# Integration by parts of delta function

I am having the worst time trying to solve this integral,

$$\int g(t)\frac{d}{df}\delta[f(t)]dt,$$

$$= \int g(t)\frac{dt}{df}\frac{d}{df}\delta[f(t)]dt.$$

This should yield,

$$-\bigg[ \frac{dt}{df}\frac{d}{dt} \bigg( \frac{g(t)}{\frac{df}{dt}} \bigg) \bigg]_{f(t)=0}$$

I have tried every possible way to solve this integral (setting various u's and dv's).

Any help would be greatly appreciated!

First Attempt:

$$u = \frac{dt}{df}\frac{d}{dt} \rightarrow du=\frac{d}{dt}\bigg(\frac{dt}{df}\frac{d}{dt}\bigg)dt$$ $$dv = g(t)\delta[f(t)]dt \rightarrow v = \bigg[\frac{g(t)}{|df/dt|}\bigg]_{f(t)=0}$$

But this is an epic fail (I think)

Second Attempt:

$$u = g(t)\frac{dt}{df} \rightarrow du=\frac{d}{dt}\bigg(g(t)\frac{df}{dt}\bigg)dt$$ $$dv = \frac{d}{dt}\delta[f(t)]dt \rightarrow v = \delta[f(t)]$$

But I don't think this yields the correct solution either...

-
I just wanted to say that I really appreciate that you showed your work. So that you know, in your first attempt, setting $u = \frac{dt'}{df} \frac{d}{dt'}$ is weird because $\frac{d}{dt'}$ is neither a regular function nor a number, but an operator that takes in functions. In the second, you took $g(t')$ out of its derivative without noting that it was differentiated. –  mixedmath Mar 14 '13 at 2:09
@Shinobii: can you please change the notation? I am assuming that you mean for $t'$ to be a variable name. Is that correct. If you change all of those to $t$, nothing will change in the result. Is my assumption correct? –  Amzoti Mar 14 '13 at 2:36
You are indeed correct, I can change it for aesthetics. –  Shinobii Mar 14 '13 at 3:41

Heuristically we have

\begin{align*} \int g \frac{d\delta}{df}(f) \, dt &= \int g \frac{d\delta}{df}(f) \frac{dt}{df} \, df = \int \frac{g}{f'} \frac{d\delta}{df}(f) \, df \\ &= - \int \frac{d}{df}\bigg(\frac{g}{f'}\bigg) \delta(f) \, df \\ &= - \int \frac{dt}{df}\frac{d}{dt}\bigg(\frac{g}{f'}\bigg) \delta(f) \, df \\ &= - \left. \frac{1}{f'}\bigg(\frac{g}{f'}\bigg)' \right|_{f=0}. \end{align*}

But this is just a deceptive heuristics and we have to justify (and also calibrate) this result in mathematical language. To this end, suppose that $f$ be smooth, that $g \in C_{c}^{\infty}(\Bbb{R})$ and that $g$ decays fast enough not to raise any integrability issue. Also, let $x_j$ be zeros of $f$ and assume that they are simple zeros of $f$: $f'(x_j) \neq 0$. Then there exists a disjoint family of open neighborhoods $U_j$ of $x_j$ such that $f$ is invertible on $U_j$ with a local inverse, which we denote $f^{-1}$ whenever no ambiguity arises. Then

\begin{align*} \int_{\Bbb{R}} g \frac{d\delta}{df}(f) \, dt &= \int_{\Bbb{R}} g(t) \delta'(f(t)) \, dt = \sum_{j} \int_{U_j} g(t) \delta'(f(t)) \, dt \\ &= \sum_{j} \int_{f(U_j)} g(f^{-1}(u)) \left| (f^{-1}(u))' \right| \delta'(u) \, du \qquad (u = f(t)) \\ &= - \sum_{j} \mathrm{sgn}\,(f^{-1}(u))' \cdot \int_{f(U_j)} \big[ g(f^{-1}(u)) (f^{-1}(u))' \big]' \delta(u) \, du \\ &= - \sum_{j} \mathrm{sgn}\,(f^{-1}(u))' \cdot\left.\big[ g(f^{-1}(u)) (f^{-1}(u))' \big]'\right|_{u=0} \end{align*}

Here, we exploited the following property

$$\int_{\Bbb{R}} \varphi(x) \delta'(x) \, dx = - \int_{\Bbb{R}} \varphi'(x) \delta(x) \, dx = -\varphi'(0), \quad \varphi \in C_{c}^{\infty}(\Bbb{R}).$$

Now, simple calculation shows that

$$\big[ g(f^{-1}(u)) (f^{-1}(u))' \big]' = \frac{f'(t) g'(t)-g(t)f''(t)}{f'(t)^3},$$

where $t = f^{-1}(u)$. Finally, by noting that $f$ increases on $U_j$ if and only if $f^{-1}$ increases on $f(U_j)$, it follows that $\mathrm{sgn} \, (f^{-1})'(0) = \mathrm{sgn}\,f'(x_j)$. Therefore we have

\begin{align*} \int_{\Bbb{R}} g \frac{d\delta}{df}(f) \, dt &= - \sum_{j} \mathrm{sgn} \, f'(t) \cdot \left. \frac{f'(t) g'(t)-g(t) f''(t)}{f'(t)^3} \right|_{t=x_j} \\ &= - \sum_{j} \left. \frac{1}{\left| f'(t) \right|} \bigg( \frac{g(t)}{f'(t)} \bigg)' \right|_{t=x_j} \\ \end{align*}

as desired.

Edit. Following jbc's advice, I made amends to the sign problem. I realized this problem few days ago, but forgot fixing until he/she pointed it out explicitly. It is to blame my short-term memory :( And also thanks, jbc!

-
Wow, thank you very much! Very impressive answer, I completely understand now. –  Shinobii Mar 14 '13 at 12:24

I would like to add something to the above answer for two reasons. Firstly, I think that there is a sign error in the final result. Secondly, despite the title, I think that it is not about partial integration but about changing variables in distributions and this is a recurring theme on this site. Queries on this topic are usually answered with ad hoc methods, i.e., by assuming that the given object exists and then that one can manipulate it as one does in the classical case. As an empirical method, there is nothing to be said against this procedure but since this is a mathematical forum one should take account of the fact that substituting functions into distributions is rather more delicate than in the classical case---the substition $T \circ \phi$ can only be carried out under special assumptions on the distribution $T$ and the smooth function $\phi$. (We remark that most applications are for the specal case where $T$ is a $\delta$-distribution or, as in this query, its derivative).

A rigorous and elementary definition was given by Sebastiao e Silva in the 60's and is expounded in the text "An elementary introduction to the theory of ditributions" by Campos Ferreira. Briefly, if a distribution is of finite order, i.e., is an iterated derivative of a continuous function (as $\delta$ and $\delta'$ are), say $T = \tilde D^m F$ for a continuous function $F$ (where the symbol $\tilde D$ denotes the derivative in the sense of distributions), then $T\circ \phi$ is defined to be $\left (\frac 1 {\phi' }\tilde D\right )^m (F\circ \phi)$. (For the motivation for this definition and the proof that it is well-defined---the hard part---and has the properties one would want, see the above reference). One sees that this definition only makes sense if $\phi$ is a homeomorphism whose derivative doesn't vanish (one can extend it by a more careful analysis but this will not affect the following).

If one applies this to $\delta$ and its derivative, one gets the formulae $$\delta \circ \phi = \frac 1{|\phi'(a)|} \delta_a$$ and $$\delta' \circ \phi = \frac 1{|\phi'(a)^3|}\left ( \phi'(a)\delta_a'+\phi''(a) \delta_a \right )$$ where $a$ is $\phi^{-1}(0)$.

The formula requested in the query is then obtained by applying these distributions to the test function $g$ and using the rules for integrating terms involving $\delta$ and $\delta'$. (We have taken the liberty of using a different notation for derivatives). A rather more delicate analysis can be used to extend this to the case of a smooth function $\phi$ for which $\phi$ has, say, only finitely many zeroes. The formulae remain the same but one sums over the set of zeroes.

-
Hmm, perhaps you can elaborate on why you think there is an error with the sign. I cannot see any errors, but perhaps I am mistaken. Also, thanks for the reference, in this case this was a physical problem as opposed to a strictly mathematical one, but what is math without rigor! –  Shinobii Mar 17 '13 at 19:40
Oh, never mind! I see the error you were talking about. Thanks for pointing that out and thanks to sos440 for the edit! –  Shinobii Mar 20 '13 at 2:36