Hedge variance swapping by vanilla option(constant vega portfolio against underlying asset)

One book said hedging variance swaps $$I= \sqrt{\dfrac{1}{t}\int^t_0\sigma^2(S,t)}d t$$ by vanilla option，say value $V(S,E;\sigma)$(Black-Scholes fomula) where $S$ is underlying asset, $E$ is the strike price. Then he constructed a portfolio with value $$P=\int^{\infty}_0f(E)V(S,E;\sigma)d E$$ then compute the vega of portfolio: $\textrm{Vega}_P= \dfrac{\partial P}{\partial\sigma},$ then let $$\dfrac{\partial \textrm{Vega}_P }{\partial S}=0$$ obtain $f(E) = \dfrac{k}{E^2}.$

His conclusion is variance swaps can be hedged with vanilla option, using the 'one over strike squared' rule. I can not understand:

1.What's the meaning of the representation of $P$(why take the integral) i.e how do we implement this portfolio by vanilla option, hold $F(E)$ share?

2.Why we need constant vega against $S$ i.e how to hedge?

1. An integral is used because the portfolio contains an infinite number of instruments. More specifically, the idea is to hold a continuous strip of $f(E)$ units of European vanilla options struck at $E$ for each $E \in [0,\infty[$. Of course this remains a theoretical concept: to build a similar portfolio in practice, one must consider a partition of the truncated strike domain (hence a finite number of options): $$P = \sum_{i=1}^N f(E_i) V(S,E_i;\sigma)$$
2. You need a constant, non-zero, Vega against $S$ because you would like a product which is sensitive to volatility and independent of the path the asset will take, i.e. a pure volatility bet.
• You can imagine that (historically) people tried to hedge the Varswap with a single ATM option, unfortunately $S$ changes over time and the option is no longer ATM, so the Vega is now lower and the hedging is not as effective as before. How to solve this? Someone smarter than me figured out that we have to use more than one option, in fact (it turns out) an infinite number. – noob2 Mar 7 '17 at 10:13
• It seems make sense, but I still confusion that why do we hedge the variance at its largest value? By the way, does this hedge depend on the Black-Scholes ? Since, I saw the $f(E) = \dfrac{1}{E^2}$ depends on the formula of $V.$ – A.Oreo Mar 8 '17 at 4:03