Proving $\int_0^\infty\frac{\sin(x)}x\ dx=\frac{\pi}2$. Why is this step correct?

Let $x = (\lambda + \frac{1}{2}) t$. Then $$\int_0^T \frac{\sin((\lambda + \frac{1}{2}) t)}{t} \mathrm{d}t = \int_0^{(\lambda + \frac{1}{2}) T} \frac{\sin(x)}{\frac{x}{\lambda + \frac{1}{2}}}\frac{\mathrm{d}x}{(\lambda + \frac{1}{2})}=\int_0^{(\lambda + \frac{1}{2}) T} \frac{\sin(x)}{x} \mathrm{d}x $$ Take the limit: $$\lim_{\lambda\to\infty} \int_0^{(\lambda + \frac{1}{2}) T} \frac{\sin(x)}{x} \mathrm{d}x = \int_0^{\infty} \frac{\sin(x)}{x} \mathrm{d}x $$