Introduction
The iterated integral
does not converge absolutely, i.e. the integral of the
absolute value is not finite:
Fubini's theorem tells us that if the integral of the absolute
value is finite, then the order of integration does not matter;
if we integrate first with respect to x and then with respect
to y, we get the same result as if we integrate first with
respect to y and then with respect to x. The assumption
that the integral of the absolute value is finite is
"Lebesgue integrability". That the
assumption of Lebesgue integrability in Fubini's theorem
cannot be dropped can be seen by examining this particular
iterated integral. Clearly putting "dx dy" in place
of "dy dx" has the effect of multiplying the value of
the integral by −1 because of the "antisymmetry" of the
function being integrated. Therefore, unless the value of the
integral is zero, putting "dx dy" in place of
"dy dx" actually changes the value of the integral.
That is indeed what happens in this case.
How to evaluate this integral
The integral
can be evaluated via the trigonometric substitution
- y = xtan(θ),
- x2 + y2 = x2 + x2tan2(θ) = x2(1 + tan2(θ)) = x2sec2(θ).
The bounds of integration can be found thus:
The integral then becomes
![=\frac{1}{x}\int_0^{\arctan(1/x)} \cos^2(\theta)-\sin^2(\theta)\,d\theta =\frac{1}{x}\int_0^{\arctan(1/x)} \cos(2\theta)\,d\theta =\frac{1}{x}\left[\frac{\sin(2\theta)}{2} \right]_{\theta:=0}^{\theta=\arctan(1/x)}](a/../upload/math/76bdbbeb499ac4903b6db462fa37278f.png)
![=\frac{1}{x}\left[\sin(\theta)\cos(\theta) \right]_{\theta:=0}^{\theta=\arctan(1/x)} =\frac{1}{x}\sin(\arctan(1/x))\cos(\arctan(1/x)).](a/../upload/math/de363925e262529bd3912b8cd575310a.png)
Now recall the trigonometric identities
The expression above then becomes
This takes care of the "inside" integral with respect to y;
now we do the "outside" integral with respect to x:
![\int_0^1\frac{1}{1+x^2}\,dx =\left[\arctan(x)\right]_0^1 =\arctan(1)-\arctan(0)=\frac{\pi}{4}.](a/../upload/math/59b4e4a0a3a90c576f2bc88aee7f682f.png)
Thus we have
and
The moral of the story
When
then the two iterated integrals
may have different finite values.