How to prove the below equality?

Question

We have $$I = \int_{\Bbb R} 3\sqrt{\mid x\mid}e^{-3\mid x-2\mid}dx$$ How to prove the following? $$I=\int_0^{+\infty} 3\sqrt{x+2}e^{-3x}dx+\int_0^{+\infty} 3\sqrt{\mid 2- x\mid}e^{-3x}dx$$

Answer 1

Perform the change of variables $t = x-2$, so that $$ I = \int_{\Bbb R} 3 \sqrt{|t+2|}e^{-3|t|}dt = \int_{0}^{+\infty} 3 \sqrt{|t+2|}e^{-3|t|}dt+ \int_{-\infty}^0 3 \sqrt{|t+2|}e^{-3|t|}dt$$

If $t\ge 0$, then $t+2\ge 0$ and you can remove absolute values.

If $t\le 0$, then $u = -t\ge 0$ (hence $u=|t|$) and $t+2 = 2-u$. Hence (using this new change of variables $u = -t$ on $\Bbb R^-$):

$$I = \int_{0}^{+\infty} 3 \sqrt{t+2}e^{-3t}dt+ \int_{+\infty}^0 3 \sqrt{|2-u|}e^{-3u}.(-du)$$ and thus $$I = \int_{0}^{+\infty} 3 \sqrt{t+2}e^{-3t}dt+ \int_0^{+\infty} 3 \sqrt{|2-u|}e^{-3u}.du$$

Finally, renaming $t$ and $u$ as $x$, one gets the desired result.