Bring the double integral to a single integral: $\iint\limits_{|x| + |y| \le 1} f(x + y) \, dx \, dy$

Question

Bring the double integral to a single integral: $$\iint\limits_{|x| + |y| \le 1} f(x + y) \,dx \,dy$$

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The substitution $u = x + y$ seems to be the only way since we should get rid of several variables in the function argument (also, I did $-x+y = v$ for the second variable). First, I split the integral into four: $$\iint\limits_{|x| + |y| \le 1} f(x + y) \,dx \,dy = \iint\limits_{0\le x+y\le1} f(x+y)\,dx\,dy + \iint\limits_{-1\le x+y\le 0} f(x+y)\,dx\,dy + \iint\limits_{0\le -x+y\le1} f(x+y)\,dx\,dy + \iint\limits_{-1\le -x+y\le0} f(x+y)\,dx\,dy$$ Now, with the substitution, we can deal with the first two: $$\iint\limits_{0\le x+y\le1} f(x+y)\,dx\,dy + \iint\limits_{-1\le x+y\le 0} f(x+y)\,dx\,dy = \int_0^1 \frac 12 f(u)du + \int_{-1}^0 \frac 12 f(u)du = \frac 12 \int_{-1}^1 f(u)\,du$$ where $\frac{1}{2}$ is the Jacobian. I don't know if I'm going right and do not know how to deal with the next integrals.

Answer 1

Your idea is correct but your answer is off by a factor of $2$ due to mistakes in working. When you use change of variable, the integral still remains a double integral but yes it can effectively be expressed as a single integral in this case with the below change of variable.

The region is a square formed by $\small |x|+|y| \leq 1$ or $\small -1 \leq x + y \leq 1, -1 \leq x-y \leq 1$. So, it makes sense to use substitution,

$\small x+y = u, x-y = v$. Jacobian is $\displaystyle \small\frac{1}{2}$.

The transformed region is $\small -1 \leq u \leq 1, -1 \leq v \leq 1$ and $\small f(x+y) = f(u)$. As integrand is only a function of $u$, we can convert the double integral to a single integral with respect to $du$, making use of the fact that the range of $v$ is $2$ throughout the region.

$I = \displaystyle \small \int_{-1}^{1} \int_{-1}^{1} f(u) \ |J| \ dv \ du = \int_{-1}^{1} 2 \cdot f(u) \cdot \frac{1}{2} \ du = \int_{-1}^{1} f(u) \ du$

Answer 2

The part in the first quadrant is not $0<x+y<1,$ but $\Big[ \, x>0\ \&\ y>0\ \&\ x+y<1 \, \Big].$ And similarly for the other quadrants. $$ \iint\limits_{\begin{smallmatrix} x\,>\,0 \\ y\,>\,0 \\ x\,+\,y\,<\,1 \end{smallmatrix}} \cdots = \iint\limits_{\begin{smallmatrix} \,\, 0\,<\,u\,<\,1 \\[2pt] -u\,<\,v\,<\,+u \end{smallmatrix}} \cdots =\int_0^1 \left( \int_{-u}^{+u} \cdots \, dv \right) \, du $$