What causes delta functions to appear in the derivative of the Pi function?

[3.14159253589]

What is the derivative of the pi function

 

[dimension10]

If you are talking about the prime number thing, I am not sure about the exact one but here is an approximation:

\pi(n)\approx \int_{2}^{n}\frac{dt}{\mbox{ln}(t)}

\frac{d}{dn}\pi(n)\approx \frac{d}{dn}\int_{2}^{n}\frac{dt}{\mbox{ln}(t)}

\frac{d}{dn}\pi(n)\approx \frac{d}{dn}\lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}\frac{\delta t}{\mbox{ln}(t)}

As the derivative of a sum is the sum of the derivatives,

\frac{d}{dn}\pi(n)\approx \lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}\frac{d}{dn}\frac{\delta t}{\mbox{ln}(t)}

\frac{d}{dn}\pi(n)\approx \lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}-\frac{\delta t}{n \; {\mbox{ln}}^{2}(n)}

\frac{d}{dn}\pi(n)\approx - \int_{2}^{n}\frac{dt}{n \; {\mbox{ln}}^{2}(n)}

So that is the approximate rate of change of the pi function of t as t changes.

 

[I like Serena]

This is not quite right.
The proper derivative is:
\frac{d}{dn}\pi(n)\approx \frac{d}{dn}\int_{2}^{n}\frac{dt}{\ln(t)} = \frac{1}{\ln(n)}

 

[dimension10]

This is not quite right.
The proper derivative is:
\frac{d}{dn}\pi(n)\approx \frac{d}{dn}\int_{2}^{n}\frac{dt}{\ln(t)} = \frac{1}{\ln(n)}

Then where did I make a mistake?

 

[dimension10]

According to Wolfram Alpha,

\frac{d}{dn}(\int_{2}^{n}\frac{dt}{\ln(t)})=\frac{-n+n\; \mbox{ln}(n)+2}{n \;{\mbox{ln}}^{2}(n)}

that is,

{\mbox{ln}}^{-1}(n)+(2{n}^{-1}{\mbox{ln}}^{-2}(n))-{\mbox{ln}}^{-2}(n)

But for Li(n) alone, it gives {\mbox{ln}}^{-1}(n) which is your solution.

And my solution gives - \int_{2}^{n}\frac{dt}{n \; {\mbox{ln}}^{2}(n)} which is equal to \frac{n-2}{n\; {\mbox{ln}}^{2}(n)}

I guess all three solutions are equal to each other and thus, correct?

 

[MathematicalPhysicist]

If you are talking about the prime number thing, I am not sure about the exact one but here is an approximation:

\pi(n)\approx \int_{2}^{n}\frac{dt}{\mbox{ln}(t)}

\frac{d}{dn}\pi(n)\approx \frac{d}{dn}\int_{2}^{n}\frac{dt}{\mbox{ln}(t)}

\frac{d}{dn}\pi(n)\approx \frac{d}{dn}\lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}\frac{\delta t}{\mbox{ln}(t)}

As the derivative of a sum is the sum of the derivatives,

\frac{d}{dn}\pi(n)\approx \lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}\frac{d}{dn}\frac{\delta t}{\mbox{ln}(t)}

\frac{d}{dn}\pi(n)\approx \lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}-\frac{\delta t}{n \; {\mbox{ln}}^{2}(n)}

\frac{d}{dn}\pi(n)\approx - \int_{2}^{n}\frac{dt}{n \; {\mbox{ln}}^{2}(n)}

So that is the approximate rate of change of the pi function of t as t changes.

I believe the problem is here:
\frac{d}{dn}\pi(n)\approx \lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}\frac{d}{dn}\frac{\delta t}{\mbox{ln}(t)}

The derivative operator \frac{d}{dn} should operate on the sum.
I am not even sure what it means to have delta t inside a sum where t is a dumby variable, and then taking a limit of it as it approaches zero, the notation here is quite problematic.

 

[I like Serena]

Then where did I make a mistake?

As MathematicalPhysicist already said, the first mistake is when you moved d/dn to the other side of the summation symbol.
This is not allowed, because the summation is dependent on n.

You made another mistake when you differentiated the expression dependent on t with respect to n.
Since the expression is not dependent on n, the result is zero.

According to Wolfram Alpha,

\frac{d}{dn}(\int_{2}^{n}\frac{dt}{\ln(t)})=\frac{-n+n\; \mbox{ln}(n)+2}{n \;{\mbox{ln}}^{2}(n)}

How did you get WolframAlpha to say that?
I do not get that.

I guess all three solutions are equal to each other and thus, correct?

The three solutions are not equal to each other, so they cannot all be correct.

 

[dimension10]

As MathematicalPhysicist already said, the first mistake is when you moved d/dn to the other side of the summation symbol.
This is not allowed, because the summation is dependent on n.

Thanks. I forgot about that.

You made another mistake when you differentiated the expression dependent on t with respect to n.
Since the expression is not dependent on n, the result is zero.

Oops.

How did you get WolframAlpha to say that?
I do not get that.

I think I know what happened. It must have again considered d as constant rather than an infinitesimal.There seems to be a simpler solution using the second fundamental theorem of calculus and that would yield

\frac{1}{\mbox{ln}(n)}

which is the answer given by you.

 

[zetafunction]

the derivative of the prime counting function is just the sum

\delta (x-p) taken over all primes 'p'

 

[Robert1986]

I am confused; pi is a step function and is therefore only differentiable at the "interior" of a step at which point it is 0.

 

[zetafunction]

yes but since the step function is discontinous delta function appear whenever the function has discontinuties, in the case of Pi function the discontinuities are located at the prime numbers