Improper double integral evaluation by changing the order of integration I was watching an MIT OCW recitation video about exchanging the order of integration on double integrals.
The example was:
$$\intop_{t=0}^{\frac{1}{4}}\intop_{u=\sqrt{t}}^{\frac{1}{2}}\frac{e^{u}}{u}\:\mathrm{d}u\:\mathrm{d}t$$, which easily solves to $1-\frac{\sqrt{e}}{2}$.
But something is bugging me. $\frac{e^u}{u}$ is not defined for $u=0$ and hence this is an improper integral, right?
So the question is: how do we know that this integral converges?
 A: This is not an answer at present as I need to find an appropriate definition of the improper Riemann integral on a rectangle. I have written that a certain function below is integrable as an improper Riemann integral, but this is based on some presumption of 'reasonableness' until I find the appropriate definition.
The interchange is straightforward to justify using the Lebesgue integral, so the result it true, however presumably the desire is to establish the result using the Riemann integral.
Let $f(t,u) = 1_{[\sqrt{t},{1 \over 2}]} (u) { e^u \over u}$, for $(t,u) \in R=[0,{1 \over 4}] \times (0, { 1 \over 2}]$.
Since $f(t,\sqrt{t}) = { e^{\sqrt{t}} \over \sqrt{t}}$, we see that $f$ is unbounded on $R$, hence the integral must be improper (as a Riemann integrable function must be bounded).
Note that $u \mapsto { e^u \over u}$ is decreasing on $(0,{ 1 \over 2}]$, hence
$f(t,u) \le { e^{\sqrt{t}} \over \sqrt{t}} \le {1 \over \sqrt{t}}$, for $(t,u) \in R$.
The function $(t,u) \mapsto {1 \over \sqrt{t}}$ is integrable as an improper Riemann integral.
