# $\int_0^1\int_x^1f(t) \,dt \,dx=\int_0^1t f(t) \,dt$

Lets suppose that f is continuous in $$[0,1]$$. I want to prove that $$\int_0^1\int_x^1f(t) \,dt \,dx=\int_0^1t f(t) \,dt$$ for the left part I set the following function: $$F(y)=\int_0^y f(t)\,dt$$ so from the left part I have that $$\int_0^1\int_x^1f(t)\,dt\,dx$$ $$=\int_0^1(F(1)-F(x))\,dx$$ $$=F(1)-\int_0^1F(x)\,dx$$ $$=F(1)-F(0)-\int_0^1F(x)\,dx$$ Then, I tried to solve the right part by integrating by parts but I got confussed because of the limits of integration. $$u=t$$, $$du=dt$$, $$v=\int f(t)\,dt$$, $$dv=f(t)$$.

Here is were I´am stuck, can you help me?

• My answer below is devoted to the question of what happens to the bounds of integration when the order is reversed. From there you can probably do the rest. Commented Jul 19, 2021 at 19:13

Change the order of integral.

$$\displaystyle \int_0^1\int_x^1 f(t) \ dt \ dx = \int_0^1\int_0^t f(t) \ dx \ dt$$

• quick and clean. Commented Jul 19, 2021 at 17:10
• Assuming we aren't in some pathological function where somehow Fubini's doesn't apply....hrm, I wonder if you can make such a function that it doesn't....
– Alan
Commented Jul 19, 2021 at 17:37
• @Alan good point. There may be counterexamples but in that case the function itself won't be integrable over the region is what I think. Commented Jul 19, 2021 at 17:42
• @MathLover Probably, and I'm pretty certain at the level a question like this would be asked, Fubini's wouldn't even be mentioned, just assumed
– Alan
Commented Jul 19, 2021 at 17:43

Edit : since OP doesn't precise what is the regularity of $$f$$, I went ahead and assumed it is continuous.

Let $$F(x) = \int_x^1 f(t)\,\text dt$$. Then, $$F$$ is $$C^1$$ and $$F'(x) = -f(x)$$. Integration by part gives : $$\int_0^1 F(x) \,\text dx = [xF(x)]_0^1 - \int_0^1xF'(x) \, \text dx$$ Using $$F(1) = 0$$, we get : $$\int_0^1\int_x^1f(t)\, \text dt=\int_0^1 x f(x)\,\text dx$$

• You say $F$ is $C^1,$ but $F$ is not differentiable at points where $f$ has a jump. Commented Jul 19, 2021 at 19:14

I will assume that $$f$$ is absolutely integrable so that we can apply Fubini. Using Iverson Brackets: \begin{align} \int_0^1\int_x^1f(t)\,\mathrm{d}t\,\mathrm{d}x &=\int_0^1\int_0^1f(t)[t\gt x]\,\mathrm{d}t\,\mathrm{d}x\tag1\\ &=\int_0^1\int_0^1f(t)[t\gt x]\,\mathrm{d}x\,\mathrm{d}t\tag2\\ &=\int_0^1tf(t)\,\mathrm{d}t\tag3 \end{align} Explanation:
$$(1)$$: write the inner integral using the indicator function of $$t\gt x$$
$$(2)$$: change order of integration
$$(3)$$: evaluate the integral in $$x$$

\begin{align} & \int_0^1 \left( \int_x^1 \cdots\,dt \right) \,dx \\[8pt] = {} & \iint\limits_{t,x\,:\,0\,<\,x\,<\,t\,<\,1} \cdots \,d(t,x) \\[8pt] = {} & \int_0^1 \left( \int_0^t \cdots \, dx \right) \,dt \end{align}