Editing Integration by parts (section)

===Validity for less smooth functions===
It is not necessary for <math>u</math> and <math>v</math> to be continuously differentiable. Integration by parts works if <math>u</math> is [[absolutely continuous]] and the function designated <math>v'</math> is [[Lebesgue integrable]] (but not necessarily continuous).<ref>{{cite web |title=Integration by parts| url=https://www.encyclopediaofmath.org/index.php/Integration_by_parts |website=Encyclopedia of Mathematics}}</ref> (If <math>v'</math> has a point of discontinuity then its antiderivative <math>v</math> may not have a derivative at that point.)

If the interval of integration is not [[compact space|compact]], then it is not necessary for <math>u</math> to be absolutely continuous in the whole interval or for <math>v'</math> to be Lebesgue integrable in the interval, as a couple of examples (in which <math>u</math> and <math>v</math> are continuous and continuously differentiable) will show. For instance, if

<math display="block">u(x)= e^x/x^2, \, v'(x) =e^{-x}</math>

<math>u</math> is not absolutely continuous on the interval {{closed-open|1, ∞}}, but nevertheless:

<math display="block">\int_1^\infty u(x)v'(x)\,dx = \Big[u(x)v(x)\Big]_1^\infty - \int_1^\infty u'(x)v(x)\,dx</math>

so long as <math>\left[u(x)v(x)\right]_1^\infty</math> is taken to mean the limit of <math>u(L)v(L)-u(1)v(1)</math> as <math>L\to\infty</math> and so long as the two terms on the right-hand side are finite. This is only true if we choose <math>v(x)=-e^{-x}.</math> Similarly, if

<math display="block">u(x)= e^{-x},\, v'(x) =x^{-1}\sin(x)</math>

<math>v'</math> is not Lebesgue integrable on the interval {{closed-open|1, ∞}}, but nevertheless

<math display="block">\int_1^\infty u(x)v'(x)\,dx = \Big[u(x)v(x)\Big]_1^\infty - \int_1^\infty u'(x)v(x)\,dx</math>
with the same interpretation.

One can also easily come up with similar examples in which <math>u</math> and <math>v</math> are ''not'' continuously differentiable.

Further, if <math>f(x)</math> is a function of bounded variation on the segment <math>[a,b],</math> and <math>\varphi(x)</math> is differentiable on <math>[a,b],</math> then

<math display="block">\int_{a}^{b}f(x)\varphi'(x)\,dx=-\int_{-\infty}^{\infty} \widetilde\varphi(x)\,d(\widetilde\chi_{[a,b]}(x)\widetilde f(x)),</math>

where <math>d(\chi_{[a,b]}(x)\widetilde f(x))</math> denotes the signed measure corresponding to the function of bounded variation <math>\chi_{[a,b]}(x)f(x)</math>, and functions <math>\widetilde f, \widetilde \varphi</math> are extensions of <math>f, \varphi</math> to <math>\R,</math> which are respectively of bounded variation and differentiable.{{cn|date=August 2019}}