What is the value of the integral $\int_0^\infty 0dx$? I know that this question might be very simple, but I want to know: what is the value of following integral?
$$\int_0^\infty0dx$$
 A: Hint:
$$\forall t>0: \int_0^t 0 \mathrm{d}x=0$$
A: A more intuitive angle: recall that
$$\int_a^b f(x)dx$$
can be analogized as the area between $f$ and the $x$ axis on the interval $[a,b]$. Take $a=0, b \to \infty,$ and $f(x)=0$. Then we have
$$\int_0^\infty 0 dx$$
or, essentially, the area between the line $y=0$ (the $x$-axis) and the $x$-axis, for $x\geq0$. In that light, it should be immediately clear there is $0$ area, and thus the integral is $0$.
A: The trick with such things is always: look at the definition, and just apply it carefully.


*

*By definition, $\int_0^\infty f(x)dx$ is just $\lim_{r\rightarrow\infty} \int_0^r f(x)dx$, so we just need to calculate $\int_0^r 0dx$ for an arbitrary real $r$.

*But $\int_0^r 0dx$ is always $0$ (you should already be comfortable with this).

*So $\int_0^\infty f(x)dx=\lim_{r\rightarrow\infty} \int_0^r f(x)dx=\lim_{r\rightarrow\infty}0=0$.

The reason $\int_0^\infty 0dx$ may seem mysterious at first is that it feels like the old "paradox": "What happens when you add infinitely many infinitely small quantities together?" But this "paradox" is also present in integration all the time: $\int_a^bf(x)dx$ is intuitively "the sum of the areas of infinitely many infinitely thin rectangles." So this is a problem that we've already resolved, and the fact that $\int_0^\infty0dx=0$ should be no more mysterious than the general behavior of integration (in fact, it should be less mysterious).
A: By one definition, the integral is the infinite limit of the sum of the areas of rectangles whose base is $\Delta x$ and whose height is $f(x_i)$. In this case $f(x_i)=0$. So $f(x_i)\Delta x=0$. Summing these, you get, of course $0$ and the limit is $0$.
