Is this integral indeterminate?

Question

What is:

$$\int_{-\infty}^{+\infty} x \, \mathrm{d}x ?$$

Is it $0$ or is it indeterminate ($+\infty - \infty$)?

Answer 1

It is indeterminate. But it is quite interesting to see why it is indeterminate. If you defined it to be the limit of the integral with extremes $a$ and $-a$ for $ a \to \infty$, then it would be zero. But that is not correct because, long story short, when you take a limit you don’t want the result to depend on how you arrived at that “limit point”. For example, consider it as a function of two variables:

$$ F(a,b)= \int_{a} ^b x dx $$

Then your integral would be the limit of this function as $a \to -\infty$ and $b \to \infty$. Check yourself that this limit doesn’t exist because it depends on how you approach to $\infty$ and $-\infty$.

Of course there are generalisations, or situations where you have to approximate things in a certain way (see the principal value distribution, for instance), but generally speaking that integral is not definite

Answer 2

You have to use a measure theory (Lebesgue for example) but this is divergent .

So you can use the distributions theory (as the Dirac distribution for example) and for a odd function, the result is $0$

Answer 3

This integral is improper but does not exist. For an integral of that kind to exist there should be a real $a$ such as both $\int\limits_{-\infty}^{a} x \, \mathrm{d}x$ and $\int\limits_{a}^{+\infty} x \, \mathrm{d}x $ are real numbers. But as we see : $\int\limits_{a}^{+\infty} x \, \mathrm{d}x =\lim_{N\to \infty}{N^2/2-a^2/2}=\infty$ ....

Answer 4

Suppose that we define $$ \int_{-\infty}^\infty x dx =\lim_{a\rightarrow \infty}\Big(\int_{-a}^a xdx\Big) $$ because by symmetry we would like the integral to be $0$. For the same reason, we expect that: $$ \int_{-\infty}^\infty (x+1) dx =0 $$ since it is also symmetric for $x=-1$. Let us apply the same definition of above: \begin{multline} \int_{-\infty}^\infty (x+1) dx =\lim_{a\rightarrow \infty}\Big(\int_{-a}^a (x+1)dx\Big)=\lim_{a\rightarrow \infty}\Big(\int_{-a}^a xdx + \int_{-a}^a dx\Big) =\lim_{a\rightarrow \infty}\Big(\frac{x^2}{2}\Big|_{-a}^a+x|_{-a}^a\Big)\\=\lim_{a\rightarrow \infty}2a=\infty\,, \end{multline}

So I think that the correct answer is that it is not convergent, and as spotted before the two distinct extremes are necessary.

Answer 5

https://en.wikipedia.org/wiki/Improper_integral :

A limitation of the technique of improper integration is that the limit must be taken with respect to one endpoint at a time. Thus, for instance, an improper integral of the form $$\int_{-\infty}^{\infty}f(x)dx$$ can be defined by taking two separate limits; to wit $$\int_{-\infty}^{\infty}f(x)dx = \lim_{a \rightarrow - \infty} \lim_{b \rightarrow \infty} \int_{-a}^{b}f(x)dx$$provided the double limit is finite. It can also be defined as a pair of distinct improper integrals of the first kind: $$\lim _{a\to -\infty }\int _{a}^{c}f(x)\,dx+\lim _{b\to \infty }\int _{c}^{b}f(x)\,dx$$ where c is any convenient point at which to start the integration. This definition also applies when one of these integrals is infinite, or both if they have the same sign.
An example of an improper integrals where both endpoints are infinite is the Gaussian integral $ \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}$. An example which evaluates to infinity is $ \int _{-\infty }^{\infty }e^{x}\,dx$. But one cannot even define other integrals of this kind unambiguously, such as $ \int _{-\infty }^{\infty }x\,dx$, since the double limit is infinite and the two-integral method $$ \lim _{a\to -\infty }\int _{a}^{c}x\,dx+\lim _{b\to \infty }\int _{c}^{b}x\,dx$$ yields $ \infty -\infty$. In this case, one can however define an improper integral in the sense of Cauchy principal value: $$\int _{-\infty }^{\infty }x\,dx=\lim _{b\to \infty }\int _{-b}^{b}x\,dx=0.$$

Note that the Cauchy Principal Value fails to have some basic properties that we would expect of integrals, such as subtitution of variables. If we define $u=x+1$, then $x = u-1$ and $dx = du$, and the limits stay the same (since the value of $u$ when $x$ is $\pm \infty$ is $\pm \infty$), so we have $$\int_{-\infty}^{\infty}(u-1)du$$, and the PV of that will not be zero.

Answer 6

The improper integral $$ \int_{-\infty}^{\infty}x\ dx $$ is not defined, whereas the principal value integral $$ \textrm{p. v.}\int_{-\infty}^{\infty}x \ dx $$ exists and equals zero (as is the case for any odd function). Thus, the answer depends on your definitions.

Answer 7

@ Milo Brandt : Thanks for your remark . The distribution is in fact relevant but it's a little difficult to describe in few words .

Answer 8

In mathematics we have to define or postulate existence of objects before use them . Often we have to extend the theory . For example the theory of distribution can legitimate and explain your result . It will be too long to develop the theory so i only try to do the taste . let f(x) = x or f is an arbitrary continuous function from R to R . f is a continuous function , so the integral of f is define on every [a,b] . For the integral , you can take the Riemann integral for the taste . If g is a continuous function ( in fact infinitely derivable but we don't need it for the taste )and if g is 0 outside some [a,b] , we say that g is define with a compact support . Let D the set of all continuous functions from R to R with arbitrary compact supports . We can show that D is a real vector space . We define a linear form T on the space D , associate to f like this : If g € D we set T(g) = integral of f*g on R .T is call a distribution on R associate to f . We say that the distribution T is odd is T(g) = 0 for all even function g of D . We can show that is f is a continuous odd function on R , then the distribution T , associate to f , is a odd distribution . We now define a sequence . Let n € N * . Let gn a even continuous function on R so that gn(x) = 1 on [ -n ; n ] et gn(x) = 0 outside [ -(n+1), n+1] ( you can draw the graph of gn ) . Let h(x) = 1 on R . We can show that the limit ( in some sense but too long to define ) of the sequence gn is h . f(x) = x is a odd function so the distribution T associate to f is a odd distribution , but the function gn are even functions so T(gn) =0 . We can show that the linear form T is continuous on D in some sense , so that lim T(gn) = T(h) , so T ( h ) = 0 .

Or in the theory of distributions , the distribution T associate to a continuous function f on R is : T(h) = integral on R of f(x)*h(x) dx , with abusive notation of course because h is not with compact support !! So integral on R of x.dx equal 0 But BE CAREFULL !! it's only a notation , and with bad use , we can do big mistakes, as for example square root of -1 = i , but the square root is not a continuous function on C the set of complex numbers
Sorry for this long answer but without enough rigorous details of the theory of distributions !