Calculus Improper Integral I have the following integral
$\int_0^\infty\frac{\partial}{\partial\alpha}(1-2x)dx$,
with $\alpha$ independent of $x$. Is this defined at all? If yes, is the answers zero? I keep on reading that a definite integral of zero is zero but I am not convinced. Does that follow from
$\int_0^\infty0dx=0\lim_{b\rightarrow\infty}\int_0^bdx=0\lim_{b\rightarrow\infty}x]_0^b=0$?
My problem with this calculation is that I have to multiply zero by infinity which I believe is not defined. Thank you for your time.
 A: In Lebesgue's theory, zeros times infinity equals zero, so your integral is well defined and its value is indeed 0: $$\int_{0}^{\infty} 0.dx = 0.\mu({\mathbb{R})}= 0$$
Careful, you can't tell $\lim{f(x).g(x)}$ with $\lim_{x\to\infty} f(x) = \infty$ and $\lim_{x\to\infty} g(x) = 0$ in general tough:
For instance


*

*for $g(x)=x^2$ and $f(x)= \frac{1}{x}$, $\lim_{x\to\infty} f(x).g(x) = \infty$

*for $g(x)=x$ and $f(x)= \frac{1}{x}$, $\lim_{x\to\infty} f(x).g(x) = 1$

*for $g(x)=x$ and $f(x)= \frac{1}{x^2}$, $\lim_{x\to\infty} f(x).g(x) = 0$

A: \begin{align*}
\int_0^\infty\frac{\partial}{\partial\alpha}(1-2x)dx
&=\int_0^\infty0dx\\
&=\lim_{b\rightarrow\infty}\int_0^b0dx\\
&=\lim_{b\rightarrow\infty}0\\
&=0
\end{align*}
You do not need to multiply zero by an infinity.
Instead you only need to know the limit of a constant function $f(b)=\int_0^b0dx=0$ is $0$ when $b\rightarrow\infty$.
A: Yes, the answer is $0$ and you are not multiplying $0$ by $\infty$
Note that  $$\int_0^\infty0dx=$$
$$\lim_{b\rightarrow\infty}\int_0^b 0dx=$$
$$\lim_{b\rightarrow\infty}0=0$$
