'll probably prefer to use a computer or programmable calculator. Differentiate the series in (b) and show that the derivative is actually equal to 1 ln(f). [ . Use the series for ex.]

Explanation / Answer

1)

1/ln(t) >= 1/t for t>=2 so the integral diverges to infinity by comparison test with the diverging p-integral int(2..+infty) dt/t

The other part also diverges to infinity so we can use l'hospital rule and using the second fundamental theorem of calculus the derivative of the integral is 1/ln(t)

The derivative of t/ln(t) is (ln(t)-1)/ln(t)^2

So the limit is the same than the limit of :

(1/ln(t)) / (ln(t)-1)/ln(t)^2 = ln(t)/(ln(t)-1) = 1/(1-1/ln(t))

The latter as obvious limit 1 so we are done.

2) Using wolfram alpha use the syntax :

ln(ln(t)) + ln(t) + sum ln(t)^n/(n*n!) for n=2 to infinity

for t=2 we get 0.4679

for t=10 we get 5.5883

for t=100 we get 29.5489

for t=1000 we get 177.032

for t=10^4 we get 1245.56

for t=10^6 we get 78627

for t=10^9 we get 5084923.5

So the integral for each x is approximatively :

x=10 is 5.5883-0.4679 = 5.1204

x=100 => 29.5489-0.4679 = 29.081

x=1000 => 176.564

x=10^4 => 1245.09

x=10^6 => 78626.5

x=10^9 => 5084923

Name f(x) = x/ln(x)

pi(10) = 4 and f(10)=4.343

pi(100) =25 and f(100) = 21.715

pi(1000) =168 and f(1000)= 144.765

pi(10^4) = 1229 and f(10^4) = 1085.74

pi(10^6)= 78498 and f(10^6) = 72382.4

pi(10^9) = 50847534 and f(10^9) = 4825493

So we see that the integral is closer to pi(x) than f(x) !

3) By derivating we get ( call the stuff S )

S' = 1/(t*ln(t)) + 1/t + 1/(t*ln(t)) sum(n>=2) ln(t)^(n) / n!

S' = 1/(t*ln(t)) + 1/t + 1/(t*ln(t)) * ( e^(ln(t)) - ln(t) - 1) using the hint

S' = 1/(t*ln(t)) + 1/t + 1/(t*ln(t)) * ( t - ln(t) - 1)

S' = 1/(ln(t)) as desired.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.