Let $H\in$ $]0,1[$. A fractional Brownian motion $\left(B_H(t)\right)_{t\geq 0}$ can be represented as ${1\over C(H)}\int_\mathbb{R}\left((t-s)_+^{H-{1\over2}}-(-s)_+^{H-{1\over2}}\right)dB(s)$ where $B(t)$ is the standard Brownian motion and $C(H)=\left(\int_0^{\infty}\left((1+s)^{H-{1\over2}}-(-s)^{H-{1\over2}}\right)^2ds+{1\over 2H}\right)^{1\over2}$
I would like to understand the following equation:
Let $X(t)$ be the integral above, then: $\mathbb{E}\left[X(t)^2\right]={1\over C(H)^2}\left[\int_{-\infty}^0\left((t-s)^{H-{1\over2}}-(-s)^{H-{1\over 2}}\right)^2ds+\int_0^t(t-s)^{2H-1}ds\right]={t^{2H}\over C(H)^2}\left[\int_{-\infty}^0\left((1-s)^{H-{1\over2}}-(-s)^{H-{1\over2}}\right)^2ds+\int_0^1(1-s)^{2H-1}ds\right]=t^{2H}$
My problem is what's happening in the brackets. The second summand is ok, but the first one, I don't know why it holds $\int_{-\infty}^0\left((t-s)^{H-{1\over2}}-(-s)^{H-{1\over 2}}\right)^2ds=t^{2H}\int_{-\infty}^0\left((1-s)^{H-{1\over2}}-(-s)^{H-{1\over2}}\right)^2ds$
Anybody help me to understand this, please.