Showing the sum over primes is equal to an integral First, note that $$\vartheta = \sum_{p \leq x} \log p$$I am trying to show $$\vartheta(x) = \pi(x)\log(x)-\int_2^x\frac{\pi(u)}{u}du$$ I am trying to show this by summation of parts. The theorem of partial summation is 

Let $f$ be a continuous and differentiable function. Suppose $\{a_n\}_{n\geq 1} \subseteq \mathbb{R}$. Set $A(t) =\sum_{n\leq t}a_n$. Then $$\sum_{n\leq x} a_n f(n) = A(x)f(x) - \int_1^x A(t)f'(t) dt$$

My proof is as follows (it seems simple enough). Let $f(n) = \log(p)$ if n is a prime. Clearly $f(n)$ is continuous and diffrentiable. Set $$a_n = \begin{cases} 1 & \text{if } n = prime  \\ 0 & \text{otherwise}\end{cases}$$
Then by summation of parts we have $$\sum_{n\leq x}log(p)\cdot a_n = A(x)\log(x) - \int_1^x\frac{A(n)}{u}du$$ where $A(t) = \sum_{p \leq x} 1 = \pi(x)$
Is this sufficient enough?

It looks reasonably good, but one thing you need to be more clear about is the definition of $f(n)$.  Currently "$f(n) = \log(p)$ when $n$ is a prime" is not an adequate definition, since it fails to define, for instance $f(2.5)$.
It sounds like you might be defining $f(n) = 0$ when $n$ is not a prime; this would not be very usable since it results in a discontinuous function.  On the other hand, $f(x) = \log x$ (already hinted at when you implicitly determined $f'(u) = 1/u$) works just fine for the argument.
Another minor nitpick: you need to justify the switch between $\int_1^x$ in one equation and $\int_2^x$ in the other.  This isn't hard because if you think about the definition, $A(t)$ happens to be $0$ for all $t < 2$.
