Why is $\int_{0}^{\infty} 0\, dx = 0$ and not indeterminate? It might seem that
\begin{align} \int_{0}^{\infty} 0\, dx &= 0\int_{0}^{\infty} dx\\
&= 0\cdot x\,\bigg\rvert_0^\infty\\
&= 0\cdot \infty
\end{align}
which is indeterminate. Right?
 A: Not quite. We are taking two limits--the integral, which itself is a limit, and the limits of definite integration. The order of these limits is important; in general, these limits do not commute. The fundamental theorem of calculus (or second fundamental theorem, depending on the textbook) says that
\begin{equation}\int_{a}^{b} f(x) \, dx = F(b) - F(a)\end{equation}
where $F(x)$ is the antiderivative of $f(x)$ evaluated at $a$ and $b$, respectively (with the usual assumptions of integrability, existence of $f(x)$ on the interval $(a,b)$, etc.). Furthermore,
\begin{align}\int_{a}^{\infty} f(x) \, dx &= \lim_{b\rightarrow\infty}\int_{a}^{b} f(x) \, dx\\
&= \lim_{b\rightarrow\infty}\left[F(b) - F(a)\right]
\end{align}
This limit has to be taken after we integrate. Therefore:
\begin{align} \int_{0}^{\infty} 0\, dt &=
\lim_{b\rightarrow\infty}\int_{0}^{b} 0\, dx\\
&= \lim_{b\rightarrow\infty}\left[0 \int_{0}^{b} \, dx\right]\\
&= \lim_{b\rightarrow\infty}[0\cdot b - 0\cdot 0]\\
&= \lim_{b\rightarrow\infty} 0\\
&= 0.
\end{align}
A: It depends.
With Riemann integration, the (improper) integral is defined as the
limit of $\int_0^L \cdots $ as $L \to \infty$ which will result in zero in this case.
With Lebesgue integration, we work in the extended reals and define $0 \cdot \infty = 0$. Then the integral is taken as the $\sup$ of
the integrals of simple, non negative functions majorised by the function $x \mapsto 0$. In this case, the function itself is the only suitable simple function and in that case we define
$\int 0 dx = 0 \cdot m[0,\infty) = 0$.
