How this:$\int_{0}^{+\infty} \sin \frac{\pi(1-x)}{2}dx=0 $ with $\pi(x)$ is counting prime function?

Question

let $\pi(x)$ be a the number of prime less than $x$ or prime counting function, I have accrossed in my computation of some integral related to zeta function those two formula :

$$\int_{0}^{+\infty} \sin \frac{\pi(1-x)}{2}dx=0 $$ $$\int_{0}^{+\infty} \cos \frac{\pi(1-x)}{2}dx=\infty $$

Now my question here how these integral can be evaluated ? and what does meant that about distribution of primes ?

Answer 1

Taking what is in the OP literally, the two results follow from the fact $$\pi(1-x)=0$$ for $x\ge0$.