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Is an object like $$\int^0_0 x^{-1} dx$$ defined?

Context: when considering functions such as

$$f(x)=\int^{x^2}_xt^{-1}dt$$

would it therefore be necessary to give a piecewise definition of $f$ if we wished to include $0$ in the domain?

A little bit of random speculation, but if the object is supposedly defined to be $0$, then we would be allowed to write something like

$$\int^0_0\sqrt{\text{banana}}\quad d(\text{apple})=0$$

???

Shuri2060
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  • If I'm not wrong, that this $f$ is not defined for any $x$ right? – Yanko Jul 30 '17 at 20:01
  • Nice question! For me $f(x)=\int_{0}^{x}t^{-1}dt$ is not a function, but someone else can say that it's a function with domain $dom(f)={0}$ and image $Im(f)={0}$. – Ixion Jul 30 '17 at 20:03
  • I assume it depends on your definition. Some would say $\int_a^a f,dx \equiv 0$ which would be valid, whereas if you leave it more general $\int_a^b f,dx \equiv F(b)-F(a)$ and consider $a=b$, then $F(b)-F(a)$ is not well-defined. Just a thought tho! – Dando18 Jul 30 '17 at 20:03
  • @yanko I put a better example – Shuri2060 Jul 30 '17 at 20:04
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    hmm, maybe you want to integrate from $x^2$ to $x$ because when $x$ is close to $0$ then $x^2<x$? – Yanko Jul 30 '17 at 20:05
  • https://en.wikipedia.org/wiki/How_many_angels_can_dance_on_the_head_of_a_pin%3F –  Jul 30 '17 at 20:05
  • you can consider the limit at 0, by the fundamental theorem this limit is equal to the limit of $log(x)-log(x^2)=-log(x)$ which is infinity – Yanko Jul 30 '17 at 20:06
  • @MathematicsStudent1122 Thanks - just what I was looking for! This should be closed then – Shuri2060 Jul 30 '17 at 20:07
  • Hm.. so there are some definitions which define it to be $0$. But then this allows for ridiculous examples such as the one I've put below? – Shuri2060 Jul 30 '17 at 20:17
  • Perhaps more ridiculously, we can have integrals such as $\int_{-1}^1\frac1x~\mathrm dx$, though this is a much different matter. – Simply Beautiful Art Jul 30 '17 at 20:28
  • The function in the integral can perfectly be undefined in a set of measure zero. $\int_{0}^{0}f(x)dx=\int_{-\infty}^{\infty}f(x)\chi_{{0}}dx$ is zero for every function $f$, even if not defined at $0$. – Peyton Jul 30 '17 at 20:34

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