improper integrals – Convergence of $ I=int_{0}^{+infty} frac{1}{sqrt{t}} cdot sin left(t+frac{1}{t}right) dt $

improper integrals – Convergence of $ I=int_{0}^{+infty} frac{1}{sqrt{t}} cdot sin left(t+frac{1}{t}right) dt $ – Mathematics Stack Exchange

calculus – How to calculate the volume of the following hyperboloid using integrals?

I need to find de volumen of the hyperboloid defined by the equation

${x^2over30^2}+{y^2over30^2}-{(z-160)^2)over120^2}=1$ and bounded by the planes $z=0$ and $z=190$.

I decided to use cylindrical coordinates and wrote the next integral:

$int^{190}_0 int^{2pi}_0int^T_0 ({r} )drdtheta dz$

Where $T$ is the radius $r=sqrt{30^2+{(z-160)^2over120^2}}$

I don’t know if I defined the integral correctly.

calculus and analysis – Problems with improper integrals in higher dimensions

Calculating numerically certain energy integrals in three and four dimensions related to
the Riesz potential
and the capacity, I try

NIntegrate( 1/Sqrt((x - p)^2 + (y - q)^2 + (z - r)^2), {x, y, z} (Element) 
Ball({0, 0, 0}, 1), {p, q, r} (Element) Ball({0, 0, 0}, 1), 
AccuracyGoal -> 3, PrecisionGoal -> 3, WorkingPrecision -> 25, 
Exclusions -> {(x - p)^2 + (y - q)^2 + (z - r)^2 == 0})

0

and a warning message
“NIntegrate::moptxn: The option SymbolicProcessing of the method FiniteElement is not one of {Method,MeshOptions}.”
As far as I understand it, this means that the NIntegrate command does not accept the set of the integration
in the form {x, y, z} (Element) Ball({0, 0, 0}, 1), {p, q, r} (Element) Ball({0, 0, 0}, 1). However, if it is so,
the input, not 0, should be returned.

My next try is

 NIntegrate( 1/((x - p)^2 + (y - q)^2 + (z - r)^2)^(1/2), {x, -1, 1}, {y, -Sqrt(1 - x^2), Sqrt(1 - x^2)}, 
{z, -Sqrt(1 - x^2 - y^2), Sqrt(1 - x^2 - y^2)}, {p, -1, 1}, {q, -Sqrt(1 - p^2), Sqrt(1 - p^2)}, 
{r, -Sqrt(1 - p^2 - q^2), Sqrt(1 - p^2 - q^2)}, AccuracyGoal -> 3, PrecisionGoal -> 3, WorkingPrecision -> 25, 
Exclusions -> {(x - p)^2 + (y - q)^2 + (z - r)^2 == 0})

-1244.482640558337558417913

and a warning
“NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 54 recursive bisections
in z near {x,y,z,p,q,r} = <<1>>. NIntegrate obtained -1244.482640558337558417913
and 3534.518334552443660338233`25. for the integral and error estimates.”
A similar issue with

NIntegrate(1/((x - p)^2 + (y - q)^2 + (z - r)^2)^(1/2)*
Boole(x^2 + y^2 + z^2 <= 1 && p^2 + q^2 + r^2 <= 1), {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, {p, -1, 1}, 
{q, -1, 1}, {r, -1, 1}, AccuracyGoal -> 3, PrecisionGoal -> 3, WorkingPrecision -> 25, 
Exclusions -> {(x - p)^2 + (y - q)^2 + (z - r)^2 == 0})

-1203.034524853306755966531

I wonder negative numbers since the integrand is positive.
Then i switch to MonteCarlo methods.

NIntegrate( 1/((x - p)^2 + (y - q)^2 + (z - r)^2)^(1/2)*
Boole(x^2 + y^2 + z^2 <= 1 && p^2 + q^2 + r^2 <= 1), {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, {p, -1, 1}, 
{q, -1, 1}, {r, -1, 1}, AccuracyGoal -> 2, PrecisionGoal -> 2, WorkingPrecision -> 25, 
Exclusions -> {(x - p)^2 + (y - q)^2 + (z - r)^2 == 0},Method -> "QuasiMonteCarlo")

21.12327556039856680489716

and a warning “NIntegrate::maxp: The integral failed to converge after 50000 integrand evaluations.
NIntegrate obtained 21.1232755603985668048971625. and 0.225183235272419193974785125. for the integral and error estimates.”
A more or less reliable result is obtained by

NIntegrate( 1/((x - p)^2 + (y - q)^2 + (z - r)^2)^(1/2)*
Boole(x^2 + y^2 + z^2 <= 1 && p^2 + q^2 + r^2 <= 1), {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, {p, -1, 1}, 
{q, -1, 1}, {r, -1, 1},  AccuracyGoal -> 2, PrecisionGoal -> 2, WorkingPrecision -> 25, 
Exclusions -> {(x - p)^2 + (y - q)^2 + (z - r)^2 == 0}, Method -> "AdaptiveMonteCarlo")

20.32729987338035891791629

without any warning. Unfortunately, AccuracyGoal -> 3, PrecisionGoal -> 3 is not achieved.
Also this works in eight dimensions:

NIntegrate(1/((x - p)^2 + (y - q)^2 + (z - r)^2 + (w - s)^2)^((4 - 2)/2)*
Boole(x^2 + y^2 + z^2 + w^2 <= 1 &&  p^2 + q^2 + r^2 + s^2 <= 1), {x, -1, 1}, {y, -1, 1}, {z,-1, 1}, 
{w, -1, 1}, {p, -1, 1}, {q, -1, 1}, {r, -1, 1}, {s, -1, 1}, AccuracyGoal -> 2, PrecisionGoal -> 2, 
WorkingPrecision -> 25,Exclusions -> {(x - p)^2 + (y - q)^2 + (z - r)^2 + (w - s)^2 == 0}, 
Method -> "AdaptiveMonteCarlo")

25.78510573458365881830859

BTW, the Exclusions option works in the above: compare with 24.68762920929857902438239 witout it.

The questions arise: what are other methods to calculate such integrals?
is a three-digit result possible with MonteCarlo methods?

definite integrals – Interchanging of limits, fundamentel theorem of calculus

I wanted to ask if the following proof for the fundamental theorem of calculus would work, as it is different from the one presented in my notes.

The fundamental theorem of calculus states that for a (real) intervall $Isubseteq mathbb{R}$ and $f: Itomathbb{R}$ continuous with $ain I$ we have that: $F: Itomathbb{R}$ defined by $F(x):=int_a^x f(t), dt$ is an integral of $f$.

I wanted to use the following theorem:

For a continuous function $f:(a,b)tomathbb{R}$ we have $int_a^b f(x), dx=lim_{ntoinfty} frac{b-a}{n}sum_{k=1}^n f(a+kleft(frac{b-a}{n}right)$.

To proof the theorem we have to show that

$F'(x)=f(x)$. Hence $lim_{hto 0} frac{int_a^{x+h} f(t),dt-int_a^{x} f(t),dt}{h}=f(x)$.

With a simple calculation you derive

$lim_{hto 0} frac{int_x^{x+h} f(t), dt}{h}$.

Now I tried to use the stated theorem:

$lim_{hto 0} displaystyle{frac{lim_{ntoinfty}frac1nsum_{k=1}^n f(x+kfrac{h}{n})}{h}}$

As we have a continuous function on a compact interval $(x, x+h)$ this is uniformly convergent. The sum is uniformly convergent too. So I interchanged the limits.

$lim_{ntoinfty}lim_{hto 0} displaystyle{frac{frac1nsum_{k=1}^n f(x+kfrac{h}{n})}{h}}$

Now as everything is continuous, and I let $hto 0$, I get

$lim_{ntoinfty} frac{1}{n}sum_{k=1}^n f(x)=f(x)$, as $n$ just cancels out.

Would this proof work?
My questions are: Does the interchanging of limits work here? Can I apply the theorem I used in this situation, where I look at the specific interval $(x,x+h)$, as $x$ isnt a fixed constant?

Thanks in advance.

calculus – Confusion in using double integrals & projection to calculate hemispherical surface area

As a preamble, if we have a surface given by $z=z(x,y)$ in $mathbb{R}^3$, i.e. $z =z(x,y) = sqrt{a^2-x^2-y^2}$ then $z- z(x,y)=0$ gives us a constant surface of a scalar field $f$, the $grad$ of which will tell us the normal vector to our surface: $nabla f = mathbf{k} -frac{∂z}{∂x}mathbf{i}-frac{∂z}{∂y}mathbf{j}$, the norm of which is $sqrt{1+left(frac{partial z}{partial x}right)^2+left(frac{partial z}{partial y}right)^2}$

A hemisphere of radius $a$ (with the $z$ axis as the axis of symmetry) is described by the equation $x^2+y^2+z^2 (=r^2) = a^2$ and has a surface area given by

$$iint dS = iint sqrt{1+left(frac{partial z}{partial x}right)^2+left(frac{partial z}{partial y}right)^2}dA$$

If you project onto the $xy$ plane.

This formula is obtained by considering an element of surface area $mathbf{dS}$ and its projected area $mathbf{dA}$, and relating the two by $dA = cos(alpha) dS = mathbf{hat{n}}cdotmathbf{k}, dS rightarrow dS = dA/mathbf{hat{n}}cdotmathbf{k} = sqrt{1+left(frac{partial z}{partial x}right)^2+left(frac{partial z}{partial y}right)^2}dA $ using $mathbf{n}$, the normal to the surface, from the preamble and $mathbf{k}$ the unit vector in the $z$ direction.
enter image description here

For the above example we can use $z =sqrt{a^2-x^2-y^2}$ to find our partial derivatives, and then evaluate the integral using plane polar co-ordinates ($0 leq theta leq 2pi, 0 leq r leq a $) to reach an answer of $2pi a^2$ as expected.

However, if we use a normal vector to this surface expressed in 3D polar co-ordinates:
$$mathbf{hat{n}} = mathbf{r}/|mathbf{r}| = (xmathbf{i} + ymathbf{j} + zmathbf{r} )/ |mathbf{r}| = sinphi costheta mathbf{i}+sinphi sintheta mathbf{j}+cosphi mathbf{k}$$

our logic above the included diagram should still hold: $dS = dA/mathbf{hat{n}}cdotmathbf{k}$ (here projecting onto the $xy$ i.e. $r,theta$ plane).
This yields $dS = dA/cosphi$ for our 3d polar case, which has got me stuck on the next step of evaluating

$$iint dS = iint secphi dA = iint sec(phi) rdrdtheta$$

It is intuitively true that as the angle $phi$ increases, our surface area element on the hemisphere becomes more and more vertical w.r.t. the $xy$ plane and so this $1/cosphi$ makes our corresponding area element larger, but how should I actually evaluate the integral?

$iint secphi dA = iint sec(phi) rdrdtheta$ is certainly ringing alarm bells, which makes me think that projecting on the $r,theta$ plane in the first place was a mistake, which is a problem not encountered when working in 3D cartesians.

Is there such a thing as projecting onto the “$theta,phi$ plane” and then integrating over those two variables? – if this were to be a valid method I believe it could solve my problem here. Please do ask for further clarification if it is needed.

probability distributions – How to solve Gaussian Integral (with list of gaussian integrals)

How can I solve the following Gaussian integral?
$$int _mathbb{R} Nleft(a|0, frac{1}{lambda^2}right) cdot Nleft(a; mu_a, sigma^2_aright) da$$.

According to Gaussian Processes for Machine Learning, p. 200, the product of two Gaussians gives another (un-normalized) Gaussian with: $$N(x; a, A) cdot N(x; b, B) = Z^{-1}N(x;c, C)$$
where $c = C(A^{-1}a + B^{-1}b)$ and $C=(A^{-1}+B^{-1})^{-1}$.

Applying this rule to the described integral gives the following unnormalized Gaussian distribution:
$$int_mathbb{R} Nleft(a; frac{1}{lambda^2 + frac{1}{sigma^2_a}} cdot (0 + frac{1}{sigma^2_a }cdot mu_a), frac{1}{lambda^2 + frac{1}{sigma^2_a}}right) = int_mathbb{R} Nleft(a; frac{mu_a}{lambda^2 cdot sigma^2_a}, frac{1}{lambda^2 + frac{1}{sigma^2_a}}right) da$$

According to Bayesian Perceptron: Towards fully Bayesian Neural Networks, equation (10) the result is $$frac{1}{t} cdot Nleft(frac{mu_a}{t};0, frac{1}{lambda^2}right)$$ with $t = sqrt{1+lambda^2 cdot sigma^2_a}$

I don’t know how to get to this solution. I tried to use some lists of gaussian integrals, such as A table of normal integrals
, but because they always refer to standard normal distribution, i don’t know how to continue. Do I have to scale the above normal distribution to have $mu=0$ and $sigma^2=1$ and then use the rule $int G'(x)dx = G(x)$ (owen, equ. 10)?

Thanks for helping!

integration – What is the correct interpretation for the integral of a differential $k$-form in terms of typical vector-calculus multiple integrals?

Preliminary context:

Prof. Kennan Crane from CMU posted a course on discrete differential geometry. I’ve been able to follow so far, until Lecture 7. Please let me give some context on how he defines the stuff before posing my question. Basically, a vector is defined in terms of its components as a linear combination of the basis vectors denoted with $left{e_i=frac{partial}{partial x_i}right}_{i=1}^n$. Moreover $k$-vectors are defined as linear combinations of $e_{i_1}wedgedotswedge e_{i_k}$ for some indices $i_1,dots,i_k$ (see Lecture 3). Then, it is explained that forms are basically a generalization of covectors (in euclidean space, the dual of vectors which can be thought as row vectors vs column vectors), which are linear transformations from vectors to scalars. Then, covectors (or 1-forms) are given as linear combinations of the basis 1-forms ${dx_i}_{i=1}^n$, and $k$-forms are linear combinations of $dx_{i_1}wedgedotswedge dx_{i_k}$. Until this point, the symbols $dx_i$ and $frac{partial}{partial x_i}$ are not related to derivatives or differentials, but are just basis vectors/forms. Prof. Crane explains that $k$-forms are meant to be applied to $k$-vectors in order to obtain a scalar measure. For example the area of the parallelepiped $uwedge v$ for two vectors $u,v$ can be measured by a 2-form $alphawedgebeta$ for two 1-forms $alpha,beta$, just scaled by the “size” of $alpha,beta$. In particular the concrete relation between forms and vectors basis is given by
$$
dx_ileft(frac{partial}{partial x_j}right) = delta_{ij}
$$

where $delta_{ij}$ is the Kronecker delta. Moreover, the application $(alphawedge beta)(u, v)=alpha(u)beta(v)-alpha(v)beta(u)$ which can be written as a determinant (and generalized for the $k$-forms).

Now comes my questions all regarding Lecture 7.

  1. At 6:23, the motivation behind the definition of the integral of a 2-form $omega$ is that it is the summation of $omega_{p_i}$ (evaluate $omega$ at the point $p_i$) applied to a pair of vectors $u_i=u({p_i}),v_i=v(p_i)$ (for some vector fields $u,v$) along with the whole region $Omega$:
    $$
    sum_{i}omega_{p_i}(u_i,v_i)to int_{Omega}omega
    $$
    However, in 8:31 we have the integral of a form $(x+xy)dxwedge dy$ over $Omega$ which is the unit square. Then, somehow it follows that
    $$
    int_{Omega} (x+xy)dxwedge dy = int_0^1int_0^1 (x+xy)dxdy
    $$

    where I don’t see to what vectors $u,v$ are we applying the form $dxwedge dy$. Mainly, I’m lost at the motivation behind the interpretation of the 2-form $dxwedge dy$ as a scalar $dxdy$ inside the integral, given that $dx$ and $dy$ were supposed to be just basis forms (covectors). The following may be too hand wavy, but I’m trying to get intuition about this: My thoughts are that we make a parallelepiped spanned by the vectors $u=(dx)frac{partial}{partial x}$ and $v=(dy)frac{partial}{partial y}$ where here $dx$ and $dy$ are small displacements in the directions $frac{partial}{partial x},frac{partial}{partial y}$. This parallelepiped has area $dxdy$, and we can measure such area by the form $dxwedge dy$. Hence, we use $dxwedge dy$ to measure this parallelepiped obtaining $dxwedge dy(u,v)=dxdy$. However, this reasoning seems artificial to me (since we didnt’t take advantage of the notation dx,dy for the basis covectors, instead we used other displacements $dxdy$ different from the basis covectors). It would be better to have defined $dx(frac{partial}{partial x})$=”$dx$” (where the second $dx$ is the displacement and not the covector) instead of $dx(frac{partial}{partial x})=1$ as we did before (move the displacements $dx$ and $dy$ to the definition of the covectors $dx$ and $dy$ instead of the parallelepiped $uwedge v$).

  2. At 14:00, one finds the integral of a 1-form $alpha=dy$ over the circular curve $S^1$ (parametrized as $gamma:(0,2pi)tomathbb{R}^2, gamma(s) = (cos(s),sin(s))$), which somehow results in:
    $$
    int_{S^1}alpha = int_0^{2pi} alpha_gamma(T(s))ds
    $$

    where $T(s)$ is the tangent vector to $gamma(s)$. Here we are clearly applying the form $alpha$ to a vector, different from my previous doubt. Howeve, different than before, instead of using $dy$ as the differential in the integral, we introduced an additional $ds$ where I don’t know where it came from.

I already know how to compute this integrals, and I know that their expressions make sense from the typical vector calculus perspective. But I want to understand what is the systematic way of translating the integral of $k$-forms to a multiple-integral. My confusion comes from the fact that it appears that very different “rules” were used to translate $int_{Omega} (x+xy)dxwedge dy = int_0^1int_0^1 (x+xy)dxdy$ as in the case $int_{S^1}alpha = int_0^{2pi} alpha_gamma(T(s))ds$.

regularization – A set of divergent integrals that I think, equal to $-gamma$

So, we take $frac{text{sgn}(x-1)}{x}$ and apply $mathcal{L}_t(t f(t))(x)$ four times. The transform is known to keep area under the curve. These integrals, I think, are equal to minus Euler-Mascheroni constant. Since they all have infinite parts that cancel each other, their values are finite. I have already applied Laplace transforms to regularize divergent integrals in a similar way.

$$int_0^infty frac{text{sgn}(x-1)}{x}dx=int_0^inftyfrac{2 e^{-x}-1}{x}dx=int_0^inftyfrac{x-1}{x (x+1)}dx=int_0^infty left(2 e^x text{Ei}(-x)+frac{1}{x}right) dx=$$
$$int_0^infty frac{x^2-2 x log (x)-1}{(x-1)^2 x} dx=-gamma$$

Yes?

enter image description here

Proof: take 2 of them and find average:

$$int frac12left(frac{x-1}{x (x+1)}+ left(2 e^x text{Ei}(-x)+frac{1}{x}right)right)dx=-gamma$$

enter image description here

Right?

Squeeze Theorem in Reimann Integrals

Consider the function defined by 𝑓(𝑥)= x^2. Use the Squeeze Theorem for Riemann integral, the set of Riemann integrable functions on (0,3).

I know the idea of squeeze theorem but im not sure how to apply it.

laplace transform – Fractional power of the operator $mathcal{L}_t[t f(t)](x)$ and equivalence of divergent integrals

I wonder whether an expression for fractional power of operator $mathcal{L}_t(t f(t))(x)$ that involves Laplace transform can be derived?

I am asking this because this operator preserves the area under the function:

$$int_0^infty f(x)dx=int_0^infty mathcal{L}_t(t f(t))(x) dx$$

But more importantly, this still works well even with divergent integrals, thus allowing us to define equivalent classes of divergent integrals. For instance, the following divergent integrals are obtained by consecutive applying this transform, and as such, can be postulated to be equal (the first one and the third one are similar up to a shift by the way):

$int_0^inftyfrac{theta (x-1)}{x}dx=int_0^inftyfrac{e^{-x}}{x}dx=int_0^inftyfrac{dx}{x+1}=int_0^inftyfrac{e^x x text{Ei}(-x)+1}{x}dx=int_0^inftyfrac{x-log (x)-1}{(x-1)^2}dx$

The following plot illustrates the comparison:

enter image description here

So, is there a way to express fractional powers of this transform?

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