# To show that $\exp(\psi(2n+1))\int_0^1x^n(1-x)^n dx$ is a positive integer, where $\psi(x)$ is the Chebyshev function, and $0 \leq x \leq 1$

I have to show that $$\exp(\psi(2n+1))\int_0^1x^n(1-x)^n dx \ \in \mathbb{N}$$ where $\psi(x)$ is the Chebyshev function, and $0 \leq x \leq 1$

The Chebyshev function is defined as the following: $$\psi(x)=\sum_{p^{\alpha}\leq x}\log p$$

Now, here's the problem I'm having. I tried to take an example to get a sense of how to go about it. So, let's take $n=3$, then we have

$$\exp(\psi(7))\int_0^1x^3(1-x)^3 dx$$ Clealy $$\exp(\psi(7)) = \prod_{p^{\alpha}\leq x}p=2\cdot3\cdot5\cdot7$$ and $$\int_0^1 x^3 (1-x)^3 \ dx = \frac{1}{140}$$

Clearly, $$2\cdot3\cdot5\cdot7\cdot\frac{1}{140} \notin \mathbb{N}$$

I have a feeling that I may be misinterpreting the Chebyshev function. Can I get some help?

• you should include also $2^2$, then it is not a problem. – Arian Mar 4 '18 at 13:17
• Yeah, sure, I realise that. But why should I? The definition of the Chebyshev function says that the summand should be primes, but not prime powers. But the sum is over prime powers. – Naweed G. Seldon Mar 4 '18 at 13:19
• So for each prime power, you add the log of that prime. Therefore, the product is $2\cdot 3\cdot 2\cdot 5\cdot 7$, with the second $2$ coming from $2^2=4$ – Mastrem Mar 4 '18 at 13:50
• Ah! Right, I was right. I was misinterpreting the Chebyshev function. – Naweed G. Seldon Mar 4 '18 at 13:53