Complicated Integral output with Unfamiliar Regularized Hypergeometric Function

I need the solution for following integral and it has output in MATHEMATICA as:

FullSimplify[  Integrate[x^(m - 1) Exp[-a x] Gamma[m, b/x], {x, 0, \[Infinity]},    Assumptions -> m > 1/2 && a > 0 && b > 0]] 

enter image description here

I really do not know the definition of the last Regularized Hypergeometric Function which has some numbers as superscripts.

Further, does anyone know how we can prove this analytically? Because there is very less integral formulas related to Regularized Hypergeometric Function.

Further, I found this output somehow numerically unstable for some large $ m$ .

Narrowing the range of an interpolating function while keeping integral the same?

Consider an interpolating function derived from some data, e.g.

data = Table[{x, Cos[2 π/1000 x] // N}, {x, -250, 250}]; fun[x_] = Piecewise[{{Interpolation[data][x], Min[data[[;; , 1]]] <= x <= Max[data[[;; , 1]]]}}] 

As it is, the interpolation shows a hump that goes to zero at x=-250 and x=250:

Plot[fun[x], {x, -500, 500}] 

enter image description here

Now I would like to change (manipulate) the interpolation function such that e.g. the function goes to zero at x=-100 and x=100 instead while the hight increases proportionally such as to keep the integral constant

Integrate[fun[x],{x,-Infinity,Infinity}] 

318.31

Here a rough sketch how the result should look like enter image description here

How can I do this with mathematica?

EDIT:

Naively, I could always do

targetA = NIntegrate[fun[x], {x, -Infinity, Infinity}]; transfA = NIntegrate[fun[x 5/2.], {x, -Infinity, Infinity}]; newfun[x_] = fun[x 5/2] targetA/transfA; Plot[{fun[x], newfun[x]}, {x, -500, 500}] 

enter image description here

But I wonder if there is a less pedestrian way of doing that?

Plotting Integral of Exponential functions

I am trying to Plot an integral equation that involves exponential function. My code is as follow,

L[\[Alpha]_] :=    NIntegrate[    1/(k + I*0.1) (     Exp[I*k*x] (Exp[Sqrt[k^2 + \[Alpha]/w^2]*w] - 1) (Exp[k*w] - 1 +         I*0.1) Sqrt[      k^2 + \[Alpha]/       w^2])/((Sqrt[k^2 + \[Alpha]/w^2] + k) (Exp[         Sqrt[k^2 + \[Alpha]/w^2]*w - Exp[k*w]]) + (Sqrt[         k^2 + \[Alpha]/w^2] -          k) (Exp[(k + Sqrt[k^2 + \[Alpha]/w^2]) w] -          1)), {k, -100, 100}]; Plot[{Re[L[10]], Re[L[100]], Re[L[500]]}, {x, -0.45, 0.45},   PlotRange -> Full].  

But this integral gives a lot of oscillations which it should not. This is fig 2 in this article “https://arxiv.org/pdf/1508.00836.pdf” that I am trying to plot. Any help will be highly appreciated.

Integral operator

Assume that $ A: L^1([0,1]) ]\to L^1[0,1]$ is an integral operator with a real Kernel G(x,y). Then its adjoint operator $ A^*:L^\infty([0,1]) ]\to L^\infty [0,1]$ is an integral operator with a real Kernel G(y,x).

My question is. Whether the norms of $ A$ and $ A^*$ coincide?

Formula for exponential integral over a cone

While reading ‘Computing the Volume, Counting Integral points, and Exponential Sums’ by A. Barvinok (1993), I came across the following:

“Moreover, let $ K$ be the conic hull of linearly independent vectors $ u_{1}, …, u_{n}$ so $ K = co(u_{1}, …, u_{n})$ . Denote by $ K^{*} = \{x\in \mathbb{R}^{n}: <x, y> \le 0 \text{ for all } y\in K\}$ the polar of K. Then for all $ c \in \text{Int} K^{*}$ we have

\begin{equation} \int_{K}exp(<c, x>)dx = |u_{1} \land … \land u_{n}|\prod_{i=1}^{n}<-c_{i}, u_{i}>^{-1} \end{equation}

To obtain the last formula we have to apply a suitable linear transformation to the previous integral. “

I have tried proving this but I can’t find relevant links to help me. Also, I’m unsure what $ |u_{1} \land … \land u_{n}|$ is supposed to mean. Would greatly appreciate if someone could point me to a relevant resource or provide proof. Thanks!

Density of integral values of a rational function

Let $ \mathbf{x} = (x_1, \cdots, x_n)$ , and consider a rational function $ F : \mathbb{R}^n \rightarrow \mathbb{R}$ be given by

$ $ \displaystyle F(\mathbf{x}) = \sum_{i = 1}^m \frac{Q_i(x_1, \cdots, x_{n-1})}{R_i(x_1, \cdots, x_{n-1})} x_n^i,$ $

where $ Q_i, R_i$ are non-constant polynomials with integer coefficients. Moreover we may assume that $ R_i > 0$ for all $ \mathbf{x} \in \mathbb{R}^n$ , so $ F$ is well-defined everywhere.

I am trying to understand the counting function

$ $ N_F(X) = \# \{\mathbf{x} \in \mathbb{Z}^n : \lVert \mathbf{x} \rVert_\infty \leq X, F(\mathbf{x}) \in \mathbb{Z} \}.$ $

In particular, given $ x_1, \cdots, x_{n-1}$ one can always find $ x_n \in \mathbb{Z}$ such that $ F(\mathbf{x}) \in \mathbb{Z}$ , but the smallest possible choice of such $ x_n$ could very well be extremely large. Thus it is perhaps best to consider all of the $ x_i$ ‘s as varying at once.

The above observation also gives a somewhat straightforward upper bound for $ N_F(X)$ . In particular, having chosen $ x_1, \cdots, x_{n-1}$ the resulting function $ f(x) = F(x_1, \cdots, x_{n-1}, x)$ is then a rational polynomial in a single variable, and we can clear its denominator to obtain $ g(x)$ say, and then the question is equivalent asking for the density of $ x \in [-X,X]$ such that $ g(x)$ satisfies a certain congruence. However the modulus, equal to $ \text{LCM}_{1 \leq i \leq m} R_i(x_1, \cdots, x_{n-1})$ is typically much larger than $ X$ , since the $ R_i$ ‘s are assumed to be positive definite and in particular not linear. Therefore usually there is at most one root of $ g(x)$ in $ [-X,X]$ . It thus follows that $ N_F(X) = O\left(X^{n-1}\right)$ . I am looking for a bound that beats this, and perhaps close to what one might expect to be the exact asymptotic order.

Double integral differentiation

Could somebody tell me how do I get the right side from the left side – which rule of differentiation of the double integral produces this?

\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x} \int_{v=x}^\infty &\int_{u=0}^x f(v, u)\cdot g (v)\cdot g(u)\, \mathrm{d} u\, \mathrm{d} v =\ &=\int_0^\infty g(v)\;\mathrm{d}v\cdot \frac{\mathrm{d}}{\mathrm{d}x} \int_0^x f (v, u)\; g(u)\; \mathrm{d}u + \int_x^\infty \;g(u)\;\mathrm{d}u\cdot \frac{\mathrm{d}}{\mathrm{d}x} \int_0^x f (v, u)\; g(v)\; \mathrm{d}v\end{align*}

My ignorance is probably quite profound, and this is something really simple, but I cannot see. It’s been quite a while since I last dabbled in double integrals.

When I type

d/dx(integral from v=x to infinity integral from u=0 to x f(u,v)g(u)g(v) du dv)

into WolframAlpha, I get

\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x} \Bigg(\int_{v=x}^\infty \int_{u=0}^x &f(v, u)\, g (v)\, g(u)\, \mathrm{d} u\, \mathrm{d} v\Bigg) =\ &=\int_x^\infty g(v) g(x) f(x,v) \mathrm{d}v – \int_0^x g(u) g(x) f(u,x) \mathrm{d}u,\end{align*}

which is correct, but I first need to understand how to arrive at the first formula, in order to apply the Leibnitz rule.

If someone knows how to align these equations here beautifully, do tell.