## Integral (analytic) solving take too long

I’m it my first use mathematica. I need to solved integral in analytical form, this integral:
Integrate(1/Sqrt(x^2 + (y – h – t)^2 + z^2), {t, -c, c})
x,y,h,z – some real parameters.

## integration – Is that integral $iiintlimits_{G}frac{dxdydz}{(x^2+2y^{2}+3z^{2})^{3/4}}$ finite?

Let $$G={(x,y,z)in mathbb{R}^{3}:x^{4}+y^{6}+z^{8}<10}$$
We want to check if $$iiintlimits_{G}frac{dxdydz}{(x^2+2y^{2}+3z^{2})^{3/4}}$$ is finite. My first idea is to try to calculate integral on surface $$B(0,epsilon)$$, for random $$epsilon$$ and find the $$lim_{epsilon to 0^{+}}iiintlimits_{B(0,epsilon)}frac{dxdydz}{(x^2+2y^{2}+3z^{2})^{3/4}}$$, but i’m not sure about this. Is my claim right and if not how can we deside if integral is finite or not ?

## contact geometry – Maximal dimension guaranteed for integral manifolds of hyperplane distributions

I’ve run across the following question while going through the basic definitions in contact geometry. Often, a contact structure is described as maximally non-integrable, but defined as locally being the kernel of a contact form. I’ve seen justifications for the term “maximally non-integrable” and feel that I reasonably understand them, but I have some lingering questions.

If $$alpha$$ is a non-vanishing 1-form on $$mathbb R^n$$ and $$M$$ is a submanifold, then $$alpha|_{TM}=0quadLongrightarrowquad dalpha|_{TM}=0.$$ Thus if $$pin M$$ and $$dalpha|_{ker alpha_p}$$ has rank $$2r$$, then $$text{codim},Mgeq r+1$$ (morally, we lose one dimension to $$alpha$$ and $$r$$ dimensions to $$dalpha$$). Thus if $$M$$ is to be “maximally non-integrable,” it makes sense to demand that $$dalpha|_{keralpha}$$ be non-degenerate. If this is the case, then the contact Darboux theorem tells us that Legendrian submanifolds exist through any point. But what about the converse direction? If $$keralpha$$ does not admit integral manifolds of dimension $$geq n/2$$, how do we (or is it possible to) prove that $$dalpha|_{keralpha}$$ is non-degenerate?

After some searching of the web, it seemed that I should look at “exterior differential systems” and “Pfaffian systems.” I briefly perused some introductions and found the Pfaff theorem: If $$dalpha|_{keralpha}$$ has constant rank $$2r$$ near $$p$$, then we get $$alpha=dx_1+x_2dx_3+dots+x_{2r}dx_{2r+1}$$ in some coordinate chart about $$p$$. Unfortunately, the above “converse” does not assume that $$alpha$$ has constant rank. Is this reparable? Without constant rank, we will no longer have a normal form, but perhaps it’s still possible to show that $$alphawedge (dalpha)^r=0$$ near $$p$$ implies the existence of an integral manifold through $$p$$ of codimension $$r+1$$? This would certainly imply that any hyperplane distribution on $$mathbb R^n$$ admits a $$text{floor}(n/2)$$-dimensional integral submanifold through any point.

I would also be interested in hearing about the higher codimension case. If I were to make a guess by strict analogy, I might posit that a $$k$$-plane bundle on $$mathbb R^n$$ admits integral submanifolds through any point of dimension $$text{ciel}(k/2)$$. But I’ve also heard that Pfaffian systems get a whole lot more complicated in codimension $$>1$$, so maybe not…

## integral inequality $int_{0}^{t} int_{0}^{tau} f(s)f(tau) leq C t int_{0}^{t} f(s) ^{2}ds$

How can I prove that
$$int_{0}^{t} int_{0}^{tau} f(s)f(tau)dsdtau leq C t int_{0}^{t} f(s) ^{2}ds$$

## calculus – a complicated limits or integral

While I’m working on the proof of Bertrand’s Theorem, I stuck at a limits calculation. I want to prove:
$$lim_{Deltarightarrow 0}int_{-Delta}^{Delta}dfrac{mathrm{d}omega}{(r+omega)^2sqrt{Delta^2-omega^2}}=dfrac{pi}{r^2}$$
With the help of software Mathematica, I got:
$$int_{-Delta}^{Delta}dfrac{mathrm{d}omega}{(r+omega)^2sqrt{Delta^2-omega^2}}=dfrac{pi}{(1-dfrac{Delta^2}{r^2})^{frac{3}{2}} r^2}$$
but I still have no clue how to calculate the limit or the integral. Thanks in advance if you can offer some help.

## calculus – Evaluate the indefinite integral with u substitution

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## complex analysis – Cauchy’s integral theorem proof from Green’s theorem

I am studying Cauchy’s integral theorem from shaum’s outline,the theorem states that

Let $$f(z)$$ be analytic in a region R and on its boundary $$C$$. Then $$oint_{C}f(z)dz=0$$

After the statement of this theorem,it is written there:

This fundamental theorem, often called Cauchy’s integral theorem or simply Cauchy’s theorem, is valid for
both simply and multiple-connected regions. It was first proved by use of Green’s theorem with the added
restriction that $$f'(z)$$ be continuous in R. However, Goursat gave a proof which removed
this restriction. For this reason, the theorem is sometimes called the Cauchy –Goursat theorem when one desires to emphasize the removal of this restriction.

My Question:If a function is analytic in a region then we know that all its derivatives are analytic in that region and hence they are continuous,then why this added restriction of continuity was required while proving the theorem with the help of Green’s theorem?Whether this fact was not known at that time(the fact that if a function is analytic then it is smooth) and Goursat used it in his proof?

## calculus and analysis – Why can’t I evaluate this integral and obtain a closed-form solution?

I have the following spherical density distribution:

$$rho(x, z) = frac{1}{sqrt{x^2 + z^2}left(1+sqrt{x^2+z^2}right)^2}$$

which I have broken into a “line of sight” dimension $$z$$ and a “transverse” dimension $$x$$. Integrating this profile along the line of sight gives the projected 2d density $$Sigma$$:

$$Sigma(x) = 2int_0^inftyrho(x,z)dz$$

I wish to compute this for any generic upper bound $$zeta$$, i.e.

$$Sigma(x; zeta) = 2int_0^zetarho(x,z)dz$$

(that is, $$zeta=infty$$ corresponds to the case of projecting the entire distribution to the transverse plane, while $$zeta corresponds to a projection which is truncated in the $$z$$-dimension).

It turns out this has to be solved piecewise; the solution for $$x>1$$, via Mathematica 11.3, is

$$left.int_0^zetarho(x, z)dzrightrvert_{x>1} = frac{zeta left(sqrt{x^2+zeta^2}-1right)}{left(x^2-1right) left(x^2+zeta^2-1right)}+frac{tan ^{-1}left(frac{zeta}{sqrt{left(x^2-1right) left(x^2+zeta^2right)}}right)-tan ^{-1}left(frac{zeta}{sqrt{x^2-1}}right)}{left(x^2-1right)^{3/2}}$$

However, I am unable to obtain the solution for the case $$x<1$$. I currently only have access to Mathematica 12.0, rather than 11.3, and it is failing on this integral for both cases, even the one above. Performing

Assuming({x < 1, ζ (Element) Reals, ζ > 0},
FullSimplify(Integrate(1/(Sqrt(x^2 + z^2) (1 + Sqrt(x^2 + z^2))^2), {z, 0, ζ})))


returns a HyperGeometric function, though I suspect that the $$x<1$$ case should not be much more complicated than $$x>1$$. Can anyone confirm? Or see any issue?

## complex analysis – how to calculate the integral $int_{0}^{+infty}displaystylefrac{x^betacos(ax)}{x^2-b^2}dx$

How to calculate the following definite integrals:
$$int_{0}^{+infty}displaystylefrac{x^betacos(ax)}{x^2-b^2}dx$$
and
$$int_{0}^{+infty}displaystylefrac{x^betasin(ax)}{x^2-b^2}dx$$
Assuming a and b > 0 and |$$beta$$|<1.

I found the following integral from the table of integrals:
$$int_{0}^{+infty}displaystylefrac{x^betacos(ax-frac{betapi}{2})}{x^2-b^2}dx=-frac{pi}{2}b^{beta-1}sin(ab-frac{pibeta}{2})$$
but couldn’t find those two. Do they exist? if yes, how can I find the answer?

Thank you all!

## calculus and analysis – Evaluate a certain three-dimensional constrained integral

The result of the three-dimensional integration

Integrate(9081072000 (Subscript((Lambda), 1) - Subscript((Lambda),
2))^2 (Subscript((Lambda), 1) - Subscript((Lambda),
3))^2 (Subscript((Lambda), 2) - Subscript((Lambda), 3))^2 (-1 +
2 Subscript((Lambda), 1) + Subscript((Lambda), 2) +
Subscript((Lambda), 3))^2 (-1 + Subscript((Lambda), 1) +
2 Subscript((Lambda), 2) + Subscript((Lambda), 3))^2 (-1 +
Subscript((Lambda), 1) + Subscript((Lambda), 2) +
2 Subscript((Lambda), 3))^2 Boole(Subscript((Lambda), 1) > Subscript((Lambda), 2) &&
Subscript((Lambda), 2) > Subscript((Lambda), 3) &&
Subscript((Lambda), 3) >
1 - Subscript((Lambda), 1) - Subscript((Lambda), 2) -
Subscript((Lambda), 3) &&
Subscript((Lambda), 1) - Subscript((Lambda), 3) <
2 Sqrt(Subscript((Lambda),
2) (1 - Subscript((Lambda), 1) - Subscript((Lambda), 2) -
Subscript((Lambda), 3)))), {Subscript((Lambda), 3), 0, 1}, {Subscript((Lambda), 2), 0, 1}, {Subscript((Lambda), 1), 0, 1}),


that is,

for the two-qubit Hilbert-Schmidt absolute separability probability
apparently can be expressed as

$$begin{equation} label{HSabs} frac{29902415923}{497664}+frac{-3217542976+5120883075 pi -16386825840 tan ^{-1}left(sqrt{2}right)}{32768 sqrt{2}} = end{equation}$$
$$begin{equation} frac{32(29902415923 – 24433216974 sqrt{2})+248874917445 sqrt{2}(5 pi – 16 tan ^{-1}left(sqrt{2}right))}{2^{16} cdot 3^5} approx 0.00365826 end{equation}$$

QuantumComputingStackExchangeQuestion

Can this be explicitly confirmed using Mathematica?