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Consider an improper integral:

$$ \int_{\infty}^{-\infty}f(x) = \int_{-\infty}^{n}f(x) + \int_{n}^{\infty}f(x) $$

According to Paul's math notes:

Note as well that this requires BOTH of the integrals to be convergent in order for this integral to also be convergent. If either of the two integrals is divergent then so is this integral.

Why is this necessarily the case? Imagine an $f(x)$ that was symmetric diagonally over the y and x axis, diverging in an equal but opposite way on right and left. Wouldn't the total area under the curve equal 0 due to the symmetry?

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    $\begingroup$ What you are thinking would be: $\int_{-\infty}^{\infty} f = \lim_{n \to \infty} \int_{-n}^{n} f$. The two definitions are not the same. $\endgroup$
    – J126
    Jan 25 at 22:47
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    $\begingroup$ Think of it this way: why is $\infty-\infty$ undefined and not equal to zero? $\endgroup$
    – Zeekless
    Jan 26 at 0:01

2 Answers 2

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The short answer is: because that's the definition. We have to be clear about what the definitions are and what follows from them. $\int_{-\infty}^{\infty}$ equals $\int_{-\infty}^{n}+\int_{n}^{\infty}$ in case the two integrals on the right converge, and in case at least one of those integrals diverge, then $\int_{-\infty}^{\infty}$ diverges -- and we should make it clear that's the definition, not a theorem, so one can't ask why it is true.

That said, a legitimate question still stands which is why we chose such a definition. First of all, it should be noted that a definition which is closer to what you expect (where integrating any odd function such as $f(x)=x$ on the whole real line always equals $0$) does exist, but it goes by a different notation. It is called Cauchy's principal value and it is usually denoted by something like $PV\int_{-\infty}^{\infty}$. This definition goes as follows: $$PV\int_{-\infty}^{\infty} f(x)dx = \lim_{t\to\infty} \int_{-t}^t f(x)dx$$ so for instance $$PV\int_{-\infty}^{\infty} xdx = \lim_{t\to\infty} \int_{-t}^t xdx = \lim_{t\to\infty} 0 = 0$$

This definition is indeed useful in some contexts; and the usual definition is useful in other contexts.

Both of the definitions make intuitive sense: the PV definition makes some sense in a geometric way as you have noted, and the regular definition also plays well with intuition because if any one part diverges it does make sense to say that the whole diverges. Of course there is no point arguing about which of the two is "correct"-- they are definitions. The real point is that they are both useful, depending on what one is trying to say.

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  • $\begingroup$ Thanks. When/why is the Cauchy principal value not helpful extension of improper integrals-- i.e. in which cases would it motivate towards the standard definition? $\endgroup$
    – Ben G
    Jan 25 at 23:25
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    $\begingroup$ @bgcode Notice the excellent point in Bananach's answer: the PV definition doesn't always make geometric sense. We have $PV\int_{-\infty}^{\infty}xdx=0$ but also $PV\int_{-\infty}^{\infty}x+1 dx=\infty$ so that we got a different value just by moving the function one unit to the left, which doesn't make sense and certainly doesn't seem like a desirable property for an integral. With the usual definition that won't happen. PV treats $0$ as a special point, the regular definition treats all points as equals. 99% of the time you'll see the regular integral, only in special cases does PV come in. $\endgroup$
    – Snaw
    Jan 26 at 0:14
  • $\begingroup$ @bgcode see also the answers here math.stackexchange.com/q/2450848/998310 and this might also be useful sites.pitt.edu/~jwheeler/Principal%20Values.pdf $\endgroup$
    – Snaw
    Jan 26 at 0:25
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It's defined this way because it would often be unhelpful to work with a notion of integral that changes when a function is translated along the $x$ axis.

Imagine you defined $$ \int f(x) dx=\lim_{K\to\infty} \int_{0}^{K} f(x)dx +\int_{-K}^{0}f(x) dx. $$ Then you would have $\int f(x)dx=0$, but $\int \tilde{f}(x) dx$ would be undefined, where $f(x) =\sin(x)$ and $\tilde{f}(x)=f(x+\pi/2)$.

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    $\begingroup$ Yes it should, thanks $\endgroup$
    – Bananach
    Jan 25 at 23:51
  • $\begingroup$ But ~f could be defined in term of f? You could do some rearrangement to make it defined? Compare this to not having an answer for either f or ~f's integral.. $\endgroup$
    – Ben G
    Jan 30 at 21:44

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