# Iterated Integral With Odd Upper-Bound

Let $$f:[0,1]\rightarrow \mathbb{R}$$ be an Lebesgue integrable function on $$[0,1]$$. I would like to compute the following integral but I believe that I may be wrong when applying Fubini's theorem $$\int_0^{\sqrt{t}}\int_0^{\sqrt{v}} f(u)dudv.$$

Here's what I have computed (using the change of variables formulate formula and Fubini I "computed" that) \begin{aligned} \int_0^{\sqrt{t}}\int_0^{\sqrt{v}} f(u)dudv = & \int_0^{\sqrt{t}}\int_0^{v} f(\sqrt{u}) \frac1{2\sqrt{u}}dudv\\ = & \int_0^{t} \int_0^{\sqrt{u}}f(\sqrt{v}) \frac1{2\sqrt{v}}dvdu\\ = & \frac1{2}\int_0^{t} \int_0^{u}f(\sqrt{\sqrt{v}}) \frac1{\sqrt{\sqrt{v}}} \frac1{2\sqrt{v}}dvdu\\ = & \frac1{4}\int_0^{t} \int_0^{u}f(\sqrt[4]{v}) v^{-\frac{3}{4}}dvdu. \end{aligned} Is this correct? I feel as though something fishy happened when interchanging the integrals wrt. the bounds of integration.

• Since the integrals have no dependence on $v$, why not reduce it to $\sqrt{t}$ times a single integral over $u$? – Michael Jul 11 at 14:20
• On your work, I cannot follow it. For example your first substitution seems to be missing a factor of $2$: $$\int_0^{\sqrt{t}}f(u)du=\int_0^{t}\frac{f(\sqrt{w})}{2\sqrt{w}}dw$$ – Michael Jul 11 at 14:28
• Even putting in the factor of 2, I cannot follow your second equal sign at all. Are you trying to do multiple steps at once, such as doing an operation and then deciding to randomly switch the names of your variables? It is easier to read if you only do one step at a time. – Michael Jul 11 at 14:36
• @Michael Hi, I clarified some of the notation, I wanted second to depend on the first and I added a couple steps. Let me know if it's better. – N00ber Jul 11 at 14:54
• It seems you also changed the problem from $\int_{0}^{\sqrt{t}}\int_{0}^{\sqrt{t}}$ to $\int_{0}^{\sqrt{t}}\int_{0}^\sqrt{v}$. Nevertheless I still cannot follow your second equation, how do you justify that? It looks like you are randomly swapping $u$ and $v$. – Michael Jul 11 at 16:29

$$\int_0^{\sqrt{t}}\int_0^{v} f(\sqrt{u}) \frac1{2\sqrt{u}}dudv = \int_0^{t} \int_0^{\sqrt{u}}f(\sqrt{v}) \frac1{2\sqrt{v}}dvdu$$
The region of integration is the triangle whose vertices are $$(u=0,v=0)$$, $$(u=0,v = \sqrt{t})$$, and $$(u=\sqrt{t},v=\sqrt{t})$$. To cover this triangle using the other order of integration, you have to integrate $$v$$ from $$u$$ to $$\sqrt{t}$$, then $$u$$ from $$0$$ to $$\sqrt{t}$$. Additionally, $$f$$ is still a function of $$u$$ regardless of the order of integration. So you should have $$\begin{multline} \int_0^{\sqrt{t}}\!\!\!\int_0^{\sqrt{v}}\! f(u)dudv = \int_0^{\sqrt{t}}\!\!\!\int_0^{v}\! \frac{ f(\sqrt{u})}{2\sqrt{u}}dudv = \int_0^{\sqrt{t}}\!\!\!\int_u^{\sqrt{t}}\!\frac{f(\sqrt{u}) }{2\sqrt{u}}dvdu \\ = \int_0^{\sqrt{t}}\!\frac{f(\sqrt{u}) }{2\sqrt{u}}\left[\int_u^{\sqrt{t}}\!dv\right]du = \int_0^{\sqrt{t}}\!\frac{f(\sqrt{u})}{2\sqrt{u}}(\sqrt{t}-u)du = \int_0^{t^{1/4}}\!f(x)(\sqrt{t}-x^2)dx \end{multline}$$
• @N00ber : Eyeballfrog's point Additionally, $f$ is still a function of $u$ regardless of the order of integration nicely identifies the main issue in that second equality you wrote. It may be a helpful reminder to note that (under the mild conditions required for Fubini-Tonelli) we have $$\int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty} g(u,v)du\right]dv = \int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty} g(u,v)dv\right]du$$ and in your case $g(u,v) = f(u) 1_{\{(u,v):0<v<\sqrt{t}, 0<u<\sqrt{v}\}}$. In particular we keep it as $g(u,v)$ and do not write $g(v,u)$. – Michael Jul 11 at 17:08