Formulating Fubini's Theorem for Lebesgue Integrable Functions

[evinda]

Hello!

I want to use Fubini's theorem for $L^1(\mathbb{R}^n \times \mathbb{R}^m)$ functions in order to find the integral $\int_0^{+\infty} \frac{\sin x}{x} \left( \frac{1-e^{-x}}{x}-e^{-x} \right) dx$.

There is a hint that we should find $\int_{(0,+\infty) \times (0,1)} y \sin x e^{-xy} dx dy$.$\int_{(0,+\infty) \times (0,1)} y \sin x e^{-xy} dx dy= \int_0^{1} \int_0^{+\infty} y \sin x e^{-xy} dx dy=\int_0^{+\infty} \int_0^{1} y \sin x e^{-xy} dy dx$

$\int_0^1 y e^{-xy} dy=-\frac{e^{-x}}{x}-\frac{e^{-x}}{x^2}+\frac{1}{x^2}$

So, $\int_{(0,+\infty) \times (0,1)} y \sin x e^{-xy} dx dy=\int_0^{\infty} \sin x \left( -\frac{e^{-x}}{x}-\frac{e^{-x}}{x^2}+\frac{1}{x^2} \right) dx$.

So now we got the wanted integral. How do we continue?

 

[I like Serena]

Hello!

I want to use Fubini's theorem for $L^1(\mathbb{R}^n \times \mathbb{R}^m)$ functions in order to find the integral $\int_0^{+\infty} \frac{\sin x}{x} \left( \frac{1-e^{-x}}{x}-e^{-x} \right) dx$.

There is a hint that we should find $\int_{(0,+\infty) \times (0,1)} y \sin x e^{-xy} dx dy$.$\int_{(0,+\infty) \times (0,1)} y \sin x e^{-xy} dx dy= \int_0^{1} \int_0^{+\infty} y \sin x e^{-xy} dx dy=\int_0^{+\infty} \int_0^{1} y \sin x e^{-xy} dy dx$

$\int_0^1 y e^{-xy} dy=-\frac{e^{-x}}{x}-\frac{e^{-x}}{x^2}+\frac{1}{x^2}$

So, $\int_{(0,+\infty) \times (0,1)} y \sin x e^{-xy} dx dy=\int_0^{\infty} \sin x \left( -\frac{e^{-x}}{x}-\frac{e^{-x}}{x^2}+\frac{1}{x^2} \right) dx$.

So now we got the wanted integral. How do we continue?

Hey evinda! (Smile)

How about evaluating the integral the other way around?
That is, starting with:
$$\int_0^{+\infty} y \sin x e^{-xy} dx$$
(Wondering)

 

[evinda]

Hey evinda! (Smile)

How about evaluating the integral the other way around?
That is, starting with:
$$\int_0^{+\infty} y \sin x e^{-xy} dx$$
(Wondering)

I found that $\int_0^{+\infty} y \sin x e^{-xy} dx=\frac{y}{1+y^2}$ and so $\int_0^1 \int_0^{+\infty} y \sin x e^{-xy} dx dy=\frac{1}{2}$.

Am I right?

So from Fubini's theorem, the initial integral is equal to $\frac{1}{2}$, right?

Could you also help me formulate Fubini's theorem for Lebesgue integrable funtions?