30

It has nothing to do with the conflict with Borel which developed later, and one can find a pretty explicit answer in the aforementioned letters of Lebesgue to Borel.
(These letters were first published in 1991 in Cahiers du séminaire d’histoire des mathématiques; selected letters with updated commentaries were also published later by Bru and Dugac in an ...

20

The main applications of Lebesgue integral to concrete problems of analysis found before Poincare's death are the Riesz-Fischer theorem (1907) and Fatou's work (1906). All this is somewhat remote from the main interests of Poincare. Applications of measure theory to mechanics (ergodic theory) were found later, after his death.
You cannot expect even the ...

answered Nov 5 '19 at 19:24

Alexandre Eremenko

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20

Fix a compact group $G$ and consider its category of Banach representations:
the objects are (complex) Banach spaces $X$ endowed with a $G$-action by automorphims (not necessarily isometries) such that the action maps $G\times X\to X$ are jointly continuous and the morphisms are bounded maps $X\to Y$ commuting with the $G$-actions.
Denote the trivial ...

18

This really belongs to MSE rather than to MO, but I'm too lazy to initiate the moving process, so I'll just answer.
There may be more intelligent ways to do it, but you can also integrate by parts and get what you want, say, for smooth functions with compact support, after which you should carefully pass to the limit to extend it to the corresponding ...

17

Let $f(t) = 1$ if $t \in A$ and $f(t) = 0$ otherwise. Suppose that $a > 0$. Then
$$ \begin{aligned}
\int_a^{2 a} \operatorname{card} \{n : n t \in A\} dt & = \int_a^\infty \biggl(\sum_{n = 1}^\infty f(n t) \biggr) dt \\
& = \sum_{n = 1}^\infty \int_a^{2 a} f(n t) dt \\
& = \sum_{n = 1}^\infty \frac{1}{n} \int_{n a}^{2 n a} f(s) ds \\
& = \...

answered Nov 30 '19 at 21:25

Mateusz Kwaśnicki

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16

The proposed equality is true.
Details: To find
$$l:=\int_0^1\left(\frac{\arcsin x}x\right)^3\,dx
=-\frac{1}{16} \pi \left(\pi ^2-24 \ln2\right),
$$
make the substitution $t=\arcsin x$ and repeatedly integrate by parts to kill the powers of $t$ and reduce this integral to
$$\int_0^{\pi/2}\ln\sin t\,dt=-\frac\pi2\,\ln2,
$$
by
formula 4.225.3, page 531, ...

16

The inequality is true. One integration by parts and standard Cauchy-Schwarz gives
$$\int |xf''|^2 \leq C\left(\int |f'''|^2 + x^4|f'|^2 + |f'|^2\right).$$
The second term can be handled by integrating by parts once and applying Cauchy-Schwarz:
$$\int x^4|f'|^2 \leq \delta \int x^2|f''|^2 + C_{\delta}\left(\int(x^2+x^6)f^2\right),$$
and $x^2+x^6$ is smaller ...

16

Let
$$f(a):=\int_0^\infty x^{a-1}e^{-2x}\,dx.$$
Then
$$f''(1)=\int_0^\infty \ln^2x\,e^{-2x}\,dx,$$
which is the integral in question.
On the other hand, $f(a)=2^{-a}\,\Gamma(a)$, and hence
the integral in question is
$$f''(1)=\frac{\ln^2 2}2 - \Gamma'(1)\ln2
+ \Gamma''(1)/2
=\frac{\pi^2}{12}+ \frac{(\gamma +\ln2)^2}2=1.6293\dots,$$
where $\gamma=0.57721\dots$...

15

Poincaré studied with Hermite, who famously in a letter 1893 to Stieltjes wrote „I turn with terror and horror from this lamentable scourge of continuous functions with no derivative.“ Poincaré himself is often quoted „Heretofore when a new function was invented it was for some practical end; today they are invented expressly to put at fault the reasoning of ...

13

I keep seeing this question percolate up. I think it deserves at least one more or less correct answer.
First, the 1d integrals $I(a,b) = \int_{\mathbb{R}} \exp(-\frac{1}{2}x^2 - ax - b x^4) dx$ certainly exist for $b \geq 0$ and are analytic in $a$ and $b$. But they're not do-able in elementary terms. Noting that $\partial_bI = \partial_a^4I$ and ...

12

I will only consider the temperate situation (involving the spaces $\mathscr{S}$, $\mathscr{S}'$ and $\mathscr{O}_{\rm M}$) and I will only discuss the first example as well as a "dual" version with a derivative instead of an integral to be commuted with a distribution pairing. I will show that Riemann integration (and its dual, i.e., the classical ...

answered Jan 23 at 22:57

Abdelmalek Abdesselam

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12

Let me expand the integrand in powers of $z$ and integrate over $z$,
$$I=\int_0^1 dx\int_0^1 dy\int_0^1 dz\;\frac{3}{3-z(x+\sqrt{xy}+y)}$$
$$\qquad\qquad=\sum_{n=0}^\infty\,\frac{3^{-n}}{n+1} \int_0^1 dx\int_0^1 dy\;(x+\sqrt{xy}+y)^n.$$
The integral over $x$ and $y$ is an element $c_{n}\in\mathbb{Q}$,
$$I=\sum_{n=0}^\infty\,\frac{3^{-n}}{n+1}c_n,\;\;c_n=\...

answered Dec 10 '19 at 21:55

Carlo Beenakker

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12

It doesn't seem to me that either article follows the line of reasoning as you have presented it. Indeed, we do not take the integral over the union of integrals $(x_n,(1+\varepsilon)x_n)$. We do get that the integral over those intervals is infinite, but you correctly note this does not give a contradiction. Instead, the argument goes as follows.
Let me ...

10

Yes it can, because the long square root equals $\sqrt{9x+1}-\sqrt{12 x}$. After this observation you get a standard integral, which reduces to an integral of rational function if you change the variable to $\sqrt{9/x+1}$.

answered Aug 19 '19 at 14:04

Fedor Petrov

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10

Titchmarsh’s Fourier integrals (1937, 7.6.4) has proof and attribution to Ramanujan.

10

I don't know if it's exactly what you're looking for, but line integration is the unique way to assign a real number $I(\omega,c)\in\mathbb{R}$ to every pair of a smooth $1$-form $\omega$ on a smooth manifold $M$ with boundary and smooth path $c\colon[0,1]\to M$ such that:
(adjunction) if $f\colon M\to N$ is a smooth map of smooth manifolds with boundary, $\...

10

Diestel and Uhl commented on this in Measure theory and its applications:

9

The Lebesgue-Borel conflict may provide a hint why Poincaré was, like Borel, not impressed by Lebesgue's contribution:
As we hinted earlier, all was not well between Borel and Lebesgue and
their long-standing friendship deteriorated until it finally
collapsed, at Lebesgue's instigation, in 1917. The evidence we have is
provided by letters preserved ...

answered Nov 5 '19 at 9:24

Carlo Beenakker

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9

Not quite an answer, but too long for a comment.
Let me make my life easier a bit and take $\Omega=\mathbb{R}^N$ while increasing the exponent slightly. Namely, I will assume that
$$
I:=\ \int_{\mathbb{R}^{2N}}\ \frac{|f(x)-f(y)|}{|x-y|^{\alpha}}\ d^Nx d^Ny
$$
is finite, where $\alpha>N+1$.
Then we have, just by the triangle inequality involving the ...

answered May 13 at 21:55

Abdelmalek Abdesselam

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8

I am currently working on a similar system. But your properties (2) and (3) would not work and need change.
Instead, the following properties would work much better:
$$\int_a^c f(x) dx=\int_a^b f(x)dx+\int_b^c f(x)dx\tag{1}$$
$$\int_a^b (f(x)+g(x)) dx=\int_a^b f(x)dx+\int_a^b g(x)dx\tag{2}$$
$$\int_a^b c f(x) dx =c \int_a^b f(x) dx\tag{3}$$
$$\int_{-\...

8

The integral is twice Catalan's constant
$$
G = L(1,\chi_4) = 1 - \frac1{3^2} + \frac1{5^2} - \frac1{7^2} + \frac1{9^2} - + \cdots.
$$
This constant can be computed efficiently to high precision, even though no further "closed form" is known or expected.
I guessed this as follows.
Since ${\rm gd}^{-1}(x) = \int_0^x \sec t \, dt$,
your integral is $\int_0^{\...

answered Sep 10 '19 at 3:28

Noam D. Elkies

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8

It is clear that $I(0)=0$, hence by Leibniz's integration rule, it suffices to show that
$$\int_{-\infty}^\infty\frac{-2f'(x)\sin\phi}{1+2f'(x)(1-\cos\phi)}\,dx=2\phi,\qquad |\phi|<\pi.$$
We can assume, without loss of generality, that $0<\phi<\pi$. By a bit of algebra, we can rewrite the last equation as
$$\int_{-\infty}^\infty\frac{e^x\sin\phi}{e^{...

7

Coarea formula will do this for you. It is a "Fubini formula" relating the integral of a function $u$ on a Riemann manifold $(M_0,g_0)$ to the integrals along the fibers of a smooth map $F:(M_0,g_0)\to (M_1,g_1)$. The map $F$ need not be a submersion. See this note for details.

dg.differential-geometry integration smooth-manifolds geometric-measure-theory multivariable-calculus

answered Nov 19 '19 at 19:28

Liviu Nicolaescu

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7

The property Σ=Σ_1 amounts to (X,Ε,μ) being locally determined.
A measure space (X,Σ,μ) is locally determined if μ is semifinite
and A∈Σ if and only if A∩F∈Σ for all F∈Σ such that μ(F) is finite.
See Fremlin, Measure Theory, Definition 211H.
Almost all measurable spaces that arise in practice (e.g., from Radon measures)
are strictly localizable and ...

6

Let $r$ be less than half the distance from $K$ to the complement of $B_1(0)$. Start with a ball of radius $r$ centered at each point of $K$. By https://en.wikipedia.org/wiki/Besicovitch_covering_theorem there is a subcover of $K$ by balls that is a union of $C_n$ collections $A_i$, where each $A_i$ consists of pairwise disjoint balls of radius $r$. For each ...

6

Consider the integral over $\frac\pi{n}[k,k+1]$, where $k=-n,\ldots,n-1$. Denote $x=\frac{k\pi+y}n$, you get $$I_k:=\frac1n\int_{0}^{\pi}\frac{|\sin \frac{k\pi+y}n|}{|\cos y|^{2/n}}dy.$$
Denote
$$
\varepsilon_k:=I_k-\frac1n\int_0^\pi \left|\sin \frac{k\pi+y}n\right|dy=
\frac1n \int_0^\pi \left|\sin \frac{k\pi+y}n\right|\left(|\cos y|^{-2/n}-1\right)dy.
$$
We ...

answered Nov 12 '19 at 8:10

Fedor Petrov

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6

Another approach appears as a comment on your question, so this is just a rip-off trying to make things tidier but surely there are other ways to Titchmarsh and this to prove it
$
\int_{\mathbb{R}}\frac{dx}{\Gamma(\alpha+x)\Gamma(\beta-x)}\\
=\frac{1}{\Gamma(\alpha+\beta-1)}\int_{\mathbb{R}}\frac{\Gamma(\alpha+\beta-1)dx}{\Gamma(\alpha+x)\Gamma(\beta-x)}\\
=...

6

As you've pointed out, the only parameter that matters here is the angle $\theta$ between $x$ and $y$. To see how, consider instead the Gaussian integral:
$$
I(x,y)=\frac{1}{(2\pi)^{(d+1)/2}}\int_{u\in\mathbb{R}^{d+1}}\max(0,x^Tu)\cdot\max(0,y^Tu)\exp(-\frac{1}{2}u\cdot u)du
$$
The integral you are interested in is obtained by changing to $(d+1)$ dimensional ...

6

These identities, and many more, follow from a theorem in Integrals of polylogarithmic functions, recurrence relations, and associated Euler sums,
For example,
There are also variations with $\log x$ factor, such as

5

Yes, Risch's algorithm, given an elementary univariate function, will compute an elementary anti-derivative, if one exists. "Failure" of the algorithm implies that no elementary anti-derivative exists.
No, the algorithm can handle transcendental constants just fine. The problem arises when you don't know the full transcendence relationships between your ...

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