Is $f(x,t) = \sin (x) e^{-tx}$ in $L^1([0, \infty) \times [0, \infty))$?

Question

I have the integral $\displaystyle\int_0^\infty \int_0^\infty \sin (x) e^{-tx} \ dt \ dx.$ I want to switch the order of integration using Fubini's Theorem, which requires that the integrand be absolutely integrable.

I tried the following estimate: $|\sin (x) e^{-tx}| \leq e^{-tx}$.

However, $\displaystyle\iint_{[0, \infty) \times [0, \infty)} e^{-tx} = \infty$.

Answer 1

If $f(x,t) = \sin (x) e^{-tx}$ would belong to $L^1([0, \infty) \times [0, \infty))$, $$\int_o^\infty \left( \int_0^\infty \vert f(x,t) \vert dt \right) dx$$ would be finite.

However $$\int_o^\infty \left( \int_0^\infty \vert \sin x \ e^{-xt} \vert dt \right) dx = \int_0^\infty \frac{\vert \sin x \vert }{x} dx$$ diverges. So $f$ does not belong to $L^1$.

Answer 2

$\int e^{-tx}\mathrm{d}t = \frac 1 x$ , so your $\iint$ collapse to $\int \frac {\sin(x)} x \mathrm{d}x$