I know that this question might be very simple, but I want to know: what is the value of following integral?
$$\int_0^\infty0dx$$
I know that this question might be very simple, but I want to know: what is the value of following integral?
$$\int_0^\infty0dx$$
On
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 $0$.
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).