Antiderivative of $g(x)dg(x)$

Question

I am reading a book by Shreve "Stochastic Calculus for Finance II" and after computing a stochastic integral $\int_{0}^{T}W(t)dW(t)$ where $W(t)$ is a Brownian motion he compares it to the integral $$\int_{0}^{T}g(t)dg(t) = \int_{0}^{T}g(t)g^\prime (t)dt = 0.5g^2(T),$$ where $g(t)$ is a differentiable function with $g(0)=0$. I don't get the fact that $\int g(t)g^\prime (t)dt = 0.5g^2(t)$. For me the right hand side is equal to $\int g(t)dt$, without the $g^\prime (t)$ term. Thanks in advance.

Answer 1

You seem to be confusing $\int g(t)\, dt$ with $\int g(t)\, dg(t)$. It might help to make the substitution explicitly.

Let $y = g(t)$. Then $dg(t) = dy = g'(t) dt$, and so substituting $y=g(t)$ in the integral gives $$\int_{0}^{T} g(t)\, dg(t) = \int_{t=0}^{t=T} y\, dy = \left[ \dfrac{y^2}{2} \right]_{t=0}^{t=T} = \left[ \dfrac{g(t)^2}{2} \right]_0^T = \dfrac{g(T)^2}{2}$$

Answer 2

Just as $\int g^\prime dt=g+C$, $\int f^\prime(g)g^\prime dt=f(g)+C$ by the chain rule. The case at hand is $f=g^2/2$.