Max of $2$ independent random variables

Let's prove it backwards:

$$mathbb{E}[T]=int_{-infty}^{infty}G(t)[1-G(t)]dt=int_{-infty}^{infty}G(t)dt-int_{-infty}^{infty}G^2(t)dt=$$

$$tG(t)Bigg|_{-infty}^{infty}-int_{-infty}^{infty}tg(t)dt-tG^2(t)Bigg|_{-infty}^{infty}+int_{-infty}^{infty}2t g(t)G(t)dt=$$

$$underbrace{t[G(t)-G^2(t)]Bigg|_{-infty}^{infty}}_{=0}-underbrace{mathbb{E}[X_1]}_{=0}+int_{-infty}^{infty}2t g(t)G(t)dt$$

1. To prove that the first addend is zero you have only to calculate the limits...it's easy

2. the second addend is zero as per initial statement

3. the third addend is the result. It matches with the same result you correctly calculated by another way...the proof is finished.

Edit: further details on limit calculus:

$$lim_{t rightarrow +infty}frac{t}{frac{1}{G-G^2}} rightarrow-frac{(G-G)^2}{g-2gG}=0$$

Let $X^+ = max(X,0)$ and $X^- = max(-X,0)$. Then $X^+ = displaystyle int_0^{infty} mathbb{1}(X>t)dt = max(a ge : X ge a)$ and $X^- = displaystyle int_{-infty}^0 mathbb{1}(X le t) dt = min(a le 0: X > a)$.

Now we have: $$mathbb{E}(X) = mathbb{E}(X^+ - X^-) = mathbb{E}(X^+) - mathbb{E}(X^-) = ldots = int_0^{infty} (1-F(x))dx - int_{-infty}^0F(x)dx$$

Now for $F(x) = G^2(x)$ we have:

$$mathbb{E}(X) = mathbb{E}(max(X_1,X_2)) = int_0^{infty}(1-G(x))(1+G(x)) dx - int_{-infty}^0 G^2(x)dx$$

From this point it's easy to get your fact.

$Hint$: we have that $$mathbb{E}(X) = int_0^infty (1-G(x))G(x) dx + int_0^infty (1-G(x))dx -int_0^infty G^2(x)dx +$$ $$+int_{-infty}^0(1-G(x))G(x)dx -int_{-infty}^0(1-G(x))G(x)dx = $$ $$=int_{-infty}^{infty} (1-G(x))G(x)dx + R$$

And you need to prove: $$R = int_0^{infty}(1-G(x) )dx - int_0^{infty} G^2(x)dx-int_{-infty}^{0} (1-G(x))G(x)dx equiv 0$$