Mathematica does not recognize this case:

```
Integrate((T^a/(Gamma(a))) * (E^(-Tx)) (x^(a - 1)), {x, 0, Infinity})
Integrate::idiv: Integral of (E^-Tx T^a x^(-1+a))/Gamma(a) does not converge on {0,(Infinity)}.
```

This should converge because

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# Tag: gamma

## Mathematica erroneously says this gamma integral does not converge

## coding theory – Variance Gamma parameter estimation in R studio

## analytic number theory – Is every modular function on $Gamma$ univalent?

## Fitting data to left skewed gamma distribution

## simplifying expressions – Is there any way to prioritize Gamma function?

## How to calculate gamma function values for quaternions with Mathematica?

## Logic – If a set of sentences $ Gamma $ has an infinite model, does it also have a denumerable model?

## Gamma coin round 3 air drops | Proxies-free

## Optimization – Maximize $ log (4) c + log (3) a + log (2) x $ if $ a + c + x + y = 1 $, $ (a + c) ^ 2 + (x + y) ^ 2 + 2xc leq 1-2 gamma $, $ 0 leq gamma leq 1/4 $

## Statistics – probability between two values of a gamma distribution

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Mathematica does not recognize this case:

```
Integrate((T^a/(Gamma(a))) * (E^(-Tx)) (x^(a - 1)), {x, 0, Infinity})
Integrate::idiv: Integral of (E^-Tx T^a x^(-1+a))/Gamma(a) does not converge on {0,(Infinity)}.
```

This should converge because

I am using the vgFit function in R Studio to estimate parameters for stock prices. I converted the stock prices to a vector and I get this error. My code was:

```
param<-c(0,1,0,1)
fit<-vgFit(c(x), param=param)
```

The error is:

```
Error in optim(paramStart, llsklp, NULL, method = startMethodSL, hessian = FALSE, :
function cannot be evaluated at initial parameters
```

Suppose $f$ is a modular function on $Gamma$ then it has a fundamental region $R_L$. Since $f$ is modular, it can be expressed as a rational function of $J(tau)=frac{g_2^3(tau)}{Delta(tau)}$. To prove this we actually construct a rational function

$$g(tau)=frac{prod(J(a_i)-J(tau))}{prod(J(b_i)-J(tau))}$$

where the $a_i,b_i$ are zeros and poles of $f$. It seems to me that this construction implies the possibility of having multiple poles in the fundamental region. I know for sure that $J$ has only one pole since it attains every value exactly once.

Since a modular function is univalent if and only if its fundamental region has genus $1$ and if we identify the edges of $R_Gamma$ it would become homeomorphic to $mathbb{P}^1$, any modular function that has the fundamental region $R_Gamma$ should be univalent, that is, any modular function is univalent in the closure of its fundamental region.

This really confused me and I am not sure if this is correct.

How can I fit the following two set of data to a left skewed gamma function, which I what I think should fit the data best?:

data 1 is here: https://pastebin.com/X2HTjTP7

data 2 is here: https://pastebin.com/8Rh4BHDT

Is there any other suggestion of what would be the best distribution or equation to fit the data?

A picture of how data 1 looks is here:

A picture of how data 2 looks is here:

Thank you in advanced,

In Mathematica, if you type

```
Gamma[1, 0, -a]
```

It automatically simplifies to `1 - E^a`

. I wonder though if there is a way to stop Mathematica from simplify Gamma function automatically.

I understand that I can use `Inactive`

or `HoldFrom`

. But I actually want having to remember using Inactive every time.

How can we calculate the values for that? **gamma** Function with **Quaternions** on Mathematica?

For example:

```
Gamma(5) = 24
N(Gamma(I)) = -0.15495 - 0.498016 I
N(Gamma(1 + 2 I)) = 0.151904 + 0.0198049 I
N(Gamma(Quaternion(5, 0, 0, 0))) = Gamma(Quaternion(5., 0., 0., 0.)) (* which is wrong *)
```

How can we calculate specifically?

```
Gamma(Quaternion(1, 2, 3, 4))?
```

Tries:

```
N(Gamma(FromQuaternion(Quaternion(1, 2, 0, 0)))) (* works *)
N(Gamma(FromQuaternion(Quaternion(1, 2, 3, 4)))) (* doesn't work *)
QuaternionPower(x_, y_) := E^(y ** Log(x))
QuaternionGamma(z_) := Integrate(QuaternionPower(E, -x + Log(x)*(z - 1)), {x, 0, Infinity})
N(QuaternionGamma(FromQuaternion(Quaternion(1, 2, 0, 0)))) = 0.151904 + 0.0198049 I (* works *)
N(QuaternionGamma(FromQuaternion(Quaternion(1, 2, 3, 0)))) = ConditionalExpression(Gamma((1. + 2. I) + 3. J), Re(J) > -0.333333) (* doesn't work *)
```

I think that's wrong, since we can only conclude from the Downward-Lowenheim-Skolem theorem $ Gamma $ has an enumeration model, it is not shown whether this enumeration model is infinite. Certainly ours $ Gamma $ already has an infinite model, but nothing says that its enumerable model must be identical to its infinite model.

I can't see how this claim is true, but I think it should be true?

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I want to find out how to show by hand that the maximum of

$$ log (4) c + log (3) a + log (2) x $$

when

$$ a geq 0, c geq 0, x geq 0, y geq 0, $$

$$ a + c + x + y = 1, $$

$$ (a + c) ^ 2 + (x + y) ^ 2 + 2xc leq 1-2 gamma, $$

Where $ gamma $ is a fixed constant so that $ 0 leq gamma leq 1/4 $.

Maple showed me that the maximum value is reached at $ frac { log (6)} {2} + frac { sqrt {1-4 gamma}} {2} log (3/2) $that is given by $ a = frac {1+ sqrt {1-4 gamma}} {2} $, $ b = frac {1- sqrt {1-4 gamma}} {2} $, $ c = y = 0 $.

I've tried using Lagrangian multipliers by changing the last constraint to an equality, but it gets very messy and I'm stuck.

I would also be happy if I could just show it $ log (4) c + log (3) a + log (2) x leq frac { log (6)} {2} + frac { sqrt {1-4 gamma}} { 2} log (3/2) $ for all such $ a, c, x, y $. One method I've tried is to use the concavity of $ log (x) $. The equation of the line that goes through $ (2, log (2)) $ and $ (3, log (3)) $ is $ L (x) = log (3/2) x- log (9/8) $. Then by concavity of $ log (x) $ we have $ log (x) leq L (x) $ for all integers $ x $. Then

begin {align *} & log (4) c + log (3) a + log (2) x \

& leq L (4) c + L (3) a + L (2) x \

& = log (3/2) (4c + 3a + 2x + y) – log (9/8) \

& = log (3/2) left ( frac {5 + 3} {2} c + frac {5 + 1} {2} a + frac {5-1} {2} x + frac {5- 3} {2} y right) – log (9/8) \

& = log (3/2) left ( frac {5+ sqrt {1-4 gamma}} {2} right) – log (9/8) + log (3/2) left ( frac {ax + 3 (cy) – sqrt {1-4 gamma}} {2} right) \

& = frac { log (6)} {2} + frac { sqrt {1-4 gamma}} {2} + log (3/2) left ( frac {ax + 3) ( cy) – sqrt {1-4 gamma}} {2} right). \

end {align *}

That is, to get the inequality I want, it would be enough to show it $ a-x leq 3 (y-c) + sqrt {1-4 gamma} $. However, I have not been able to prove this inequality. Any suggestions or help will be greatly appreciated.

Another option would be to use jamming techniques, but I have no experience with them. Thank you for your help.

I'm trying to solve a problem where I have a random variable X that follows a gamma distribution of $ Gamma ( theta = 5, alpha = 4) $.

I'm trying to find the probability of $ P (10 leq X leq 30) $. Of course, if this were a normal distribution, it would be easy, but I currently don't know how to deal with this problem. I am confused as to whether to convert it to one $ chi ^ 2 $ and how that helps me to solve this problem. Thank you very much

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