## Special functions – integral expressions for Bessel-like power series

I am interested in power lines of form $$f (z) = sum_ {k = 0} ^ infty frac {z ^ k} {(k!) ^ alpha}.$$ When $$alpha = 1$$is this $$exp (z)$$, To the $$alpha = 2$$ This is a Bessel function and for a larger integer $$alpha$$ We get a hypergeometric series. These special functions ($$alpha> 1$$) have integral expressions in the form of an integral of an elementary function.

Can be said of something for non-integral values $$alpha$$? If $$alpha> 1$$ is there any real number, is there any hope to write an integer expression for the sum? Has this form of series been studied anywhere in the literature in general? Any useful techniques to work with?

## Calculus – An integral estimate for small area

Consider an openly limited connected set $$(0,0) in Omega subseteq mathbb {R ^ 2}$$ and a compact set $$K subseteq Omega$$ so that $$| Omega setminus K |$$ is small enough (for example for everyone $$epsilon> 0$$. $$| Omega setminus K | leq epsilon$$the condition can be changed to others, as in the following example). Now I hope that the following estimate applies to everyone $$t> 0$$ (at least close to zero):
$$int _ { Omega setminus K} frac {dxdy} {1+ frac {| x |} {t}} leq C ( epsilon) int _ { Omega} frac {dxdy} {1 + frac {| x |} {t}}$$
Where $$C ( epsilon)$$ can be explicit and $$lim _ { epsilon rightarrow0} C ( epsilon) = 0$$Maybe you can prove it $$epsilon ^ 2$$. $$2 epsilon + epsilon ^ 2$$ or others.

My question is, can we find such an estimate?

An example to support this assumption is the following: $$Omega = (- 1,1) ^ 2$$. $$K = (- 1+ epsilon, 1- epsilon) ^ 2$$, Then
begin {align} int _ { Omega setminus K} frac {dxdy} {1+ frac {| x |} {t}} & = int _ {(- 1,1) ^ 2} frac {dxdy} {1+ frac {| x |} {t}} – int _ {(- 1+ epsilon, 1- epsilon) ^ 2} frac {dxdy} {1+ frac {| x |} {t}} . & = 4t ln (1+ frac {1} {t}) – 4 (1- epsilon) t ln (1+ frac {1- epsilon} {t}) \ & leq 4t (2 epsilon- epsilon ^ 2) ln (1+ frac {1} {t}) \ & = (2 epsilon- epsilon ^ 2) int _ {(- 1,1) ^ 2} frac {dxdy} {1+ frac {| x |} {t}} end {align}
where we use $$ln (1 + x) – delta ln (1+ delta x) leq (1- delta ^ 2) ln (1 + x)$$ to the $$0 < delta <1$$ and $$x> 0$$

## What is the integral of Arctan (x) ^ 3 * sec (x) ^ 4 dx? [closed]

I have problems calculating this integral. Can someone explain how to do it, or at least start with it?

## How do you draw an integral on mathematica?

How do I draw this integral on mathematica? I try to do this by hand, but I would like to check on mathematica if I'm doing it right.

## How would you draw the integral? [a,b] with respect to a variable x in mathematica?

How would you represent the definite integral ∫4 dx in the interval [a, b], where a = 0 and b = 4 are the upper and lower limits of mathemetmatica? The reason why I ask is that I practice graphing certain integrals by hand and want to check it for mathematica

## How do I draw a certain integral on mathematica?

I'm not sure how to put this into a good format, but I want to show this on mathematica: 𝑢 (𝑡) = ∫ (1) / (4 * pi * t) * e ^ ((- y ^ (2 )) / (4 * t)) * y (1-y ^ (2)) dy in the interval (a, b), where a = 0 and b = 1 are the upper and lower limits.

## cv.complex variables – Bromwich integral that has been converted to an integral on the real axis

I am new to complex integration and inverse Laplace transformations. I've already asked this question on math.se, but I haven't received an answer.

The author of a textbook claims that the inverse Laplace transform has expression
$$f (t) = frac {2 exp (bt)} { pi} int_0 ^ infty Re bigl ( hat {f} (b + iu) bigr) cos (ut) , mathrm {d} u.$$
He gets this formula by replacing it $$s = b + iu$$ in the Bromwich integral
$$f (t) = frac {1} {2 pi i} int_ {bi infty} ^ {b + i infty} exp (st) hat {f} (s) , mathrm {d } s.$$
However, I've checked this formula numerically and it doesn't seem to apply:

``````fhat <- function(s) 1/(s+3) # Laplace transform of exp(-3x)
b <- 5
integrand <- function(u, x){
Re(fhat(b+1i*u))*cos(x*u)
}
x <- 2
2*exp(b*x)/pi * integrate(integrand, 0, Inf, x = x, subdivisions = 10000)\$value
# -0.1124648
exp(-3*x)
# 0.002478752
``````

To the $$b = -2$$ The result is close to the expected value $$exp (-3x)$$::

``````b <- -2
2*exp(b*x)/pi * integrate(integrand, 0, Inf, x = x, subdivisions = 10000)\$value
# 0.002479138
``````

I got that $$b$$ must be chosen somewhere to the right of the singularities of $$has {f}$$ (Here $$-3$$). Am I wrong? Here the result depends on the choice $$b$$, Do I get something wrong or is something wrong in this textbook?

## Fourier Analysis – Show no lower limit for the vibration integral

Consider the vibration integral
$$I (x) = int_0 ^ 1 e ^ {ix t ^ 2} dt.$$
The standard Van der Corput lemma tells us that $$I (x) = O (| x | ^ {- 1/2})$$ to the $$| x |> 1$$,

I was looking for a lower limit like this: $$| I (x) | geq c | x | ^ {- 1/2}$$, but it turns out to be impossible because someone tells me $$I (x)$$ has infinite zeros $$x_n to infty$$, But is it true? If so, how can I prove it?

## How can you evaluate this integral more precisely and faster?

I have the following integral:

$$int _ {- a / 2} ^ {a / 2} Cf (x) sin { frac {m pi (x + 0.5a)} {a}} sin { frac {p pi (x + 0.5a)} {a}} dx$$

Where $$f (x) = 53.62 cos ^ 4 left ( frac {7 pi sqrt {x ^ 2}} {9} right) +3.04 sin ^ 4 left ( frac {7 pi sqrt {x ^ 2}} {9} right) +2.54 sin ^ 2 left ( frac {14 pi sqrt {x ^ 2}} {9} right)$$

$$m = 1,2, dots, M$$ $$p = 1,2, dots, P$$, $$C = text {constant}$$

I am using the following code

``````f(x) = 53.62 Cos((7 (Pi) Sqrt(x^2))/9)^4 +
3.04 Sin((7 (Pi) Sqrt(x^2))/9)^4 +
2.54 Sin((14 (Pi) Sqrt(x^2))/9)^2;
nL = 20; nC = 20;
K = ConstantArray(0, {nL*nC, nL*nC});
a=1;b=1;
Do(
Do(
Do(
Do(

k = (m - 1) nC + n;
l = (p - 1) nC + q;

If(n == q,
K((k, l)) = ((Pi)^2 b m^2 p^2)/(2 a^4)
NIntegrate(
f(x) Sin((m (Pi) (x + 0.5 a))/a) Sin((
p (Pi) (x + 0.5 a))/a), {x, -a/2, a/2},
Method -> "Trapezoidal"), K((k, l)) = 0)

, {q, 1, nC}),
{p, 1, nL}),
{n, 1, nC}),
{m, 1, nL});
``````

Does anyone know of any other faster and more accurate approach?

## real analysis – Mehta integral and orthogonality

This is a kind of follow-up question to an earlier question I asked about the Mehta integral, see here at mathoverflow.

The Mehta integral is the following expression:

$$frac {1} {(2 pi) ^ {n / 2}} int _ {- infty} ^ { infty} cdots int _ {- infty} ^ { infty} prod_ {i = 1} ^ ne ^ {- t_i ^ 2/2} prod_ {1 le i

In this answer, Iosif Pinellis showed that if we choose scaling $$gamma = 1 / n ^ 2$$ then the value of the integral is essentially $$n$$-independently. This is mainly due to the product $$F (t): = prod_ {1 le i contains of order $$n ^ 2$$ many terms.

Now I've examined a related object:

To simplify the notation, we are introducing a key figure $$d mu_ {n, gamma} (t): = prod_ {i = 1} ^ n e ^ {- t_i ^ 2/2} prod_ {1 le i
It's easy to see
$$begin {equation} label {eq: ortho} frac {1} {(2 pi) ^ {1/2}} int _ {- infty} ^ { infty} e ^ {- t ^ 2/2} (t ^ 2-1) dt = 0. end {equation}$$

Now it seems that again for different scaling of $$gamma$$ we find interesting phenomena for the value of

$$nu: = lim_ {n rightarrow infty} frac { int (t_1 ^ 2-1) d mu_ {n, gamma} (t)} { sqrt { int (t_1 ^ 2-1) ^ 2 d mu_ {n, gamma} (t) int 1 d mu_ {n, gamma} (t)}}$$

where I use the limit without really knowing if it exists.

Case 1:

How to guess from the other thread if we decide $$gamma = 1 / n ^ 2$$ it seems that the product $$F (t)$$ does not add to the value of the above limit and we have $$nu = 0.$$

Case 2:

If we choose $$gamma = 1 / n$$ then it seems like the value of $$nu$$ converges neither to zero nor to one.

Case 3:

If we choose to take $$gamma = 1$$ then we seem to understand that $$nu = 1.$$

Now I have to admit that I could not numerically evaluate these high-dimensional integrals and therefore did something much less reliable to get a guess about the behavior of the integral $$n$$ large:

I chose a set $$t_2, …, t_n$$ of random numbers that were sampled from a Gaussian distribution and then calculated

$$frac { frac {1} { sqrt {2 pi}} int _ {- infty} ^ { infty} (t ^ 2-1) e ^ {- t ^ 2/2} prod_ {j = 2} ^ n vert t-t_j vert ^ {2 gamma} dt} { sqrt { frac {1} { sqrt {2 pi}} int _ {- infty} ^ { infty} (t ^ 2-1) ^ 2e ^ {- t ^ 2/2} prod_ {j = 2} ^ n vertt-t_j vert ^ {2 gamma} dt frac {1 } { sqrt {2 pi}} int _ {- infty} ^ { infty} e ^ {- t ^ 2/2} prod_ {j = 2} ^ n vert t-t_j vert ^ {2 gamma} dt}}.$$

As a function of $$n$$ I then got the following diagrams for cases 1-3:

My question is: can we say analytically what happens to scaling in case 2? $$gamma = 1 / n$$ to the value of $$nu$$?