Changing lower limits integral.

$\displaystyle\int_2^\infty\dfrac1{(x-1)^3}\,\mathrm dx\quad$ Let $u=x-1 \\ \mathrm du=\mathrm dx$

$\displaystyle=\int_1^\infty\dfrac{\mathrm du}{u^3}=\lim_{R\to\infty}\int_1^R\dfrac{\mathrm du}{u^3} \\\displaystyle =\lim_{R\to\infty}\dfrac{-1}{2u^2}\Bigg|_1^R=\lim_{R\to\infty}\left(\dfrac12-\dfrac1{2R^2}\right)=\dfrac12$

This is a suggested solution to the integral (line $1$). Why do they change the lower limit at line $2$ from $2$ to $1$?

• If $x=2,u=x-1=1$ – lab bhattacharjee Nov 28 '13 at 15:55

Since $u=x-1$, the lower bound is $x=2$, so $u=1$. The upper bound is $\infty-1=\infty$.
$$\int_{2}^{\infty}\frac{dx}{(x-1)^3}=|x-1=t\Rightarrow dx=dt, x=2\Rightarrow t=2-1-1; \infty-1=\infty|=\int_{1}^{\infty}\frac{dt}{t^3}$$ $$=\lim_{m\to\infty}\int_{1}^{m}\frac{dt}{t^3}=\lim_{m\to\infty}\left(-\frac{1}{2t^2}\right)|_{1}^{m}=\lim_{m\to\infty}\left(-\frac{1}{2m^2}+\frac{1}{2}\right)=\frac{1}{2},$$ because $$\lim_{m\to\infty}\left(-\frac{1}{2m^2}\right)=0$$