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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.

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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 $f(x)= \frac{1}{x^2}$, $\lim_{x\to\infty} f(x).g(x) = 0$
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  • $\begingroup$ You don't need to talk about $0$ times infinity. The function that's identically $0$ is integrable and its integral is $0$. $\endgroup$ – Ethan Bolker May 11 '18 at 14:25
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\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 $b\rightarrow\infty$.

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Yes, the answer is $0$ and you are not multiplying $0$ by $\infty$

Note that $$\int_0^\infty0dx=$$

$$\lim_{b\rightarrow\infty}\int_0^b 0dx=$$

$$\lim_{b\rightarrow\infty}0=0$$

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