I am studying something and encountered this:  

"
Let $R(\theta,T) = \int\_{-T}^{T} \frac{(\sin \theta t)}{t}dt, S(T) = \int\_0^T\frac{(\sin x)}{x}dx$, then for $\theta > 0$ and changing variables $t=x/\theta $ shows that 

$R(\theta,T)=2\int\_0^{T\theta}\frac{\sin x}{x}dx = 2S(T\theta)$ while for $\theta<0$, $R(\theta,T) = -R(|\theta|,T)$" which I don't understand. 

If $\theta<0$, then $R(\theta,T)=2\int\_{-T|\theta|}^{0}\frac{\sin x}{x}dx =2\int\_0^{T|\theta|}\frac{\sin x}{x}dx = R(|\theta|,T)$ as $\frac{\sin x}{x}$ is an even function, right? I am missing something simple here, thanks and appreciate an explanation.