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Calculus – Understand how to integrate parts into the gamma function

The gamma function is defined as …

$formula 1$
,

I'll see how Gaussian representation The gamma function is derived and the first step is the integration of parts. No steps are shown and the following is the result of applying the integration of parts …

$form2$
,

I am confused how these values were derived. That's my attitude …

$u$
,

$dv$
,

I imagine they chose it that way u and dv, which does ______________ mean…

$v$
,

$you$
,

I'm not sure how they came you, I tried to derive u and ended at …

$ e[n ln(1-t/n)]$

and then got another answer after trying to derive it. Can someone show me how you is derived or shows me where I am wrong so that I can complete the parts through integration?

Color Correction – How can gamma be applied correctly to linear raw files from an image processing camera?

Visually, this looks absolutely correct to me. In your unmatched image on my color-calibrated monitor, the steps between the gray spots appear at irregular intervals. In the adjusted image, they are more noticeable.

You ask:

Is color saturation a natural consequence of applying a gamma value, and if so, what can be done to compensate for this effect?

Short answer:

No. This is a consequence if correct black levels and white points are not set. You should do this before you apply the gamma curve.

Long answer:

The gamma curve that you used has the following form:

Original with histogram from linear version with applicable gamma curve

The light line applies to the values displayed in the histogram. Once you've applied this curve, you'll get a histogram like this:

Curve applied

This is basically a boring, "flat" curve – that is, even though it has no linear value, it is perceptively basically like that. You can see, however, that the values are all in the middle. This is very functional and maybe that's what you want for image processing, but in general it's not what we want visually Ones. You may just want to increase the contrast by dragging the black and white dots as follows:

Increase contrast

What gives a picture like this:

much less washed out

… much less washed out.

Your camera may have an adjustable black level, and you may want to raise it slightly. You must also set a suitable white level for your camera for conversion. You may want to check the Dcraw code to see how it normally does. (Well, spoiler: dcraw sets the white level to the 99th percentile of the histogram.)

Incidentally, this result has such a histogram:

increased contrast

what you can see extends to the extremes of the histogram. Since I am working with an 8-bit image in an 8-bit area, you can see that the colors are getting sparse. For "real work" you want to work with a higher bit depth (and probably only apply a transformation instead of a series).

This is definitely a bit boring. In the visual world, we may want to apply an s-curve to increase the power:

S-curve

yielding

punchy!

… or something like that.

raw – How to apply gamma correctly to the image processing camera

I'm using a machine vision camera that produces linear pixels, according to the manufacturer. As a result, I need to apply a gamma correction to the image before I can see the result. However, after applying a gamma correction to the image, the colors appear washed out.

Before each gamma correction:

Example of a raw gift

After applying 2.2 gamma:

Example port 2.2 gamma application

Overall, the brightness in the picture looks much better. The colors in the Color Checker, however, seem very oversaturated. Is color saturation a natural consequence of applying a gamma value, and if so, what can be done to compensate for this effect?

xfce – Redshift Gamma ramp size too small: 0 Error while starting the adaption method randr

I use Linux Mint 18.3, it is Ubuntu 16.04 base

in my thinkpad t480 (laptop has nvidia card), the redshift works fine

but in my dell optiplex 3060m (desktop pc has only intel gpu), the redshift does not work:

Gamma ramp size too small: 0
Failed to start fitting method randr.

How can I fix this?

Numerics – Calculation between gamma functions

I've calculated the gamma functions in Mathematica while there is no agreed answer.

By definition, $ Gamma[alpha]= int_0 ^ infty t ^ { alpha-1} e ^ {- t} dt $, $ Gamma[alpha,z]= int_z ^ infty t ^ { alpha-1} e ^ {- t} dt $ and $ Gamma[alpha,z_1,z_2]= int_ {z_1} ^ {z_2} t ^ { alpha-1} e ^ {- t} dt $, So we should have $ Gamma[alpha]-Gamma[alpha,z]$ should be the same $ Gamma[alpha,0,z]$, This is not the case. I'll just leave it $ alpha = 200 $ and $ z $ from 1 to 5 the numbers do not fit; see below.

    α = 200;
table[{Gamma[α] - gamma[α, z]gamma[α, 0, z]}, {z, 1, 5}]{{0th * 10 ^ 359, 0.00184859}, {0. * 10 ^ 359, 1.0983 * 10 ^ 57}, {0. * 10 ^ 359.6.71225 * 10 ^ 91},
{0th * 10 ^ 359, 2.41279 * 10 ^ 116}, {0. * 10 ^ 359, 2.14999 * 10 ^ 135}}

Why this?

complex analysis – Using the Cauchy formula to solve $ int_ gamma frac {e ^ {z ^ 2}} {z ^ 2-6z} =

Calculate the integral $ I = int_ gamma f (z) dz $,from where $ f (z) = frac {e ^ {z ^ 2}} {z ^ 2-6z} $ from where $ gamma = {z || z-2 | = 3 } $

I thought about using it $ f (z_0) = frac {1} {2 pi i} int frac {f (z_0)} {z-z_0} dz $

rewrite $ int_ gamma frac {e ^ {z ^ 2}} {z ^ 2-6z} = int_ gamma frac {e ^ {z ^ 2}} {z (z-6)} = int_ gamma frac { frac {e ^ {z ^ 2}} {z}} {(z-6)} $

The scope $ gamma $ determines that this is the domain in which we want the function $ frac {e ^ {z ^ 2}} {z} $ to use contains $ z-6 $,

So the application of the Cauchy formula:

$ int f (z) dz = pi i frac {e ^ {6 ^ 2}} {3} $

The solution, however, says that the value of integral is $ frac {- pi i} {3} $

Question:

What am I doing wrong?

Thank you in advance!

Calculus and Analysis – Can this integral equation problem $ int_ gamma frac {e ^ {ik | xy |}} {4 pi | xy |} varphi (y) dy = u_ {x_0} ^ {in} (x) $ be solved?

I am not sure if Mathematica can solve integral equations in 2D / 3D. I found this page in the documentation, but this is only for 1D.

The following is what I would like to solve, it can be an electromagnetic problem, but that is beside the point. Leave the incident box $ u_ {x_0} ^ {in} (x) $ given from a point source $ x_0 $:$$
u_ {x_0} ^ {in} (x) = frac {e ^ {ik | x-x_0 |}} {4 pi | x-x_0 |},
$$
Then I have to find $ varphi in gamma $ so that$$
S_ Gamma ^ k[varphi](x) = u_ {x_0} ^ {in} (x), quad quad forall x in gamma,
$$
from where
$$
S_ Gamma ^ k[varphi](x): = int_ gamma frac {e ^ {ik | x-y |}} {4 pi | x-y |} varphi (y) dy,
$$
and
$ Gamma in mathbb {R} ^ 3 $ is the triangle defined by its vertices $$ gamma: = {v_1, v_2, v_3 },
$$
With begin {align} v_1 & = (4,0,0), \ v_2 & = (8,0,0), \ v_1 & = (6,2,0). end

The problem is in 3D, but the integration area is a 2D triangle on the $ x $–$ y $ Level, with a singular integrand though $ x = y $,

Is it possible to solve this problem with Mathematica?

Photoshop – How do I find out if TIF images have gamma correction or not?

I have several DNG image files taken with a LGE Nexus 5X with Open Camera for Android as a camera app. For further image processing, I converted the DNG files with Camera Raw 8 and Photoshop CC 2015 into TIF files. I have converted the images without changing the color or exposure settings.
As far as I know, the DNG files should be linear (without gamma correction). My question is whether the converted TIF file is still linear or not. Does Photoshop / Camera Raw perform a gamma correction on TIF files?

I would be glad if you could help me with this question.
Best,
Hans

Calculus and Analysis – Complex infinity of hypergeometric + gamma function

I solve an integral that yields a hypergeometric function + gamma function. The point is that my values of n (see code below) are integers, so n = 1,2,3,4 ..., But there is complex singularity if I want to introduce the value of n in the expression, if the integral has already been calculated.

f[r_, k_, kl_, n_] : = Sin[k*kl*r]/ (k ^ 2 * r) * k ^ -n
integration[r_, kl_, n_, m_] : =
Accepted,[{kl > 0, n > 0, r > 0, m > 0}, 
   Integrate[f[r, k, kl, n], {k, 1, m}]]// FullSimplify

function[r_] : = Integration[r, kl, n, Infinity] // FullSimplify
function[r]

The result is :

-kl (kl r) ^ n Cos[(n [Pi]) / 2]gamma[-1 - n] + (
kl HypergeometricPFQ[{-(n/2)}, {3/2, 1 - n/2}, -(1/4) kl^2 r^2]) / n

So Mathematica can solve the integral for every value of n, But if I use the value of this expression n that I want then I get a singularity (of course, due to the gamma function and the hypergeometric function)

-kl (kl r) ^ n Cos[(n [Pi]) / 2]gamma[-1 - n] + (
kl HypergeometricPFQ[{-(n/2)}, {3/2, 1 - n/2}, -(1/4) kl^2 r^2]) /
n /. n -> 2

I get

Infinity :: indet: Undefined expression of ComplexInfinity + ComplexInfinity.

But I do not underestimate, as I attribute the integral to the value of n Well, then I get an analistic expression for it

f[r_, k_, kl_, n_] : = Sin[k*kl*r]/ (k ^ 2 * r) * k ^ -n
integration[r_, kl_, n_, m_] : =
Accepted,[{kl > 0, n > 0, r > 0, m > 0}, 
   Integrate[f[r, k, kl, n], {k, 1, m}]]// FullSimplify

function[r_] : = Integration[r, kl, 2, Infinity] // FullSimplify
function[r]

I get:

(kl r Cos[kl r] +
kl ^ 3 r ^ 3 CosIntegral[kl r] + (2 - kl ^ 2 r ^ 2) Sin[kl r]) / (6 r)

And there is no singularity in this expression. So my question is: There is a way to get a general expression in terms of the variable nwithout getting a singular behavior as it happens through the gamma and the hypergeometry function? Mathematica Can this term be simplified to remove the differences?

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