Distribution of X-Y for identical independent random variables

I have X and Y which are independent and both have an exponential distribution with density function $f(x) = e^{-x}$ if $x\gt0$

I want to find the distribution of X+Y and X-Y. Let U=X+Y, V=X-Y

My approach: For U, to find the cumulative distribution, I integrated the joint density function $e^{-x}e^{-y}$ for $0 \lt x \lt u-y$ and $0 \lt y \lt u$ to get $1-e^{-u} - ue^{-u}$ which seems correct to me given that $0 \lt U \lt \infty$

But for X-Y i am struggling to find the appropriate bounds.

I tried the following: $$F_V(v) = \int_0^{\infty}\int_0^{v+y}e^{-x}e^{-y}dxdy = 1-e^{-v}/2$$

However, from my understanding, V is not bounded from below and hence this function does not make any sense as a cumulative distribution function.

The standard approach is to use convolution; using the CDF approach: $$Z=X-Y$$ $$F_Z(z)=P(X-Y \le z)=\int_a^\infty P(X-y\le z|Y=y)f_Y(y)dy$$ $$=\int_a^\infty P(X\le y+z)e^{-y}dy=\int_a^\infty \left(\int_0^{y+z} e^{-x}dx\right)e^{-y}dy$$ $$=\int_a^\infty \left(1- e^{-y-z}\right)e^{-y}dy$$

If $z\ge 0$, then $a=0$.

If $z<0$, then $a=-z=|z|$, because $Z$ takes negative value $z$ if only $Y\ge|z|$.