Inexplicable results of definite triple integral in three different symbolic maths software

Question

The results I'm getting from three different symbolic maths applications do not match my expectations. I have checked the result carefully and I still cannot explain the discrepancy. I half expect someone to find a flaw in my reasoning, because surely these three battle-tested applications cannot be wrong.

To make things simple I will be dropping the denominator in the rest of this post.

In all the applications that I have tried

$$ \int_{-1}^1 \int_{-1}^1 \int_{-1}^1 x^a y^b z^c \; dx \; dy \; dz $$

evaluates to zero, when $a = 2$, $b = 1$ and $c = 3$. Why this is so I cannot explain.

Here is my reasoning. If

$$ \int \int \int x^a y^b z^c \; dx \; dy \; dz = x^{(a + 1)}y^{(b + 1)}z^{(c + 1)} $$

Therefore the definite integral over the limits $[-1, 1]$ ought to be

$$ 1 - (-1)^{(a + 1)}(-1)^{(b + 1)}(-1)^{(c + 1)} $$

Therefore when $a = 2$, $b = 1$ and $c = 3$ the definite integral ought to be 2, not 0.

Furthermore, if it is true that

$$ 1 - (-1)^{(a + 1)}(-1)^{(b + 1)}(-1)^{(c + 1)} = (-1)^{(a + b + c)} + 1 $$

Then the definite integral ought to be 2 if $(a + b + c)$ is even.

Answer 1

The integrations should be performed separately.

$$\int_{-1}^1 \int_{-1}^1 \int_{-1}^1x^ay^bz^c\;dx\;dy\;dz$$

$$=\tfrac1{(a+1)(b+1)(c+1)}(1-(-1)^{a+1})\cdot(1-(-1)^{b+1})\cdot(1-(-1)^{c+1})$$

That's not the same thing as $$\tfrac1{(a+1)(b+1)(c+1)}(1-(-1)^{a+1 + b+1 + c+1})$$ which is what you appear to be computing. It looks like you are integrating each factor and evaluating the product of the antiderivatives. Nope.

In other words, $$\left[f(x)g(x)\cdots\right]_{x_0}^{x_1}\neq\left[f(x)\right]_{x_0}^{x_1}\cdot \left[g(x)\right]_{x_0}^{x_1}\cdots$$ in general.

Answer 2

Writing $$\int \int \int x^a y^b z^c \; dx \; dy \; dz = \frac{1}{a+1}x^{(a + 1)}\frac{1}{b+1}y^{(b + 1)}\frac{1}{c+1}z^{(c + 1)}$$ is kinda wrong (like said in the comments) since it's more like $$\int_{-1}^1 \int_{-1}^1 \int_{-1}^1 x^a y^b z^c \; dx \; dy \; dz = \left[\frac{1}{a+1}x^{(a + 1)}\right]_{-1}^1\cdot\left[\frac{1}{b+1}y^{(b + 1)}\right]_{-1}^1\cdot\left[\frac{1}{c+1}z^{(c + 1)}\right]_{-1}^1$$ See, the problem is those expressions will not yield the same result.

Answer 3

If you want to compute the $x$ integral first, then the $y$, then the $z$: $$ \int_{-1}^1 x^a y^b z^c\; dx = \left. \frac{1}{a+1} x^{a+1} y^b z^c \right|_{x=-1}^1 = \frac{\left(1 - (-1)^{a+1}\right)}{a+1} y^b z^c $$ Then integrate this from $y = -1$ to $1$, obtaining $$ \int_{-1}^1 \int_{-1}^1 x^a y^b z^c \; dx \; dy = \frac{\left(1 - (-1)^{a+1}\right)\left(1-(-1)^{b+1}\right)}{(a+1)(b+1)} z^c $$ and finally integrate this from $z=-1$ to $1$, obtaining $$ \int_{-1}^1 \int_{-1}^1 \int_{-1}^1 x^a y^b z^c \; dx \; dy\; dz = \frac{\left(1 - (-1)^{a+1}\right)\left(1-(-1)^{b+1}\right)\left(1-(-1)^{c+1}\right)}{(a+1)(b+1)(c+1)}$$ As others have noted, this is $0$ if any of $a,b,c$ are odd.