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Can someone help me understand this a bit better:

$\int (x-y)^2 dx = \int(y-x)^2dx$

as $(y-x)^2 = (x-y)^2$. Now, if I make the change $z = x-y$ in the one on the LHS I get:

$\int z^2 dz$

as $dz = dx$. Now, if I make the change $z=y-x$ in the one on the RHS I get:

$\int z^2 (-dz)$

as $dz = -dx$. This implies that

$\int z^2 dz = - \int z^2 dz = 0$

which is clearly not true. I don't understand - can someone help me understand changing the variable of integration when you have a square or a square root please?

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In the first case, $z=x-y$. In the second case $z=y-x$. It doesn't follow that $\int z^2 dz = - \int z^2 dz = 0$, because these are two different $z$'s. The confusion comes from the fact that you used the same variable name for two different values. Look at it again, but with a different variable for the second case:

$$z=x-y$$

$$u=y-x$$

$$\begin{align} \int{z^2}dz&=-\int{u^2}du \\ \frac{1}{3}z^3+C &= -\frac{1}{3}u^3+C \end{align}$$

This follows because $z=-u$.

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  • $\begingroup$ That makes sense. So what about a Gaussian integral: $\int exp[-(x-y)^2] dx = \int exp[-(y-x)^2] dx$ $\endgroup$ – John Raddison Jul 18 '13 at 20:01
  • $\begingroup$ @JohnRaddison The exact same principle applies. The Gaussian is an even function, so its integral will be an odd function, just like in the case you gave. $\endgroup$ – Ataraxia Jul 18 '13 at 20:04
  • $\begingroup$ $z = x-y$ & $u = y-x$ $\int \exp[-z^2/2] dz = - \int \exp[-u^2/2] du $ $\sqrt(2 \pi) = - \sqrt(2 \pi)$ $\endgroup$ – John Raddison Jul 18 '13 at 20:07
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You are using $z$ on both sides of the equation as if it contains the same value, when in fact it does not.

On the LHS, $z = x-y$, and on the RHS, $z = y-x$. These are two different z values. To clarify, you should use $z1 = x-y$ and $z2 = y-x$, which makes it obvious that $z1 = -z2$.

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With a definite integral you get

$$\int_a^b (x-y)^2\,\mathrm dx=\int_{a-y}^{b-y}z^2\,\mathrm dz $$ and $$\int_a^b (y-x)^2\,\mathrm dx=\int_{y-a}^{y-b}z^2\,\mathrm d(-z) =\int_{y-b}^{y-a}z^2\,\mathrm dz$$ which both give the same value $\frac13(b-y)^3-\frac13(a-y)^3$.

As you also tagged the question as measure-theory (presumably Lebesgue integral), note that in that case you need to take the absolute value of the derivative in substitution, i.e. again you get $$\int_A z^2\,\mathrm dz=\int_{-A}z^2\,\mathrm dz$$

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