Is it possible to integrate by parts using fractional derivatives?

Question

Motivation

It often happens to me that I calculate an integral and find that by doing the integration by parts the integral diverges.

Basic example

$$\int_0^{\infty}\frac{\sin(t)}{t}\mathrm{d}t=\frac{\pi}{2}$$ but $$\int_0^{\infty}\frac{\cos(t)}{t^2}\mathrm{d}t\to\infty\qquad\text{and}\qquad\int_0\cos(t)\ln(t)\mathrm{d}t\to\infty$$

In both cases, by moving an entire derivative to one or the other term, the integral diverges, so I wondered if it was possible to "lighten" the load moved.

My link

Thinking about diffusion problems (where there are second derivatives in the Laplacian), the solutions enter their "natural space" $H^1$ when writing the variational formulation, hence when you dump a derivative on the test function, and we had seen a case where it made sense to dump half a derivative on the test function to make the order of the derivatives on the test solution and the solution equal (in practice only to set it in the correct Sobolev space).

My question

Is it possible to generalize the integration by parts formula using fractional derivatives?

My idea

I had to try to find it myself by taking the fractional derivative of a product of functions and then integrating to obtain one of the two expressions, but Leibniz's formula does not apply to fractional derivatives.

$$\frac{\mathrm{d}^n}{\mathrm{d}x^n}f(x)g(x)=\sum_{k=0}^{n}\binom{n}{k}f^{(k)}(x)g^{(n-k)}(x)$$ In the case in question I thought (I don't know if it's correct) that a formula by analogy could be valid as for the series of a power of a binomial

$$\frac{\mathrm{d}^\alpha}{\mathrm{d}x^\alpha}f(x)g(x)\mathrm{d}x\overset{?}{=}\sum_{k=0}^{\infty}\binom{\alpha}{k}f^{(k)}(x)g^{(\alpha-k)}(x)$$

In general I have

$$\int f(x)g^{(n)}(x)\mathrm{d}x=\sum_{k=0}^{n-1}(-1)^k f^{(k)}g^{(n-k)}(x)+(-1)^n\int f^{(n)}(x)g(x)\mathrm{d}x$$ So maybe there is a similar formula that could have a form like this

$$\int f(x)g^{(\alpha)}(x)\overset{?}{=}\sum_{k=0}^{\infty}(-1)^k f^{(k)}g^{(\alpha-k)}(x)+c_{\alpha}\int f^{(\alpha)}(x)g(x)\mathrm{d}x$$

Link with Laplace transform

A similar formula also exists for Laplace transforms.

$$\int _{0}^{\infty }f(x)g(x)\mathrm{d}x=\int _{0}^{\infty }({\mathcal {L}}f)(s)\cdot ({\mathcal {L}}^{-1}g)(s)\mathrm{d}s$$

"Similar" in the sense that it follows the philosophy of applying one operator to one function and the inverse operator to the other function (if you think of the derivative as an operator).

The discussion could also be generalized by asking to find all the $\mathcal{A}$ operators that satisfy the property

$$\int f(x)\cdot g(x)\mathrm{d}x=C_a\int \mathcal{A}[f](t)\cdot \mathcal{A}^{-1}[g](t)\mathrm{d}t$$ but for now I'll limit myself to fractional derivatives

Disclaimer
I used Google translate to write the post and I have the doubt that It may have mistranslated the term "dump" (for me it means that I move the derivative operator from one part to another).

Integration by parts is based on the derivative product rule, so you'd need som fractional derivative product rule' $d^{\alpha}(uv)/dx=?$ for non-natural $\alpha.$'This seems unlikely. — Thomas Andrews, Mar 11 '25 at 16:58
@ThomasAndrews In fact it was my first approach but I don't know how it can be treated. I was imagining it more from the operators' point of view. — Math Attack, Mar 11 '25 at 17:09
Here's an example:$$\int_{0}^{1}\frac{\arcsin(\sqrt{x})\arcsin(\sqrt{1-x})}{\sqrt{x(1-x)}},dx=\frac{\pi}{2}\int_{0}^{1}D^{1/2}\log(1-x)\cdot D^{1/2}_{\perp}\log(x),dx.$$ — user170231, Mar 24 '25 at 17:39

score 2 · Answer 1 · answered Mar 11 '25 at 19:07

First of all, the fractional derivatives in the most natural way doesn't make sense for general (even smooth) functions. In harmonic analysis there is something called pseudodifferential operators which can redefine fractional derivatives.

Let us use Fourier transform $\hat f(\xi)=\int_{\mathbb R}f(x)e^{-2\pi ix\xi}dx$, then its inverse tranform is $\check g(x)=\int_{\mathbb R}g(\xi)e^{2\pi ix\xi}d\xi=\hat g(-x)$. Basically if you have $L=\sum_{j=1}^nc_j\frac{d^j}{dx^j}$ then $\widehat{Lf}(\xi)=\sum_{j=1}^nc_j(2\pi i\xi)^j\hat f(\xi)$. You can replace $\sum_{j=1}^nc_j(2\pi i\xi)^j$ by any (at most polynomially growth) function $P(\xi)$. The corresponding operator is usually denoted by $P(\frac1{2\pi i}\frac d{dx})$. Note that it is a function on $i\frac d{dx}$ but not $\frac d{dx}$.

In this setting we are more preferable to use fractional derivative $|\frac d{dx}|^s$ given by $|\frac d{dx}|^sf(x)=(|2\pi\xi|^s\hat f(\xi))^\vee$.

A problem of such thing is that it has no locality, meaning that the value of $|\frac d{dx}|^sf$ at $x$ depends on the whole $f:\mathbb R\to\mathbb C$ but not its local behaviour near $x$.

But if you just want to do integration by parts, when there is no boundary term, using $\int \hat f(x)g(x)dx=\int f(\xi)\hat g(\xi)d\xi$ we see that $\int P(\frac1{2\pi i}\frac d{dx})[f](x)g(x)dx=\int P(\xi)\hat f(\xi)\hat g(-\xi)d\xi=\int f(x) P(-\frac1{2\pi i}\frac d{dx})[g](x)dx$.

But if you want to keep the boundary term a lot of extra assumptions are required, like having convergent Taylor series with large enough convergent radius. Such thing is called fractional Leibniz rule.

In your case it is not likely to makes sense due to the non-locality of fractional derivatives.